Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup

Let $S'$ be the image $\psi \left({S}\right)$ of $S$. Then $\left({S', \oplus'}\right)$ is a subsemigroup of $\left({T', \oplus'}\right)$.


Construction of Inverse Completion/Identity of Quotient Structure

Let $c \in C$ be arbitrary. Then: :$\eqclass {\tuple {c, c} } \boxtimes$ is the identity of $T'$.


Construction of Inverse Completion/Invertible Elements in Quotient Structure

Every cancellable element of $S'$ is invertible in $T'$.


Construction of Inverse Completion/Generator for Quotient Structure

$T' = S' \cup \left({C'}\right)^{-1}$ is a generator for the semigroup $T'$.


Construction of Inverse Completion/Quotient Structure is Inverse Completion

$T'$ is an inverse completion of its subsemigroup $S'$.


Ring of Integers Modulo Composite is not Integral Domain

Let $m \in \Z: m \ge 2$. Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$. Let $m$ be a composite number. Then $\struct {\Z_m, +, \times}$ is not an integral domain.


Ring Zero is not Cancellable

Let $\struct {R, +, \circ}$ be a ring which is not null. Let $0$ be the ring zero of $R$. Then $0$ is not a cancellable element for the ring product $\circ$.


Congruence Relation on Ring induces Ideal

Let $\left({R, +, \circ}\right)$ be a ring. Let $\mathcal E$ be a congruence relation on $R$. Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ be the equivalence class of $0_R$ under $\mathcal E$. Then $J$ is an ideal of $R$.


Ideal induces Congruence Relation on Ring

Let $\struct {R, +, \circ}$ be a ring. Let $J$ be an ideal of $R$ Then $J$ induces a congruence relation $\EE_J$ on $R$ such that $\struct {R / J, +, \circ}$ is a quotient ring.


Vector Space has Basis between Linearly Independent Set and Spanning Set

Let $V$ be a vector space over a field $F$. Let $L$ be a linearly independent subset of $V$. Let $S$ be a set that spans $V$. Suppose that $L \subseteq S$. Then $V$ has a basis $B$ such that $L \subseteq B \subseteq S$.


Null Ring is Trivial Ring

Let $R$ be the null ring. Then $R$ is a trivial ring and therefore a commutative ring.


Null Ring is Ring

Let $R$ be the null ring. That is, let: : $R := \left({\left\{{0_R}\right\}, +, \circ}\right)$ where ring addition and the ring product are defined as: * $0_R + 0_R = 0_R$ * $0_R \circ 0_R = 0_R$ Then $R$ is a ring.


Trivial Group is Cyclic Group

The trivial group is a cyclic group and therefore abelian.


Trivial Ring is Commutative Ring

Let $\struct {R, +, \circ}$ be a trivial ring. Then $\struct {R, +, \circ}$ is a commutative ring.


Congruence Relation and Ideal are Equivalent

Let $\left({R, +, \circ}\right)$ be a ring. Let $\mathcal E$ be an equivalence relation on $R$ compatible with both $\circ$ and $+$, i.e. a congruence relation on $R$. Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ be the equivalence class of $0_R$ under $\mathcal E$. Then: : $(1a): \quad J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ is an ideal of $R$ : $(2a): \quad$ The equivalence defined by the quotient ring $R / J$ is $\mathcal E$ itself. Similarly, let $J$ be an ideal of $R$. Then: : $(1b): \quad J$ induces a congruence relation $\mathcal E_J$ on $R$ : $(2b): \quad$ The ideal of $R$ defined by $\mathcal E_J$ is $J$ itself.


Quotient Ring is Ring

Let $\struct {R, +, \circ}$ be a ring. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$. Then $R / J$ is also a ring.


Ideal is Additive Normal Subgroup

Let $\left({R, +, \circ}\right)$ be a ring. Let $J$ be an ideal of $R$. Then $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$.


Characterization of Integrable Functions

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R, f \in \MM_{\overline \R}$ be a $\Sigma$-measurable function. Then the following are equivalent: :$(1): \quad f \in \map {\LL_{\overline \R} } \mu$, that is, $f$ is $\mu$-integrable. :$(2): \quad$ The positive and negative parts $f^+$ and $f^-$ are $\mu$-integrable. :$(3): \quad$ The absolute value $\size f$ of $f$ is $\mu$-integrable. :$(4): \quad$ There exists an $\mu$-integrable function $g: X \to \overline \R$ such that $\size f \le g$ pointwise.


Quotient Ring is Ring/Quotient Ring Addition is Well-Defined

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$. Then $+$ is well-defined on $R / J$, that is: :$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies \paren {x_1 + y_1} + J = \paren {x_2 + y_2} + J$


Implicitly Defined Real-Valued Function

Let $F: \struct {\mathbf X' \subseteq \R^{n + 1} } \to \struct {\mathbb I' \subseteq \R}$ have continuous partial derivatives.  {{explain|Can the language of this be brought into line with existing definitions of implicit functions?}} Let $\tuple {\mathbf x, z}$ denote an element of $\R^{n + 1}$, where $\mathbf x \in \R^n$ and $z \in \R$. Suppose $\exists \tuple {\mathbf x_0, z_0} \in \mathbf X'$ such that: :$\map F {\mathbf x_0, z_0} = 0$ :$\dfrac \partial {\partial z} \map F {\mathbf x_0, z_0} \ne 0$ Then there exists a unique mapping of the form: :$g: \mathbf X \to \mathbb I$ where $\mathbf X \subseteq \R^n$ contains $\mathbf x_0$ and $\mathbb I$ is an open real interval containing $z_0$, such that: :$\forall \mathbf x \in \mathbf X, z \in \mathbb I: \map F {\mathbf x, z} = 0 \iff z = \map g {\mathbf x}$ and $g$ itself has continuous partial derivatives. {{proofread}}


Characteristic Function of Subset

Let $A \subseteq B \subseteq S$. Then: :$\forall s \in S: \chi_A \left({s}\right) \le \chi_B \left({s}\right)$ where $\chi$ denotes characteristic function.


Characteristic Function Determined by 0-Fiber

Let $A \subseteq S$. Let $f: S \to \set {0, 1}$ be a mapping. Denote by $\chi_A$ the characteristic function on $A$. Then the following are equivalent: :$(1): \quad f = \chi_A$ :$(2): \quad \forall s \in S: \map f s = 0 \iff s \notin A$ Using the notion of a fiber, $(2)$ may also be expressed as: :$(2'):\quad \map {f^{-1} } 0 = S \setminus A$


Ideal induced by Congruence Relation defines that Congruence

Let $\left({R, +, \circ}\right)$ be a ring. Let $\mathcal E$ be a congruence relation on $R$. Let $J = \left[\!\left[{0_R}\right]\!\right]_\mathcal E$ be the ideal induced by $\mathcal E$. Then the equivalence defined by the coset space $\left({R, +}\right) / \left({J, +}\right)$ is $\mathcal E$ itself.


Rational Numbers Null Set under Lebesgue Measure

Let $\lambda$ be $1$-dimensional Lebesgue measure on $\R$. Let $\Q$ be the set of rational numbers. Then $\lambda \left({\Q}\right) = 0$, i.e. $\Q$ is a $\lambda$-null set.


Cartesian Product of Intersections/General Case

:$\displaystyle \paren{ \prod_{i \mathop \in I} S_i } \cap \paren{ \prod_{i \mathop \in I} T_i } = \prod_{i \mathop \in I} \paren{S_i \cap T_i}$


Cartesian Product of Semirings of Sets

Let $\SS$ and $\TT$ be semirings of sets. Then $\SS \times \TT$ is also a semiring of sets. Here, $\times$ denotes Cartesian product.


Congruence Relation on Group induces Normal Subgroup

Let $\left({G, \circ}\right)$ be a group whose identity is $e$. Let $\mathcal R$ be a congruence relation for $\circ$. Let $H = \left[\!\left[{e}\right]\!\right]_\mathcal R$, where $\left[\!\left[{e}\right]\!\right]_\mathcal R$ is the equivalence class of $e$ under $\mathcal R$. Then: : $\left({H, \circ \restriction_H}\right)$ is a normal subgroup of $G$ where $\circ \restriction_H$ denotes the restriction of $\circ$ to $H$.


Normal Subgroup induced by Congruence Relation defines that Congruence

Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $\mathcal R$ be a congruence relation for $\circ$. Let $\eqclass e {\mathcal R}$ be the equivalence class of $e$ under $\mathcal R$. Let $N = \eqclass e {\mathcal R}$ be the normal subgroup induced by $\mathcal R$. {{explain|Check the above because it does not look right}} Then $\mathcal R$ is the equivalence relation $\mathcal R_N$ defined by $N$.


Quotient Structure on Group defined by Congruence equals Quotient Group

Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $\mathcal R$ be a congruence relation for $\circ$. Let $\struct {G / \mathcal R, \circ_\mathcal R}$ be the quotient structure defined by $\mathcal R$. Let $N = \eqclass e {\mathcal R}$ be the normal subgroup induced by $\mathcal R$. Let $\struct {G / N, \circ_N}$ be the quotient group of $G$ by $N$. Then $\struct {G / \mathcal R, \circ_\mathcal R}$ is the subgroup $\struct {G / N, \circ_N}$ of the semigroup $\struct {\powerset G, \circ_\mathcal P}$.


Vector Scaled by Zero is Zero Vector

Let $F$ be a field whose zero is $0_F$. Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms. Then: :$\forall \mathbf v \in \mathbf V: 0_F \circ \mathbf v = \bszero$


Zero Vector Scaled is Zero Vector

Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over a field $F$, as defined by the vector space axioms. Then: :$\forall \lambda \in \mathbb F: \lambda \circ \bszero = \bszero$ where $\bszero \in \mathbf V$ is the zero vector.


Vector Product is Zero only if Factor is Zero

Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms. Then: :$\forall \lambda \in F: \forall \mathbf v \in \mathbf V: \lambda \circ \mathbf v = \bszero \implies \paren {\lambda = 0_F \lor \mathbf v = \mathbf 0}$ where $\bszero \in \mathbf V$ is the zero vector.


Null Ring is Ideal

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$. Then the null ring $\struct {\set {0_R}, +, \circ}$ is an ideal of $R$.


Ring Epimorphism Preserves Subrings

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $S$ be a subring of $R_1$. Then $\phi \sqbrk S$ is a subring of $R_2$.


Preimage of Image of Ideal under Ring Homomorphism

Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $K = \ker \left({\phi}\right)$, where $\ker \left({\phi}\right)$ is the kernel of $\phi$.  Let $J$ be an ideal of $R_1$. Then: :$\phi^{-1} \left({\phi \left({J}\right)}\right) = J + K$


Image of Preimage of Ideal under Ring Epimorphism

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $S_2$ be an ideal of $R_2$. Then: :$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$


Vectors are Right Cancellable

Let $\struct {\mathbf V, +, \circ}_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms. Then every $\mathbf v \in \struct {\mathbf V, +}$ is right cancellable: :$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf a + \mathbf c = \mathbf b + \mathbf c \implies \mathbf a = \mathbf b$


Vectors are Left Cancellable

Let $\left({\mathbf V, +, \circ}\right)_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms. Then every $\mathbf v \in \left({\mathbf V, +}\right)$ is left cancellable: :$\forall \mathbf a, \mathbf b, \mathbf c \in \mathbf V: \mathbf c + \mathbf a = \mathbf c + \mathbf b \implies \mathbf a = \mathbf b$


Additive Inverse in Vector Space is Unique

Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over a field $F$, as defined by the vector space axioms. Then for every $\mathbf v \in \mathbf V$, the additive inverse of $\mathbf v$ is unique: :$\forall \mathbf v \in \mathbf V: \exists! \paren {-\mathbf v} \in \mathbf V: \mathbf v + \paren {-\mathbf v} = \mathbf 0$


Vector Inverse is Negative Vector

Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms. Then: :$\forall \mathbf v \in \mathbf V: -\mathbf v = -1_F \circ \mathbf v$


Non-Trivial Commutative Division Ring is Field

Let $\struct {R, +, \circ}$ be a non-trivial division ring such that $\circ$ is commutative. Then $\struct {R, +, \circ}$ is a field. Similarly, let $\struct {F, +, \circ}$ be a field. Then $\struct {F, +, \circ}$ is a non-trivial  division ring such that $\circ$ is commutative.


Division Ring has No Proper Zero Divisors

Let $\left({R, +, \circ}\right)$ be a division ring. Then $\left({R, +, \circ}\right)$ has no proper zero divisors.


Linear Transformation is Injective iff Kernel Contains Only Zero

Let $\mathbf V, \mathbf V'$ be vector spaces, with respective zeroes $\mathbf 0, \mathbf 0'$. Let $T: \mathbf V \to \mathbf V'$ be a linear transformation. Then: :$T$ is injective {{iff}} $\map \ker T = \set {\mathbf 0}$ where: :$\mathbf 0$ is the zero of the domain of $T$ :$\map \ker T$ is the kernel of $T$.


Diagonal Relation on Ring is Ordering Compatible with Ring Structure

Let $\struct {R, +, \circ, \preceq}$ be a ring whose zero is $0_R$. Then the diagonal relation $\Delta_R$ on $R$ is an ordering compatible with the ring structure of $R$.


Nagata-Smirnov Metrization Theorem

Let $T = \left({S, \tau}\right)$ be a topological space. Then $T$ is metrizable {{iff}} $T$ is regular and has a basis that is countably locally finite.


Smirnov Metrization Theorem

Let $T = \left({S, \tau}\right)$ be a topological space. Then $T$ is metrizable {{iff}} it is paracompact and locally metrizable.


Identity Mapping is Automorphism/Semigroups

Let $\left({S, \circ}\right)$ be a semigroup. Then $I_S: \left({S, \circ}\right) \to \left({S, \circ}\right)$ is a semigroup automorphism.


Inverse of Reflexive Relation is Reflexive

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is reflexive, then so is $\mathcal R^{-1}$.


Inverse of Antireflexive Relation is Antireflexive

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is antireflexive, then so is $\mathcal R^{-1}$.


Inverse of Non-Reflexive Relation is Non-Reflexive

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is non-reflexive, then so is $\mathcal R^{-1}$.


Inverse of Symmetric Relation is Symmetric

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is symmetric, then so is $\mathcal R^{-1}$.


Inverse of Asymmetric Relation is Asymmetric

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is asymmetric, then so is $\mathcal R^{-1}$.


Inverse of Antisymmetric Relation is Antisymmetric

Let $\RR$ be a relation on a set $S$. If $\RR$ is antisymmetric, then so is $\RR^{-1}$.


Inverse of Non-Symmetric Relation is Non-Symmetric

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is non-symmetric, then so is $\mathcal R^{-1}$.


Inverse of Transitive Relation is Transitive

Let $\mathcal R$ be a relation on a set $S$. Let $\mathcal R$ be transitive. Then its inverse $\mathcal R^{-1}$ is also transitive.


Inverse of Antitransitive Relation is Antitransitive

Let $\RR$ be a relation on a set $S$. If $\RR$ is antitransitive, then so is $\RR^{-1}$.


Inverse of Non-Transitive Relation is Non-Transitive

Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ is non-transitive, then so is $\mathcal R^{-1}$.


Inverse of Ordered Ring Isomorphism is Ordered Ring Isomorphism

Let $\left({S, +, \circ, \preceq}\right)$ and $\left({T, \oplus, *, \preccurlyeq}\right)$ be ordered rings. Let $\phi: S \to T$ be an ordered ring isomorphism. Then $\phi^{-1}: T \to S$ is also an ordered ring isomorphism.


Feit-Thompson Theorem

All finite groups of odd order are solvable. That is, every non-abelian group finite simple group is of even order.


Orthogonal Group is Subgroup of General Linear Group

Let $k$ be a field. Let $\operatorname O \left({n, k}\right)$ be the $n$th orthogonal group on $k$. Let $\operatorname{GL} \left({n, k}\right)$ be the $n$th general linear group on $k$. Then $\operatorname O \left({n, k}\right)$ is a subgroup of $\operatorname{GL} \left({n, k}\right)$.


Orthogonal Group is Group

Let $k$ be a field. The $n$th orthogonal group on $k$ is a group.


Composite of Ordered Ring Isomorphisms is Ordered Ring Isomorphism

Let $\left({S_1, +_1, \circ_1, \preccurlyeq_1}\right), \left({S_2, +_2, \circ_2, \preccurlyeq_2}\right), \left({S_3, +_3, \circ_3, \preccurlyeq_3}\right)$ be ordered rings. Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring isomorphisms. Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring isomorphism.


Composite of Ordered Ring Monomorphisms is Ordered Ring Monomorphism

Let $\left({S_1, +_1, \circ_1, \preccurlyeq_1}\right), \left({S_2, +_2, \circ_2, \preccurlyeq_2}\right), \left({S_3, +_3, \circ_3, \preccurlyeq_3}\right)$ be ordered rings. Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring monomorphisms. Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring monomorphism.


Rescaling is Linear Transformation

Let $\left({R, +, \cdot}\right)$ be a commutative ring. Let $\left({V, +, \circ}\right)_R$ be an $R$-module. Then for any $r \in R$, the rescaling: :$m_r: V \to V, v \mapsto r \circ v$ is a linear transformation.


Determinant of Rescaling Matrix

Let $R$ be a commutative ring. Let $r \in R$. Let $r \, \mathbf I_n$ be the square matrix of order $n$ defined by: :$\sqbrk {r \, \mathbf I_n}_{i j} = \begin{cases} r & : i = j \\ 0 & : i \ne j \end{cases}$ Then: :$\map \det {r \, \mathbf I_n} = r^n$ where $\det$ denotes determinant.


Inverse of Rescaling Matrix

Let $R$ be a commutative ring with unity. Let $r \in R$ be a unit in $R$. Let $r \, \mathbf I_n$ be the $n \times n$ rescaling matrix of $r$. Then $\left({r \, \mathbf I_n}\right)^{-1} = r^{-1} \, \mathbf I_n$.


Powers of Ring Elements/General Result

:$\forall m, n \in \Z: \forall x \in R: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$.


Characteristic of Division Ring is Zero or Prime

Let $\struct {D, +, \circ}$ be a division ring. Let $\map {\operatorname {Char} } D$ be the characteristic of $D$. Then $\map {\operatorname {Char} } D$ is either $0$ or a prime number.


Magdy's Exchanger

'''Magdy's Exchanger''' is a simple method to get a semi-algebraic formula for inverse trigonometric functions using the help of inverse hyperbolic functions that has a real logarthmic formula hence we will have an example to get the inverse sine function. First: :$\displaystyle \arcsin x  = \int \frac 1 {\sqrt{1 - x^2}} \, \mathrm d x$ Using Integration by Parts: :$\displaystyle \int \frac 1 {\sqrt{1-x^2}} \,\mathrm d x = -\frac {\sqrt {1-x^2}} x - \int \frac{\sqrt{1-x^2}} {x^2} \, \mathrm d x$ The problem is that the integration on the {{RHS}} will inverse itself if we solve it by parts. Then the result will be: :$0 = 0$ Magdy's method will come into effect now. As we know the inverse hyperbolic function that is most similar to the inverse sine function is the inverse sinh function. The idea is to get the last integration in the {{RHS}} from the inverse sinh function. So we use parts again on the derivative of inverse sinh as follows: :$\displaystyle \operatorname{arcsinh} x = \int \frac 1 {\sqrt{1 + x^2}} \, \mathrm d x$ {{begin-eqn}} {{eqn | l = \operatorname{arcsinh} x       | r = \int \frac 1 {\sqrt{1 + x^2} } \, \mathrm d x       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ }} {{eqn | r = \frac {\sqrt {1 + x^2} } x + \int \frac {\sqrt {1 + x^2} } {x^2} \,\mathrm d x       | c = Integration by Parts }} {{eqn | r = \ln \left({x + \sqrt{x^2 + 1} }\right)       | c = Definition of Inverse Sinh Function }} {{end-eqn}} By Primitive of Root of $\dfrac {\sqrt {x^2 + a^2} } {x^2}$: :$\displaystyle \int \frac {\sqrt {1 + x^2} } {x^2} \, \mathrm d x = \ln \left({x + \sqrt{x^2 + 1} }\right) - \frac {\sqrt {1 + x^2} } x$ Taking the derivativeof both sides: :$\displaystyle \frac{\sqrt{1 + x^2} } {x^2} = \frac {1 + x^2 + x \sqrt{x^2 + 1} } {x^3 + x^2 \sqrt {x^2 + 1} } = A$ Squaring both sides: :$\dfrac {1 + x^2} {x^4} = A^2$ Now the most important step. As we want to have $1 - x^2$ in the numerator we will have the fraction on the {{LHS}}} as: :$\dfrac {1 - x^2 + 2 x^2} {x^4}$ which does not affect the value of the fraction. Now the {{LHS}} is: :$\dfrac {1 - x^2} {x^4} + \dfrac 2 {x^2}$ then the wanted fraction will be alone on the {{LHS}}: :$\dfrac {1 - x^2} {x^4} = A^2 - \dfrac 2 {x^2}$ Taking the root of both sides: :$\dfrac {\sqrt{1 - x^2} } {x^2} = \sqrt {A^2 - \dfrac 2 {x^2} }$ then integrating both sides: :$\displaystyle \int \frac {\sqrt{1 - x^2} } {x^2} \, \mathrm d x = \int \sqrt{A^2 - \frac 2 {x^2}} \, \mathrm d x$ Now the {{RHS}} is complicated but evaluates as follows: :$\displaystyle \int \sqrt{A^2 - \frac 2 {x^2} } \, \mathrm d x = \frac {x^2 \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } \ln \left({2 \sqrt {x^2 - 1} + 2 x}\right)} {\sqrt {x^2 - 1} } - x \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } + C$ Substituting the last result in the first equation of the inverse sine we get: :$\displaystyle \arcsin x = - \frac {\sqrt {1 - x^2} } x - \frac {x^2 \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } \ln \left({2 \sqrt {x^2 - 1} + 2 x}\right)} {\sqrt {x^2 - 1}} +  x \sqrt {\frac 1 {x^4} - \frac 1 {x^2} } + C$ Hence we get a semi-algebraic formula for the inverse sine function. {{qed}}


Sigma-Algebra Generated by Complements of Generators

Let $\Sigma$ be a $\sigma$-algebra on a set $X$. Let $\mathcal G$ be a generator for $\Sigma$. Then: :$\mathcal{G}' := \left\{{X \setminus G: G \in \mathcal G}\right\}$ the set of relative complements of $\mathcal G$, is also a generator for $\Sigma$.


Primary Decomposition Theorem

Let $K$ be a field. Let $V$ be a vector space over $K$. Let $T: V \to V$ be a linear operator on $V$. Let $\map p x \in K \sqbrk x$ be a polynomial such that: :$\map \deg p \ge 1$ :$\map p T = 0$ where $0$ is the zero operator on $V$. {{explain|Link to definition of $K \sqbrk x$}} Let $\map {p_1} x, \map {p_2} x, \ldots, \map {p_r} x$ be distinct irreducible monic polynomials. Let $c \in K \setminus \set 0$ and $a_1, a_2, \ldots, a_r, r \in \Z_{\ge 1}$ be constants. We have that: :$\map p x = c \map {p_1} x^{a_1} \map {p_2} x^{a_2} \dotsm \map {p_r} x^{a_r}$ The primary decomposition theorem then states the following : :$(1): \quad \map \ker {\map {p_i} T^{a_i} }$ is a $T$-invariant subspace of $V$ for all $i = 1, 2, \dotsc, r$ :$(2): \quad \displaystyle V = \bigoplus_{i \mathop = 1}^r \map \ker {\map {p_i} T^{a_i} }$ {{explain|Link to definition of $\bigoplus$ in this context.}} {{improve|Rather than dragging the unwieldy $\map \ker {\map {p_i} T^{a_i} }$ all around the page, suggest that a symbol e.g. $\kappa_i$ be used for it instead. There are many more places where the exposition is repetitive and could benefit from being broken into more modular units. And there are still places where the logical flow is compromised by being wrapped backwards upon itself with "because" and "indeed" and nested "if-thens", although I have done my best to clean out most of these.}}


Sigma-Algebra Extended by Single Set

Let $\Sigma$ be a $\sigma$-algebra on a set $X$. Let $S \subseteq X$ be a subset of $X$. For subsets $T \subseteq X$ of $X$, denote $T^c$ for the set difference $X \setminus T$. Then: :$\sigma \left({\Sigma \cup \left\{{S}\right\}}\right) = \left\{{\left({E_1 \cap S}\right) \cup \left({E_2 \cap S^c}\right): E_1, E_2 \in \Sigma}\right\}$ where $\sigma$ denotes generated $\sigma$-algebra.


Subring Module/Special Case

Let $S$ be a subring of the ring $\struct {R, +, \circ}$. Let $\circ_S$ be the restriction of $\circ$ to $S \times R$. Then $\struct {R, +, \circ_S}_S$ is an $S$-module. If $\struct {R, +, \circ}$ has a unity, $1_R$, and $1_R \in S$, then $\struct {R, +, \circ_S}_S$ is a unitary $S$-module.


Trivial Module is Module

Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$. Let $\struct {R, +_R, \circ_R}$ be a ring. Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that: :$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$ Then $\struct {G, +_G, \circ}_R$ is a module.


Null Module is Module

Let $\left({R, +_R, \circ_R}\right)$ be a ring. Let $G$ be the trivial group. Let $\left({G, +_G, \circ}\right)_R$ be the null module. Then $\left({G, +_G, \circ}\right)_R$ is a module.


Submodule of Module of Polynomial Functions

Let $K$ be a commutative ring with unity. Let $P \left({K}\right)$ be the set of all polynomial functions on $K$. Consider the set $P_m \left({K}\right)$ of all the polynomial functions: :$\displaystyle \sum_{k \mathop = 0}^{m-1} \alpha_k {I_K}^k$ for some $m \in \N^*$ where: :$\left \langle {\alpha_k} \right \rangle_{k \in \left[{0 \,.\,.\, m-1}\right]}$ is any sequence of $m$ terms of $K$. Then $P_m \left({K}\right)$ is a submodule of $P \left({K}\right)$.


Condition on Equality of Generated Sigma-Algebras

Let $X$ be a set, and let $\mathcal G$, $\mathcal H$ be sets of subsets of $X$. Suppose that: :$\mathcal G \subseteq \mathcal H \subseteq \sigma \left({\mathcal G}\right)$ where $\sigma$ denotes generated $\sigma$-algebra. Then: :$\sigma \left({\mathcal G}\right) = \sigma \left({\mathcal H}\right)$


Sigma-Algebra is Monotone Class

Let $\Sigma$ be a $\sigma$-algebra on a set $X$. Then $\Sigma$ is also a monotone class.


Trace Sigma-Algebra of Generated Sigma-Algebra

Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$. Let $A \subseteq X$ be a subset of $X$. Then the following equality holds: :$A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$ where :$\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$ :$\map \sigma {A \cap \GG}$ denotes the smallest $\sigma$-algebra on $A$ that contains ${A \cap \GG}$ :$A \cap \map \sigma \GG$ denotes the trace $\sigma$-algebra on $A$ :$A \cap \GG$ is a shorthand for $\set {A \cap G: G \in \GG}$


Intersection is Subset/General Result

Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\mathbb S \subseteq \mathcal P \left({S}\right)$. Then: : $\displaystyle \forall T \in \mathbb S: \bigcap \mathbb S \subseteq T$ === Family of Sets === In the context of a family of sets, the result can be presented as follows: {{:Intersection is Subset/Family of Sets}}


Union of Relative Complements of Nested Subsets

Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions. Then: :$\complement_T \left({S}\right) \cup \complement_S \left({R}\right) = \complement_T \left({R}\right)$ where $\complement$ denotes relative complement. Phrased via Set Difference as Intersection with Relative Complement: :$\left({T \setminus S}\right) \cup \left({S \setminus R}\right) = T \setminus R$ where $\setminus$ denotes set difference.


Homomorphic Image of R-Module is R-Module

Let $\left({R, +_R, \times_R}\right)$ be a ring. Let $\left({G, +_G, \circ_G}\right)_R$ be an $R$-module. Let $\left({H, +_H, \circ_H}\right)_R$ be an $R$-algebraic structure. Let $\phi: G \to H$ be a homomorphism. Then the homomorphic image of $\phi$ is an $R$-module.


Evaluation Linear Transformation is Linear Transformation

Let $R$ be a commutative ring. Let $G$ be an $R$-module. Let $G^*$ be the algebraic dual of $G$. Let $G^{**}$ be the algebraic dual of $G^*$. Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$. For each $x \in G$,  $x^\wedge: G^* \to R$ is defined as: :$\forall t' \in G^*: \map {x^\wedge} {t'} = \map {t'} x$ Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$ defined as: :$\forall x \in G: \map J x = x^\wedge$ where for each $x \in G$, $x^\wedge: G^* \to R$ is defined as: :$\forall t' \in G^*: \map {x^\wedge} {t'} = \map {t'} x$   Then: :$(1): \quad x^\wedge \in G^{**}$ :$(2): \quad J$ is a linear transformation.


Evaluation Isomorphism is Isomorphism

Let $R$ be a commutative ring. Let $G$ be a unitary $R$-module whose dimension is finite. Then the evaluation linear transformation $J: G \to G^{**}$ is an isomorphism.


Zero Matrix is Identity for Hadamard Product

Let $\struct {S, \cdot}$ be a monoid whose identity is $e$. Let $\map {\MM_S} {m, n}$ be an $m \times n$ matrix space over $S$. Let $\mathbf e = \sqbrk e_{m n}$ be the zero matrix of $\map {\MM_S} {m, n}$. Then $\mathbf e$ is the identity element for Hadamard product.


Sum of Ideals is Ideal/General Result

Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$. Then: : $J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$. where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product.


Integer is Expressible as Product of Primes

Let $n$ be an integer such that $n > 1$. Then $n$ can be expressed as the product of one or more primes.


Expression for Integer as Product of Primes is Unique

Let $n$ be an integer such that $n > 1$. Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.


Modulo Addition is Closed/Integers

Let $m \in \Z$ be an integer. Then addition modulo $m$ on the set of integers modulo $m$ is closed: :$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m \in \Z_m$.


Set is Subset of Union/General Result

Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $\mathbb S \subseteq \mathcal P \left({S}\right)$. Then: : $\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$


Intersection Distributes over Union/General Result

Let $S$ and $T$ be sets. Let $\powerset T$ be the power set of $T$. Let $\mathbb T$ be a subset of $\powerset T$. Then: :$\displaystyle S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$


Union Distributes over Intersection/General Result

Let $S$ and $T$ be sets. Let $\powerset T$ be the power set of $T$. Let $\mathbb T$ be a subset of $\powerset T$. Then: :$\displaystyle S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$


Scalar Multiple of Simple Function is Simple Function

Let $\struct {X, \Sigma}$ be a measurable space. Let $f: X \to \R$ be a simple function, and let $\lambda \in \R$. Then the pointwise scalar multiple $\lambda f: X \to \R$ of $f$ is also a simple function.


Image of Composite Mapping

Let $f: S \to T$ and $g: R \to S$ be mappings. Then: :$\operatorname{Im} \left({f \circ g}\right) = f \left[{\operatorname{Im} \left({g}\right)}\right]$ where: : $f \circ g$ is the composition of $g$ and $f$ : $\operatorname{Im}$ denotes image : $f \left[{\cdot}\right]$ denotes taking image of a subset under $f$.


Scalar Multiple of Integrable Function is Integrable Function

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R$ be a $\mu$-integrable function, and let $\lambda \in \R$. Then $\lambda f: X \to \overline \R$, the pointwise $\lambda$-multiple of $f$, is also $\mu$-integrable. That is, the space of integrable functions $\LL^1_{\overline \R}$ is closed under pointwise $\R$-scalar multiplication.


Left Distributive Law for Natural Numbers

The operation of multiplication is left distributive over addition on the set of natural numbers $\N$: :$\forall x, y, n \in \N_{> 0}: n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$


Right Distributive Law for Natural Numbers

The operation of multiplication is right distributive over addition on the set of natural numbers $\N_{> 0}$: :$\forall x, y, n \in \N_{> 0}: \paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$


Standard Machinery

Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $\mathcal{L}^1_{\overline \R} \left({\mu}\right)$ be the space of $\mu$-integrable functions. Let $P \left({f_1, \ldots, f_n}\right)$ be a proposition, where the variables $f_i$ denote $\mu$-measurable functions $f_i: X \to \overline{\R}$. Let every occurrence of an $f_i$ be of the form: :$\displaystyle \int \Phi \left({f_i}\right) \, \mathrm d\mu$ for a suitable index set $I$ and multilinear mapping $\Phi: \mathcal{L}^1_{\overline \R} \left({\mu}\right)^I \to \mathcal{L}^1_{\overline \R} \left({\mu}\right)$. Denote with $\chi \left({\Sigma}\right)$ the set of characteristic functions of elements of $\Sigma$, i.e.: :$\chi \left({\Sigma}\right) := \left\{{\chi_E: X \to \R: E \in \Sigma}\right\}$ Then the following are equivalent: :$(A): \quad \forall f_1, \ldots, f_n \in \chi \left({\Sigma}\right): P \left({f_1, \ldots, f_n}\right)$ :$(B): \quad \forall f_1, \ldots, f_n \in \mathcal{L}^1_{\overline \R} \left({\mu}\right): P \left({f_1, \ldots, f_n}\right)$ that is, it suffices to verify $P$ holds for characteristic functions. {{WIP|Not completely valid as stated; need to restrict to $E$ with $\mu(E)<+\infty$ and impose more conditions on $P$}}


Third Principle of Mathematical Induction

Let $\map P n$ be a propositional function depending on $n \in \N$. If: :$(1): \quad \map P n$ is true for all $n \le d$ for some $d \in \N$ :$(2): \quad \forall m \in \N: \paren {\forall k \in \N, m \le k < m + d: \map P k} \implies \map P {m + d}$ then $\map P n$ is true for all $n \in \N$.


Cross-Relation is Equivalence Relation

{{Cross-Relation Context Definition}} Then $\boxtimes$ is an equivalence relation on $\struct {S_1 \times S_2, \oplus}$.


Cross-Relation on Natural Numbers is Equivalence Relation

Let $\struct {\N, +}$ be the semigroup of natural numbers under addition. Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$. The relation $\boxtimes$ defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ is an equivalence relation on $\struct {\N \times \N, \oplus}$.


Equivalence Classes of Cross-Relation on Natural Numbers

Let $\struct {\N, +}$ be the semigroup of natural numbers under addition. Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$. Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ Then $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtimes$, where: :$\eqclass {\tuple {x_1, y_1} } \boxtimes = \eqclass {\tuple {x_2, y_2} } \boxtimes \iff \tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2}$ The equivalence class $\eqclass {\tuple {x, y} } \boxtimes$ is more often denoted more simply as $\eqclass {x, y} {}$.


Addition of Cross-Relation Equivalence Classes on Natural Numbers is Cancellable

Let $\struct {\N, +}$ be the semigroup of natural numbers under addition. Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$. Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ Let $\eqclass {x, y} {}$ denote the equivalence class of $\tuple {x, y}$ under $\boxtimes$. {{WIP|Introduce the language of the Definition:Quotient Set.}} The operation $\oplus$ on these equivalence classes is cancellable, in the sense that: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {} \oplus \eqclass {c_1, d_1} {}       | r = \eqclass {a, b} {} \oplus \eqclass {c_2, d_2} {}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {c_1, d_1} {}       | r = \eqclass {c_2, d_2} {}       | c =  }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = \eqclass {c_1, d_1} {} \oplus \eqclass {a, b} {}       | r = \eqclass {c_2, d_2} {} \oplus \eqclass {a, b} {}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {c_1, d_1} {}       | r = \eqclass {c_2, d_2} {}       | c =  }} {{end-eqn}}


Cross-Relation Equivalence Classes on Natural Numbers are Cancellable for Addition

Let $\struct {\N, +}$ be the semigroup of natural numbers under addition. Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$. Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ Let $\eqclass {x, y} {}$ denote the equivalence class of $\tuple {x, y}$ under $\boxtimes$. {{wtd|Introduce the language of the Definition:Quotient Set.}} The operation $\oplus$ on these equivalence classes is cancellable, in the sense that: {{begin-eqn}} {{eqn | l = \eqclass {a_1, b_1} {}       | r = \eqclass {a_2, b_2} {}       | c =  }} {{eqn | l = \eqclass {c_1, d_1} {}       | r = \eqclass {c_2, d_2} {}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {a_1, b_1} {} \oplus \eqclass {c_1, d_1} {}       | r = \eqclass {a_2, b_2} {} \oplus \eqclass {c_2, d_2} {}       | c =  }} {{end-eqn}}


Cross-Relation is Congruence Relation

{{Cross-Relation Context Definition}} The cross-relation $\boxtimes$ is a congruence relation on $\struct {S_1 \times S_2, \oplus}$.


Elements of Cross-Relation Equivalence Class

{{Cross-Relation Context Definition}} Let $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ be the $\boxtimes$-equivalence class of $\left({x, y}\right)$, where $\left({x, y}\right) \in S_1 \times S_2$. Then: $\forall x, y \in S_1, a, b \in S_2:$ : $(1): \quad \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes = \left[\!\left[{\left({y \circ b, b}\right)}\right]\!\right]_\boxtimes \iff x = y$ : $(2): \quad \left[\!\left[{\left({x \circ a, y \circ a}\right)}\right]\!\right]_\boxtimes = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$


Equivalence Class of Equal Elements of Cross-Relation

{{Cross-Relation Context Definition}} Then: :$\forall c, d \in S_1 \cap S_2: \tuple {c, c} \boxtimes \tuple {d, d}$


Length of Concatenation

Let $S$ and $T$ be words, and let $ST$ be their concatenation. Then: :$\left\vert{ST}\right\vert = \left\vert{S}\right\vert + \left\vert{T}\right\vert$ where $\left\vert{S}\right\vert$ denotes the length of $S$.


Relation is Connected and Reflexive iff Total

Let $S$ be a set. Let $\mathcal R$ be a relation on $S$. Then: :$\mathcal R$ is both a connected relation and a reflexive relation {{iff}}: :$\mathcal R$ is a total relation.


Total Ordering is Total Relation

Let $S$ be a set. Let $\mathcal R \subseteq S \times S$ be a total ordering. Then $\mathcal R$ is a total relation.


Natural Number Addition Commutativity with Successor/Proof 2

Let $\N$ be the natural numbers. Then: :$\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$


Natural Number Commutes with 1 under Addition

Let $n \in \N_{> 0}$ be a natural number. Then $n$ commutes with $1$ under the operation of addition: :$\forall n \in \N_{> 0}: n + 1 = 1 + n$


Natural Number Addition is Commutative/Proof 3

The operation of addition on the set of natural numbers $\N_{> 0}$ is commutative: :$\forall x, y \in \N_{> 0}: x + y = y + x$


Ordering on Natural Numbers is Trichotomy

Let $\N$ be the natural numbers. Let $<$ be the (strict) ordering on $\N$. Then exactly one of the following is true: :$(1): \quad a = b$ :$(2): \quad a > b$ :$(3): \quad a < b$ That is, $<$ is a trichotomy on $\N$.


Ordering on 1-Based Natural Numbers is Transitive

Let $\N_{> 0}$ be the 1-based natural numbers. Let $<$ be the strict ordering on $\N_{>0}$. Then $<$ is a transitive relation.


Trichotomy is Antireflexive

Let $\RR$ be a trichotomy. Then $\RR$ is an antireflexive relation.


Ordering on 1-Based Natural Numbers is Total Ordering

Let $\N_{> 0}$ be the $1$-based natural numbers. Let $<$ be the strict ordering on $\N_{>0}$. Then $<$ is a (strict) total ordering.


Left Cancellable Commutative Operation is Right Cancellable

Let $\struct {S, \circ}$ be an algebraic structure. Let $\circ$ be left cancellable and also commutative. Then $\circ$ is also right cancellable.


Right Cancellable Commutative Operation is Left Cancellable

Let $\struct {S, \circ}$ be an algebraic structure. Let $\circ$ be right cancellable and also commutative. Then $\circ$ is also left cancellable.


Natural Number Addition is Cancellable

Let $\N$ be the natural numbers. Let $+$ be addition on $\N$. Then: :$\forall a, b, c \in \N: a + c = b + c \implies a = b$ :$\forall a, b, c \in \N: a + b = a + c \implies b = c$ That is, $+$ is cancellable on $\N$.


Uniqueness of Measures

Let $\left({X, \Sigma}\right)$ be a measurable space. Let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a generator for $\Sigma$; i.e., $\Sigma = \sigma \left({\mathcal G}\right)$. Suppose that $\mathcal G$ satisfies the following conditions: :$(1):\quad \forall G, H \in \mathcal G: G \cap H \in \mathcal G$ :$(2):\quad$ There exists an exhausting sequence $\left({G_n}\right)_{n \in \N} \uparrow X$ in $\mathcal G$ Let $\mu, \nu$ be measures on $\left({X, \Sigma}\right)$, and suppose that: :$(3):\quad \forall G \in \mathcal G: \mu \left({G}\right) = \nu \left({G}\right)$ :$(4):\quad \forall n \in \N: \mu \left({G_n}\right)$ is finite Then $\mu = \nu$. Alternatively, by Countable Cover induces Exhausting Sequence, the exhausting sequence in $(2)$ may be replaced by a countable $\mathcal G$-cover $\left({G_n}\right)_{n \in \N}$, still subject to $(4)$.


Uniqueness of Measures/Proof 2


Natural Number Multiplication is Cancellable

Let $\N$ be the natural numbers. Let $\times$ be multiplication on $\N$. Then: :$\forall x, y, z \in \N_{>0}: x \times z = y \times z \implies x = y$ :$\forall x, y, z \in \N_{>0}: x \times y = x \times z \implies y = z$ That is, $\times$ is cancellable on $\N_{>0}$.


Natural Number Addition is Cancellable for Ordering

Let $\N$ be the natural numbers. Let $<$ be the strict ordering on $\N$. Let $+$ be addition on $\N$. Then: :$\forall a, b, c \in \N_{>0}: a + c < b + c \implies a < b$ :$\forall a, b, c \in \N_{>0}: a + b < a + c \implies b < c$ That is, $+$ is cancellable on $\N$ for $<$.


Natural Number Multiplication is Cancellable for Ordering

Let $\N$ be the natural numbers. Let $\times$ be multiplication on $\N$. Let $<$ be the strict ordering on $\N$. Then: :$\forall a, b, c \in \N: a \times c < b \times c \implies a < b$ :$\forall a, b, c \in \N: a \times b < a \times c \implies b < c$ That is, $\times$ is cancellable on $\N$ for $<$.


Successor to Natural Number

Let $\N_{> 0}$ be the 1-based natural numbers: :$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$ Let $<$ be the ordering on $\N_{> 0}$: :$\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$ Let $a \in \N_{>0}$. Then there exists no natural number $n$ such that $a < n < a + 1$.


Natural Number is Not Equal to Successor

Let $\N_{> 0}$ be the 1-based natural numbers: :$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$ Then: :$\forall n \in \N_{> 0}: n \ne n + 1$


Biconditional is Commutative

==== Formulation 1 ==== {{:Biconditional is Commutative/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional is Commutative/Formulation 2}}


Biconditional is Associative

==== Formulation 1 ==== {{:Biconditional is Associative/Formulation 1}} ==== Formulation 2 ==== {{:Biconditional is Associative/Formulation 2}}


Biconditional is Reflexive

: $\vdash p \iff p$


Biconditional is Transitive

The biconditional operator is transitive:


Ordinals are Totally Ordered

The ordinals are totally ordered.


Successor Set of Transitive Set is Transitive

Let $S$ be a transitive set. Then its successor set $S\,^+ = S \cup \left\{{S}\right\}$ is also transitive.


Successor Set of Ordinal is Ordinal

Let $S$ be an ordinal. Then its successor set $S^+ = S \cup \set S$ is also an ordinal.


Ordinals are Well-Ordered

The ordinals are well-ordered.


Set Intersection Preserves Subsets/Corollary

Let $A, B, S$ be sets. Then: :$A \subseteq B \implies A \cap S \subseteq B \cap S$


Set Intersection Preserves Subsets/Families of Sets/Corollary

Let $I$ be an indexing set. Let $\family {B_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$. Let $A$ be a set such that $A \subseteq B_\alpha$ for all $\alpha \in I$. Then: :$\displaystyle A \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$


Intersection is Largest Subset/General Result

Let $T$ be a set. Let $\powerset T$ be the power set of $T$. Let $\mathbb T$ be a subset of $\powerset T$. Then: :$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$ === Family of Sets === In the context of a family of sets, the result can be presented as follows: {{:Intersection is Largest Subset/Family of Sets}}


Union is Smallest Superset/General Result

Let $S$ and $T$ be sets. Let $\powerset S$ denote the power set of $S$. Let $\mathbb S$ be a subset of $\powerset S$. Then: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$ === Family of Sets === In the context of a family of sets, the result can be presented as follows: {{:Union is Smallest Superset/Family of Sets}}


Mapping is Involution iff Bijective and Symmetric

Let $S$ be a set. Let $f: S \to S$ be a mapping on $S$. Then $f$ is an involution {{iff}} $f$ is both a bijection and a symmetric relation.


Let $S$ be a set. Let $T$ be a subset of $S$. Let $\mathcal P \left({T}\right)$ be the power set  of $T$. Let $\mathbb T \subseteq \mathcal P \left({T}\right)$.  Then: ==== Complement of Intersection ==== {{:De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Intersection}} ==== Complement of Union ==== {{:De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union}}


Absorption Laws (Set Theory)

These two results together are known as the '''Absorption Laws''', corresponding to the equivalent results in logic.


Cardinality of Set Union/Corollary

Let $S_1, S_2, \ldots, S_n$ be finite sets which are pairwise disjoint. Then: :$\displaystyle \card {\bigcup_{i \mathop = 1}^n S_i} = \sum_{i \mathop = 1}^n \card {S_i}$ Specifically: : $\card {S_1 \cup S_2} = \card {S_1} + \card {S_2}$


Image of Intersection under Relation/General Result

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $\powerset S$ be the power set of $S$. Let $\mathbb S \subseteq \powerset S$. Then: :$\displaystyle \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$


Image of Intersection under One-to-Many Relation/General Result

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $\powerset S$ be the power set of $S$. Then: :$\displaystyle \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$  {{iff}} $\RR$ is one-to-many.


Image of Union under Relation/General Result

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $\powerset S$ be the power set of $S$. Let $\mathbb S \subseteq \powerset S$. Then: :$\displaystyle \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$


Preimage of Union under Mapping/General Result

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $\powerset T$ be the power set of $T$. Let $\mathbb T \subseteq \powerset T$. Then: :$\displaystyle f^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$


Preimage of Union under Relation/General Result

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $\powerset T$ be the power set of $T$. Let $\mathbb T \subseteq \powerset T$. Then: :$\displaystyle \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$ where $\RR^{-1} \sqbrk X$ denotes the preimage of $X$ under $\RR$.


Image of Intersection under Mapping/General Result

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $\powerset S$ be the power set of $S$. Let $\mathbb S \subseteq \powerset S$. Then: :$\displaystyle f \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$


Preimage of Intersection under Mapping/General Result

Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $\mathcal P \left({T}\right)$ be the power set of $T$. Let $\mathbb T \subseteq \mathcal P \left({T}\right)$. Then: :$\displaystyle f^{-1} \left[{\bigcap \mathbb T}\right] = \bigcap_{X \mathop \in \mathbb T} f^{-1} \left[{X}\right]$


Bijective Relation has Left and Right Inverse

Let $\mathcal R \subseteq S \times T$ be a relation on a cartesian product $S \times T$. Let: :$I_S$ be the identity mapping on $S$ :$I_T$ be the identity mapping on $T$. Let $\mathcal R^{-1}$ be the inverse relation of $\mathcal R$. Let $\mathcal R$ be a bijection. Then: :$\mathcal R^{-1} \circ \mathcal R = I_S$ and :$\mathcal R \circ \mathcal R^{-1} = I_T$ where $\circ$ denotes composition of relations.


Left and Right Inverse Relations Implies Bijection

Let $\mathcal R \subseteq S \times T$ be a relation on a cartesian product $S \times T$. Let: :$I_S$ be the identity mapping on $S$ :$I_T$ be the identity mapping on $T$. Let $\mathcal R^{-1}$ be the inverse relation of $\mathcal R$. Let $\mathcal R$ be such that: :$\mathcal R^{-1} \circ \mathcal R = I_S$ and :$\mathcal R \circ \mathcal R^{-1} = I_T$ where $\circ$ denotes composition of relations. Then $\mathcal R$ is a bijection.


Bijection has Left and Right Inverse

Let $f: S \to T$ be a bijection. Let: : $I_S$ be the identity mapping on $S$ : $I_T$ be the identity mapping on $T$. Let $f^{-1}$ be the inverse of $f$. Then: : $f^{-1} \circ f = I_S$ and: : $f \circ f^{-1} = I_T$ where $\circ$ denotes composition of mappings.


Symmetry in Space Implies Conservation of Momentum

The total derivative of the action $S_{12}$ from states $1$ to $2$ with regard to position is equal to the difference in momentum from states $1$ to $2$: :$\dfrac {\d S_{1 2} } {\d x} = p_2 - p_1$ {{MissingLinks|Although we do have a page Definition:State, it refers to a concept in game theory and not physics.}}


Mappings in Product of Sets are Surjections

Let $S$ and $T$ be sets. Let $\struct {P, \phi_1, \phi_2}$ be a product of $S$ and $T$. Then $\phi_1$ and $\phi_1$ are surjections.


Cartesian Product is Set Product

Let $S$ and $T$ be sets. Let $S \times T$ be the Cartesian product of $S$ and $T$. Let $\pr_1: S \times T \to S$ and $\pr_2: S \times T \to T$ be the first and second projections respectively on $S \times T$. Then $\struct {S \times T, \pr_1, \pr_2}$ is a set product.


Positive Rational Number as Power of Number with Power of Itself

Every positive rational number can be written either as: : $a^{a^a}$ for some irrational number $a$ or as: : $n^{n^n}$ for some natural number $n$.


Rational Number as Power of Number with Itself

Every rational number in the interval $\openint {\paren {\dfrac 1 e}^{\frac 1 e} }{+\infty}$ can be written either as: : $a^a$ for some irrational number $a$ or as: : $n^n$ for some natural number $n$.


Integral of Integrable Function is Homogeneous

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R$ be a $\mu$-integrable function. Let $\lambda \in \R$. Then: :$\displaystyle \int \lambda f \rd \mu = \lambda \int f \rd \mu$ where $\lambda f$ is the pointwise $\lambda$-multiple of $f$.


Pointwise Sum of Integrable Functions is Integrable Function

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f, g: X \to \overline \R$ be $\mu$-integrable functions. Suppose that their pointwise sum $f + g$ is well-defined. Then $f + g$ is also a $\mu$-integrable function. That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise addition.


Integral of Integrable Function is Additive

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f, g: X \to \overline \R$ be $\mu$-integrable functions. Suppose that their pointwise sum $f + g$ is well-defined. Then: :$\displaystyle \int f + g \rd \mu = \int f \rd \mu + \int g \rd \mu$


Pointwise Minimum of Integrable Functions is Integrable Function

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f, g: X \to \overline \R$ be $\mu$-integrable functions. Then $\map \min {f, g}$, the pointwise minimum of $f$ and $g$, is also a $\mu$-integrable function. That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise minimum.


Pointwise Maximum of Integrable Functions is Integrable Function

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f, g: X \to \overline \R$ be $\mu$-integrable functions. Then $\map \max {f, g}$, the pointwise maximum of $f$ and $g$, is also a $\mu$-integrable function. That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise maximum.


Integral of Integrable Function is Monotone

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f, g: X \to \overline \R$ be $\mu$-integrable functions. Suppose that $f \le g$ pointwise. Then: :$\displaystyle \int f \rd \mu \le \int g \rd \mu$


Triangle Inequality for Integrals

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R$ be a $\mu$-integrable function. Then: :$\displaystyle \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$


Space of Integrable Functions is Vector Space

Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $\mathcal{L}^1 \left({\mu}\right)$ be the space of real-valued $\mu$-integrable functions. Then $\mathcal{L}^1 \left({\mu}\right)$, endowed with pointwise $\R$-scalar multiplication and pointwise addition, forms a vector space over $\R$.


Measure with Density is Measure

Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $f: X \to \overline{\R}_{\ge 0}$ be a positive $\mu$-measurable function. Then the $f \mu$, the measure with density $f$ with respect to $\mu$ is a measure.


Integrable Function Zero A.E. iff Absolute Value has Zero Integral

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R$ be a $\mu$-integrable function. Then the following are equivalent: :$f = 0$ almost everywhere :$\displaystyle \int \size f \rd \mu = 0$


Integral of Integrable Function over Null Set

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R$ be a $\mu$-integrable function. Let $N$ be a $\mu$-null set. Then: :$\displaystyle \int_N f \rd \mu = 0$ where $\displaystyle \int_N$ signifies an integral over $N$.


Integrable Function is A.E. Real-Valued

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \overline \R$ be a $\mu$-integrable function. Then $\map f x \in \R$ for almost all $x \in X$.


Integrable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$. Let $f, g: X \to \overline \R$ be $\GG$-integrable functions. Suppose that, for all $G \in \GG$: :$\displaystyle \int_G f \rd \mu = \int_G g \rd \mu$ Then $f = g$ $\mu$-almost everywhere.


Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$. Suppose that $\mu \restriction_\GG$, the restriction of $\mu$ to $\GG$, is $\sigma$-finite. Let $f, g: X \to \overline \R$ be $\GG$-measurable functions. Suppose that, for all $G \in \GG$: :$\displaystyle \int_G f \rd \mu = \int_G g \rd \mu$ Then $f = g$ $\mu$-almost everywhere.


Intersection is Largest Subset/Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$. Then for all sets $X$: :$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$ where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.


Countable Set is Null Set under Lebesgue Measure

Let $S \subseteq \R$ be a countable set. Then $\lambda \left({S}\right) = 0$, where $\lambda$ is Lebesgue measure. That is, $S$ is a $\lambda$-null set.


Intersection is Subset/Family of Sets

Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$. Then: :$\displaystyle \forall \beta \in I: \bigcap_{\alpha \mathop \in I} S_\alpha \subseteq S_\beta$ where $\displaystyle \bigcap_{\alpha \mathop \in I} S_\alpha$ is the intersection of $\family {S_\alpha}_{\alpha \mathop \in I}$.


Union is Smallest Superset/Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$. Then for all sets $X$: :$\ds \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$ where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.


Fatou's Lemma for Integrals/Integrable Functions

Let $\sequence {f_n}_{n mathop \in \N} \in \LL^1$, $f_n: X \to \R$ be a sequence of $\mu$-integrable functions. Let $\displaystyle \liminf_{n mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit inferior of the $f_n$. Suppose that there exists an $\mu$-integrable $f: X \to \R$ such that for all $n \in \N$, $f \le f_n$ pointwise. Then: :$\displaystyle \int \liminf_{n \mathop \to \infty} f_n \rd \mu \le \liminf_{n \mathop \to \infty} \int f_n \rd \mu$ where: :the integral sign denotes $\mu$-integration :the {{RHS}} limit inferior is taken in the extended real numbers $\overline \R$.


Fatou's Lemma for Integrals/Positive Measurable Functions

Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions. Let $\displaystyle \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit inferior of the $f_n$. Then: :$\displaystyle \int \liminf_{n \mathop \to \infty} f_n \rd \mu \le \liminf_{n \mathop \to \infty} \int f_n \rd \mu$ where: :the integral sign denotes $\mu$-integration :the {{RHS}} limit inferior is taken in the extended real numbers $\overline \R$.


Reverse Fatou's Lemma/Integrable Functions

Let $\sequence {f_n}_{n \mathop \in \N} \in \LL^1$, $f_n: X \to \R$ be a sequence of $\mu$-integrable functions. Let $\displaystyle \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit superior of the $f_n$. Suppose that there exists an $\mu$-integrable $f: X \to \R$ such that for all $n \in \N$, $f_n \le f$ pointwise. Then: :$\displaystyle \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int \limsup_{n \mathop \to \infty} f_n \rd \mu$ where: :the integral sign denotes $\mu$-integration :the {{RHS}} limit inferior is taken in the extended real numbers $\overline \R$.


Set is Subset of Union/Family of Sets

Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$. Then: :$\displaystyle \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$ where $\displaystyle \bigcup_{\alpha \mathop \in I} S_\alpha$ is the union of $\family {S_\alpha}$.


Reverse Fatou's Lemma/Positive Measurable Functions

Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions. Suppose that there exists a positive measurable function $f: X \to \overline \R$ such that: :$\displaystyle \int f \rd \mu < +\infty$ :$\forall n \in \N: f_n \le f$ where $\le$ signifies a pointwise inequality. Let $\displaystyle \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit superior of the $f_n$. Then: :$\displaystyle \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int \limsup_{n \mathop \to \infty} f_n \rd \mu$ where: :the integral sign denotes $\mu$-integration :the {{LHS}} limit superior is taken in the extended real numbers $\overline \R$.


Monotone Convergence Theorem (Measure Theory)

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\sequence {u_n}_{n \mathop \in \N} \in \map {\LL^1} \mu$, $u_n: X \to \R$ be a increasing sequence of $\mu$-integrable functions. Let $\displaystyle \sup_{n \mathop \in \N} u_n: X \to \overline \R$ be the pointwise supremum of the $u_n$. Then $\displaystyle \sup_{n \mathop \in \N} u_n$ is $\mu$-integrable {{iff}}: :$\displaystyle \sup_{n \mathop \in \N} \int u_n \rd \mu < +\infty$ and, in that case: :$\displaystyle \int \sup_{n \mathop \in \N} u_n \rd \mu = \sup_{n \mathop \in \N} \int u_n \rd \mu$


Intersection is Associative/Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets. Let $\displaystyle I = \bigcap_{\lambda \mathop \in \Lambda} I_\lambda$. Then: :$\displaystyle \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bigcap_{i \mathop \in I_\lambda} S_i}$


Union is Associative/Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets. Let $\displaystyle I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$ denote the union of  $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$. Then: :$\displaystyle \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$


Lebesgue's Dominated Convergence Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\sequence {f_n}_{n \mathop \in \N} \in \map {\LL^1} \mu$, $f_n: X \to \R$ be a sequence of $\mu$-integrable functions. Suppose that for some $\mu$-integrable $g: X \to \R$, it holds that: :$\forall n \in \N: \size {f_n} \le g$ pointwise Suppose that the pointwise limit $f := \displaystyle \lim_{n \mathop \to \infty} f_n$ exists almost everywhere. Then $f$ is $\mu$-integrable, and: :$\displaystyle \lim_{n \mathop \to \infty} \int \size {f_n - f} \rd \mu = 0$ :$\displaystyle \lim_{n \mathop \to \infty} \int f_n \rd \mu = \int \lim_{n \mathop \to \infty} f_n \rd \mu$


Continuity under Integral Sign

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $U$ be a non-empty open set of a metric space. Let $f: U \times X \to \R$ be a mapping satisfying: :$(1): \quad$ For all $\lambda \in U$, the mapping $x \mapsto \map f {\lambda, x}$ is $\mu$-integrable :$(2): \quad$ For $\mu$-almost all $x \in X$, the mapping $\lambda \mapsto \map f {\lambda, x}$ is continuous :$(3): \quad$ There exists a $\mu$-integrable $g: X \to \R$ such that: ::::$\forall \tuple {\lambda, x} \in U \times X: \size {\map f {\lambda, x} } \le \map g x$ Then the mapping $h: U \to \R$ defined by: :$\displaystyle \map h \lambda := \int \map f {\lambda, x} \map {\rd \mu} x$ is continuous.


Differentiability under Integral Sign

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\openint a b$ be a non-empty open interval. Let $f: \openint a b \times X \to \R$ be a mapping satisfying: :$(1): \quad$ For all $t \in \openint a b$, the mapping $x \mapsto f \left({t, x}\right)$ is $\mu$-integrable :$(2): \quad$ For all $x \in X$, the mapping $t \mapsto \map f {t, x}$ is differentiable :$(3): \quad$ There exists a $\mu$-integrable $g: X \to \R$ such that: ::::$\displaystyle \forall \tuple {t, x} \in \openint a b \times X: \size {\frac \partial {\partial t} \map f {t, x} } \le \map g x$ Then the mapping $h: \openint a b \to \R$ defined by: :$\displaystyle \map h t := \int \map f {t, x} \map {\rd \mu} x$ is differentiable, and its derivative is: :$\displaystyle D_t \map h t = \int \frac \partial {\partial t} \map f {t, x} \map {\rd \mu} x$


Riemann-Lebesgue Theorem

Let $f: \closedint a b \to \R$ be a bounded mapping. Let $\mu$ be a one-dimensional Lebesgue measure. {{explain|Link to definition of one-dimensional in this specific context}} Then $f$ is Darboux integrable {{iff}} the set of all discontinuities of $f$ is a $\mu$-null set.


Gamma Function is Continuous on Positive Reals

Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers. Then $\Gamma$ is continuous.


Gamma Function is Smooth on Positive Reals

Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers. Then $\Gamma$ is smooth. {{expand|I'm quite sure it's analytic on all of its domain}}


Log of Gamma Function is Convex on Positive Reals

Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers. Let $\ln$ denote the natural logarithm function. Then the composite mapping $\ln \circ \operatorname \Gamma$ is a convex function.


Pratt's Lemma

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let: :$\sequence {g_n}_{n \mathop \in \N}$ :$\sequence {G_n}_{n \mathop \in \N}$ :$\sequence {f_n}_{n \mathop \in \N}$ be sequences of $\mu$-integrable functions. Let the pointwise limits: :$\displaystyle f := \lim_{n \mathop \to \infty} f_n$ :$\displaystyle g := \lim_{n \mathop \to \infty} g_n$ :$\displaystyle G := \lim_{n \mathop \to \infty} G_n$ exist. Let $g$ and $G$ be $\mu$-integrable. Suppose that, for all $x \in X$ and $n \in \N$: :$\map {g_n} x \le \map {f_n} x \le \map {G_n} x$ Finally, suppose the following hold: :$\displaystyle \lim_{n \mathop \to \infty} \int g_n \rd \mu = \int g \rd \mu$ :$\displaystyle \lim_{n \mathop \to \infty} \int G_n \rd \mu = \int G \rd \mu$ Then: :$\displaystyle \lim_{n \mathop \to \infty} \int f_n \rd \mu = \int f \rd \mu$ and the latter is finite.


Hölder's Inequality

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $p, q \in \R$ such that $\dfrac 1 p + \dfrac 1 q = 1$. Let $f \in \map {\LL^p} \mu, f: X \to \R$, and $g \in \map {\LL^q} \mu, g: X \to \R$, where $\LL$ denotes Lebesgue space. Then their pointwise product $f g$ is $\mu$-integrable, that is: :$f g \in \map {\LL^1} \mu$ and: :$\size {f g}_1 = \displaystyle \int \size {f g} \rd \mu \le \norm f_p \cdot \norm g_q$ where the $\norm {\, \cdot \,}_p$ signify $p$-seminorms.


Hölder's Inequality/Equality

:$\displaystyle \int \size {f g} \rd \mu = \norm f_p \cdot \norm g_q$ holds {{iff}}, for almost all $x \in X$: :$\dfrac {\size {\map f x}^p} {\norm f_p^p} = \dfrac {\size {\map g x}^q} {\norm g_q^q}$


Condition for Membership of Equivalence Class

Let $\RR$ be an equivalence relation on a set $S$. Let $\eqclass x \RR$ denote the $\RR$-equivalence class of $x$. Then: :$\forall y \in S: y \in \eqclass x \RR \iff \tuple {x, y} \in \RR$


Cartesian Product is Set Product/Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets. For all $j \in I$, let $\pr_i: \displaystyle \prod_{j \mathop \in I} \family {S_j} \to S_i$ be the $i$th projection from $\displaystyle \prod_{j \mathop \in I} \family {S_j}$ to $S_i$. Then $\struct {\displaystyle \prod_{j \mathop \in I} \family {S_j}, \family {\pr_i}_{i \mathop \in I} }$ is a set product.


Preimage of Intersection under Relation/General Result

Let $S$ and $T$ be sets. Let $\mathcal R \subseteq S \times T$ be a relation. Let $\mathcal P \left({T}\right)$ be the power set of $T$. Let $\mathbb T \subseteq \mathcal P \left({T}\right)$. Then: :$\displaystyle \mathcal R^{-1} \left[{\bigcap \mathbb T}\right] \subseteq \bigcap_{X \mathop \in \mathbb T} \mathcal R^{-1} \left[{X}\right]$


Image of Union under Relation/Family of Sets

Let $S$ and $T$ be sets. Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$. Let $\RR \subseteq S \times T$ be a relation. Then: :$\displaystyle \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$ where $\displaystyle \bigcup_{i \mathop \in I} S_i$ denotes the union of $\family {S_i}_{i \mathop \in I}$.


Image of Intersection under Relation/Family of Sets

Let $S$ and $T$ be sets. Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$. Let $\RR \subseteq S \times T$ be a relation. Then: :$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $\family {S_i}_{i \mathop \in I}$.


Preimage of Union under Relation/Family of Sets

Let $S$ and $T$ be sets. Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$. Let $\RR \subseteq S \times T$ be a relation. Then: :$\displaystyle \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$ where: :$\displaystyle \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \mathop \in I}$ :$\RR^{-1} \left[{T_i}\right]$ denotes the preimage of $T_i$ under $\RR$.


Image of Intersection under Mapping/Family of Sets

Let $S$ and $T$ be sets. Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$. Let $f: S \to T$ be a mapping. Then: :$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} f \sqbrk {S_i}$ where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $\family {S_i}_{i \mathop \in I}$.


Image of Union under Mapping/Family of Sets

Let $S$ and $T$ be sets. Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$. Let $f: S \to T$ be a mapping. Then: :$\ds f \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$ where $\ds \bigcup_{i \mathop \in I} S_i$ denotes the union of $\family {S_i}_{i \mathop \in I}$.


Minkowski's Inequality/Lebesgue Spaces

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $p \in \R$, $p \ge 1$. Let $f, g: X \to \R$ be $p$-integrable, that is, elements of Lebesgue $p$-space $\map {\mathcal L^p} \mu$. Then their pointwise sum $f + g: X \to \R$ is also $p$-integrable, and: :$\norm {f + g}_p \le \norm f_p + \norm g_p$ where $\norm {\, \cdot \, }_p$ denotes the $p$-seminorm.


Lebesgue Space is Vector Space

Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $\mathcal L^p \left({\mu}\right)$ be Lebesgue $p$-space for $\mu$. Then $\mathcal L^p \left({\mu}\right)$ is a vector subspace of $\mathcal M \left({\Sigma}\right)$, the space of $\Sigma$-measurable functions. In particular, it is a vector space.


Triangle Inequality for Series/Lebesgue Spaces

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $p \in \R$, $p \ge 1$. Let $\sequence {f_n}_{n \mathop \in \N} \in \map {\LL^p} {\mu}$ be a sequence of $p$-integrable functions, that is, a sequence in Lebesgue $p$-space. Suppose that for all $n \in \N$, $f_n \ge 0$ holds pointwise. {{expand|This condition could possibly be weakened/altered, using Lebesgue's Dominated Convergence Theorem in place of Beppo Levi's Theorem}} Then: :$\displaystyle \norm {\sum_{n \mathop = 1}^\infty f_n}_p \le \sum_{n \mathop = 1}^\infty \norm {f_n}_p$ where $\norm {\, \cdot \,}_p$ denotes the $p$-seminorm.


Riesz-Fischer Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $p \in \R$, $p \ge 1$. The Lebesgue $p$-space $\map {\LL^p} \mu$, endowed with the $p$-norm, is a complete metric space.


Pointwise Convergent Bounded Sequence in Lebesgue Space Converges in Norm

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $p \in \R_{\ge 1}$. Let $\sequence {f_n}_{n \mathop \in \N}, f_n: X \to \R$ be a sequence in Lebesgue $p$-space $\LL^p \left({\mu}\right)$. Suppose that the pointwise limit $f := \displaystyle \lim_{n \mathop \to \infty} f_n$ exists $\mu$-almost everywhere. Suppose that for some $g \in \map {\LL^p} \mu$, the pointwise inequality $\size {f_n} \le g$ holds for all $n \in \N$. Then $f \in \map {\LL^p} \mu$, and: :$\displaystyle \lim_{n \mathop \to \infty} \norm {f - f_n}_p = 0$ where $\norm {\, \cdot \,}_p$ denotes the $p$-seminorm.


Riesz's Convergence Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $p \in \R$, $p \ge 1$. Let $\sequence {f_n}_{n \mathop \in \N}, f_n: X \to \R$ be a sequence in Lebesgue $p$-space $\map {\LL^p} \mu$. Suppose that the pointwise limit $f := \displaystyle \lim_{n \mathop \to \infty} f_n$ exists $\mu$-almost everywhere, and that $f \in \map {\LL^p} \mu$. Then the following are equivalent: :$(1): \quad \displaystyle \lim_{n \mathop \to \infty} \norm {f - f_n}_p = 0$ :$(2): \quad \displaystyle \lim_{n \mathop \to \infty} \norm {f_n}_p = \norm f_p$ where $\norm {\, \cdot \,}_p$ denotes the $p$-seminorm.


Preimage of Union under Mapping/Family of Sets

Let $S$ and $T$ be sets. Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$. Let $f: S \to T$ be a relation. Then: :$\displaystyle f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$ where: :$\displaystyle \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \mathop \in I}$ :$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.


Image of Intersection under One-to-Many Relation/Family of Sets

Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Then $\RR$ is a one-to-many relation {{iff}}: :$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$  where $\family {S_i}_{i \mathop \in I}$ is ''any'' family of subsets of $S$.


Preimage of Intersection under Mapping/Family of Sets

Let $S$ and $T$ be sets. Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$. Let $f: S \to T$ be a mapping. Then: :$\displaystyle f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$ where: :$\displaystyle \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{i \mathop \in I}$. :$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.


Let $\mathbb T$ be a set of sets, all of which are subsets of a universe $\mathbb U$. Then: ==== Complement of Intersection ==== {{:De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Intersection}} ==== Complement of Union ==== {{:De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Union}}


Let $S$ be a set. Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$. Then: ==== Complement of Intersection ==== {{:De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Intersection}} ==== Complement of Union ==== {{:De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Union}}


Simple P-Integrable Functions Dense in Lebesgue Space

Let $\left({X, \Sigma, \mu}\right)$ be a measure space, and let $p \in \R$, $p \ge 1$. Let $\mathcal L^p \left({\mu}\right)$ be Lebesgue $p$-space for $\mu$. Let $\mathcal E \left({\Sigma}\right) \cap \mathcal L^p \left({\mu}\right)$ be the space of $\Sigma$-simple, $p$-integrable functions. Then $\mathcal E \left({\Sigma}\right) \cap \mathcal L^p \left({\mu}\right)$ is everywhere dense in $\mathcal L^p \left({\mu}\right)$.


Let $S$ and $T$ be sets. Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$. Then: ==== Difference with Intersection ==== {{:De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection}} ==== Difference with Union ==== {{:De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Union}}


Let $\family {S_i}_{i \mathop \in I}$ be a family of sets, all of which are subsets of a universe $\Bbb U$. Then: ==== Complement of Intersection ==== {{:De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection}} ==== Complement of Union ==== {{:De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Union}}


Jensen's Inequality (Measure Theory)

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $f: X \to \R$ be a $\mu$-integrable function such that $f \ge 0$ pointwise.


Jensen's Inequality (Measure Theory)/Convex Functions

Let $V: \hointr 0 \infty \to \hointr 0 \infty$ be a convex function. Then for all positive measurable functions $g: X \to \R$, $g \in \map {\mathcal M^+} \Sigma$: :$\map V {\dfrac {\int g \cdot f \rd \mu} {\int f \rd \mu} } \le \dfrac {\int \paren {V \circ g} \cdot f \rd \mu} {\int f \rd \mu}$ where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.


Jensen's Inequality (Measure Theory)/Concave Functions

Let $\Lambda: \hointr 0 \infty \to \hointr 0 \infty$ be a concave function. Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$: :$\dfrac {\int \paren {\Lambda \circ g} \cdot f \rd \mu} {\int f \rd \mu} \le \map \Lambda {\dfrac {\int g \cdot f \rd \mu} {\int f \rd \mu} }$ where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.


Power Set is Lattice

Let $S$ be a set. Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the relational structure defined on $\mathcal P \left({S}\right)$ by the subset relation $\subseteq$. Then $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is a lattice.


Divisor Relation induces Lattice

Let $\struct {\Z_{> 0}, \divides}$ be the ordered set comprising: :The set of positive integers $\Z_{> 0}$ :The divisor relation $\divides$ defined as: ::$a \divides b := \exists k \in \Z_{> 0}: b = ka$ Then $\struct {\Z_{> 0}, \divides}$ is a lattice.


Generator for Product Sigma-Algebra

Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces. Let $\GG_1$ and $\GG_2$ be generators for $\Sigma_1$ and $\Sigma_2$, respectively. Then $\GG_1 \times \GG_2$ is a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$.


Uniqueness of Product Measures

Let $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$ be measure spaces. Let $\GG_1$ and $\GG_2$ be generators for $\Sigma_1$ and $\Sigma_2$, respectively. Suppose that $\GG_1$ and $\GG_2$ are closed under intersection. Suppose further that there are exhausting sequences $\sequence {G_{1, n} }_{n \mathop \in \N}$ and $\sequence {G_{2, n} }_{n \mathop \in \N}$ in $\GG_1$ and $\GG_2$, respectively, such that: :$\forall n \in \N: \map \mu {G_{1, n} } < \infty$ :$\forall n \in \N: \map \nu {G_{2, n} } < \infty$ Then there is at most one measure $\rho$ on the product space $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2}$ such that: :$\forall G_1 \in \GG_1, G_2 \in \GG_2: \map \rho {G_1 \times G_2} = \map \mu {G_1} \, \map \nu {G_2}$ That is, there can be at most one product measure on $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2}$.


Existence of Product Measures

Let $\left({X, \Sigma_1, \mu}\right)$ and $\left({Y, \Sigma_2, \nu}\right)$ be $\sigma$-finite measure spaces. Then there exists a $\sigma$-finite measure $\rho$ on the product space $\left({X \times Y, \Sigma_1 \otimes \Sigma_2}\right)$ such that: :$\forall E_1 \in \Sigma_1, E_2 \in \Sigma_2: \rho \left({E_1 \times E_2}\right) = \mu \left({E_1}\right) \nu \left({E_2}\right)$ This $\rho$ is called product measure and is also denoted $\mu \times \nu$.


Tonelli's Theorem

Let $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$ be $\sigma$-finite measure spaces. Let $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$. Let $f: X \times Y \to \overline \R_{\ge 0}$ be a positive $\Sigma_1 \otimes \Sigma_2$-measurable function. Then: :$\displaystyle \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\d \mu} x \map {\d \nu} y = \int_X \int_Y \map f {x, y} \map {\d \nu} y \map {\d \mu} x$


Condition for Power Set to be Totally Ordered

Let $\mathcal P \left({S}\right)$ be the power set of a set $S$. Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the set $\mathcal P \left({S}\right)$ ordered by $\subseteq$. Then $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is totally ordered iff $S$ is either the empty set or a singleton.


Fubini's Theorem

Let $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$ be $\sigma$-finite measure spaces. Let $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$. Let $f: X \times Y \to \R$ be a $\Sigma_1 \otimes \Sigma_2$-measurable function. Suppose that: :$\displaystyle \int_{X \times Y} \size f \map {\rd} {\mu \times \nu} < \infty$ Then $f$ is $\mu \times \nu$-integrable, and: :$\displaystyle \int_{X \times Y} f \map {\rd} {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\rd \mu} x \map {\rd \nu} y = \int_X \int_Y \map f {x, y} \map {\rd \nu} y \map {\rd \mu} x$ {{expand|Implement a version of this for real functions}}


Product Sigma-Algebra Generated by Projections

Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces. Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X \times Y$. Let $\pr_1: X \times Y \to X$ and $\pr_2: X \times Y \to Y$ be the first and second projections, respectively. Then: :$\Sigma_1 \otimes \Sigma_2 = \map \sigma {\pr_1, \pr_2}$ where $\sigma$ denotes generated $\sigma$-algebra.


Measurable Mappings to Product Measurable Space

Let $\left({X, \Sigma}\right)$, $\left({X_1, \Sigma_1}\right)$ and $\left({X_2, \Sigma_2}\right)$ be measurable spaces. Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X_1 \times X_2$. Let $\operatorname{pr}_1: X_1 \times X_2 \to X_1$ and $\operatorname{pr}_2: X_1 \times X_2 \to X_2$ be the first and second projections, respectively. A mapping $f: X \to X_1 \times X_2$ is $\Sigma \, / \, \Sigma_1 \otimes \Sigma_2$-measurable {{iff}}: :$\operatorname{pr}_i \circ f: X \to X_i$ is $\Sigma \, / \, \Sigma_i$-measurable, for $i = 1, 2$


Measurable Mappings from Product Measurable Space

Let $\left({X, \Sigma}\right)$, $\left({X_1, \Sigma_1}\right)$ and $\left({X_2, \Sigma_2}\right)$ be measurable spaces. Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X_1 \times X_2$. Let $f: X_1 \times X_2 \to X$ be a $\Sigma_1 \otimes \Sigma_2 \, / \, \Sigma$-measurable mapping. Then: :$\forall x_1 \in X_1: f \left({x_1, \cdot}\right): X_2 \to X$ is $\Sigma_2 \, / \, \Sigma$-measurable :$\forall x_2 \in X_2: f \left({\cdot, x_2}\right): X_1 \to X$ is $\Sigma_1 \, / \, \Sigma$-measurable


Integral of Survival Function

Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space. Let $f: X \to \R_{\ge 0}$ be a positive $\Sigma$-measurable function. Let $F \left({f}\right): \R \to \R$ be the survival function of $f$. Then: :$\displaystyle \int f \, \mathrm d \mu = \int_{\left({0 \,.\,.\, \infty}\right)} F \left({f}\right) \, \mathrm d \lambda$ where $\lambda$ is Lebesgue measure.


Cardinal Zero is Less than Cardinal One

The zero cardinal $0$ is less than one: :$0 < 1$


Idempotent Non-Trivial Quasigroup is Not a Loop

Let $\left({S, \circ}\right)$ be an idempotent quasigroup whose underlying set $S$ comprises more than one element. Then $\left({S, \circ}\right)$ is not an algebra loop, that is, it has no identity element.


B-Algebra is Right Cancellable

Let $\struct {X, \circ}$ be a $B$-algebra. Then $\circ$ is right-cancellable for $X$. That is: :$\forall x, y, z \in X: x \circ z = y \circ z \implies x = y$


Right Regular Representation of 0 is Bijection in B-Algebra

Let $\left({X, \circ}\right)$ be a $B$-algebra. Then the right regular representation of $\left({X, \circ}\right)$ with respect to $0$ is a bijection.


Product of Cardinals is Commutative

Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: : $\mathbf a \mathbf b = \mathbf b \mathbf a$ where $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.


B-Algebra Identity: xy=x(0(0y))

Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x,y \in X: x \circ y = x \circ\left({0 \circ\left({ 0 \circ y}\right)}\right)$


B-Algebra Identity: x (y z) = (x (0 z)) y

Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x,y,z \in X: x \circ \left({y \circ z}\right) = \left({x \circ \left({0 \circ z}\right)}\right) \circ y$


B-Algebra Identity: xy = 0 iff x = y

Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x,y \in X: x \circ y = 0 \iff x = y$


First Power of Element in B-Algebra

Let $\left({X, \circ}\right)$ be a $B$-algebra. Then: :$\forall x \in X: x^1 = x$ where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.


B-Algebra Power Law

Let $\left({X, \circ}\right)$ be a B-algebra. Let $n, m \in \N$ such that $n \ge m$. Then: :$\forall x \in X: x^n \circ x^m = x^{n-m}$ where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.


B-Algebra is Left Cancellable

Let $\struct {X, \circ}$ be a $B$-algebra. Then $\circ$ is a left cancellable operation.


B-Algebra is Quasigroup

Let $\left({X, \circ}\right)$ be a $B$-algebra. Then $\left({X, \circ}\right)$ is a quasigroup.


Quasigroup is not necessarily B-Algebra

Let $\struct {S, \circ}$ be a quasigroup. Then $\struct {S, \circ}$ is not necessarily a $B$-algebra.


Group is Quasigroup

Let $\left({G, \circ}\right)$ be a group. Then $\left({G, \circ}\right)$ is a quasigroup.


B-Algebra Power Law with Zero

:$\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n-m}$


Product of Powers in B-Algebra

Let $\left({X, \circ}\right)$ be a $B$-algebra. Let $x \in X$ and $m, n \in \N$. Then: :$x^m \circ x^n = \begin{cases} x^{m-n} & : m \ge n \\ 0 \circ x^{n-m} & : n > m \end{cases}$


Superset of Co-Countable Set

Every superset of a co-countable set is co-countable.


Group Induces B-Algebra

Let $\left({G, \circ}\right)$ be a group whose identity element is $e$. Let $*$ be the binary operation on $G$ defined as: :$\forall a, b \in G: a * b = a \circ b^{-1}$ where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$. Then the algebraic structure $\left({G, *}\right)$ is a $B$-algebra.


Identity of Cardinal Product is One

Let $\mathbf a$ be a cardinal. Then: : $\mathbf 1 \mathbf a = \mathbf a$ where $\mathbf 1 \mathbf a$ denotes the product of the (cardinal) one and $\mathbf a$. That is, $\mathbf 1$ is the identity element of the product operation on cardinals.


Zero of Cardinal Product is Zero

Let $\mathbf a$ be a cardinal. Then: : $\mathbf 0 \mathbf a = \mathbf 0$ where $\mathbf 0 \mathbf a$ denotes the product of the (cardinal) zero and $\mathbf a$. That is, $\mathbf 0$ is the zero element of the product operation on cardinals.


Sum of Cardinals is Commutative

Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: :$\mathbf a + \mathbf b = \mathbf b + \mathbf a$ where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.


B-Algebra is Commutative iff x(xy)=y

Let $\left({X, \circ}\right)$ be a $B$-algebra. Then $\left({X, \circ}\right)$ is commutative {{iff}}: :$\forall x, y \in X: x \circ \left({x \circ y}\right) = y$


Identity of Cardinal Sum is Zero

Let $\mathbf a$ be a cardinal. Then: : $\mathbf a + \mathbf 0 = \mathbf a$ where $\mathbf a + \mathbf 0$ denotes the sum of the zero cardinal and $\mathbf a$. That is, $\mathbf 0$ is the identity element of the sum operation on cardinals.


Commutative B-Algebra Implies (zy)(zx)=xy

Let $\struct {B, \circ}$ be a commutative $B$-algebra.  Then: :$\forall x, y, z \in X: \paren {z \circ y} \circ \paren {z \circ x} = x \circ y$


Cardinal Product Distributes over Cardinal Sum

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals. Then: : $\mathbf a \left({\mathbf b + \mathbf c}\right) = \mathbf a \mathbf b + \mathbf a \mathbf c$ where: : $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$. : $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.


Ordering of Cardinals Compatible with Cardinal Product

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals. Then: : $\mathbf a \le \mathbf b \implies \mathbf a \mathbf c \le \mathbf b \mathbf c$ where $\mathbf a \mathbf c$ denotes the product of $\mathbf a$ and $\mathbf c$.


Ordering of Cardinals Compatible with Cardinal Sum

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals. Then: :$\mathbf a \le \mathbf b \implies \mathbf a + \mathbf c \le \mathbf b + \mathbf c$ where $\mathbf a \mathbf c$ denotes the sum of $\mathbf a$ and $\mathbf c$.


Group is B-Algebra Iff All Elements Self-Inverse

Let $\left({G, \circ}\right)$ be a group whose identity element is $e$. Then $\left({G, \circ}\right)$ is also a $B$-algebra {{iff}}: :$\forall g \in G: g = g^{-1}$ That is, {{iff}} all elements of $G$ are self-inverse.


Condition for Existence of Cardinal Sum

Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: : $\mathbf a \le \mathbf b \iff \exists \mathbf c: \mathbf a + \mathbf c = \mathbf b$ where $\mathbf c$ is also a cardinal.


Cardinal One is Cancellable for Cardinal Sum

Let $\mathbf a$ and $\mathbf b$ be cardinals. Then: :$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 \implies \mathbf a = \mathbf b$ where $\mathbf 1$ is (cardinal) one.


Mapping from Finite Set to Itself is Injection iff Surjection

Let $S$ be a finite set. Let $f: S \to S$ be a mapping. Then $f$ is injective {{iff}} $f$ is surjective.


Finite Cardinals form Infinite Set

The finite cardinals form a set which is infinite.


Increasing Function has Countable Discontinuities

Let $f: \R \to \R$ be an increasing real function. Then $f$ has at most countably many discontinuities.


Integral of Increasing Function Composed with Measurable Function

Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space. Let $f: X \to \R_{\ge 0}$ be a positive measurable function. Let $\phi: \R_{\ge 0} \to \R_{\ge 0}$ be a continuously differentiable, increasing function such that $\phi \left({0}\right) = 0$. Then: :$\displaystyle \int \phi \circ f \rd \mu = \int_0^\infty \phi' \left({t}\right) F \left({t}\right) \rd t$ where: : $F$ is the survival function of $f$ : $\displaystyle \int_0^\infty$ denotes an improper integral.


Minkowski's Inequality for Double Integrals

Let $\struct {X, \Sigma, \mu}$ and $\struct {Y, \Sigma', \nu}$ be $\sigma$-finite measure spaces. Let $\struct {X \times Y, \Sigma \otimes \Sigma', \mu \times \nu}$ be their product measure space. Let $f: X \times Y \to \overline \R$ be a $\Sigma \otimes \Sigma'$-measurable function. Then, for all $p \in \R$ with $p \ge 1$: :$\displaystyle \paren {\int_X \paren {\int_Y \size {\map f {x, y} } \map {\rd \nu} y}^p \map {\rd \mu} x}^{1/p} \le \int_Y \paren {\int_X \size {\map f {x, y} }^p \map {\rd \mu} x}^{1/p} \map {\rd \nu} y$


Integral with respect to Pushforward Measure

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\struct {X', \Sigma'}$ be a measurable space. Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping. Let $f: X' \to \overline \R$ be a $\map T \mu$-integrable function, where $\map T \mu$ denotes the pushforward measure of $\mu$ under $T$. Then $f \circ T: X \to \overline \R$ is $\mu$-integrable, and: :$\displaystyle \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$


Integrable Function under Pushforward Measure

Let $\struct {X, \Sigma, \mu}$ be a measure space. Let $\struct {X', \Sigma'}$ be a measurable space. Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping. Let $f: X' \to \overline \R$ be a mapping. Then the following are equivalent: :$(1): \quad f$ is $\map T \mu$-integrable :$(2): \quad f \circ T$ is $\mu$-integrable where $\map T \mu$ is the pushforward measure of $\mu$ under $T$.


0 in B-Algebra is Left Cancellable Element

Let $\left({X, \circ}\right)$ be a $B$-Algebra. Then: :$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$


B-Algebra Identity: 0(0x)=x

Let $\struct {X, \circ}$ be a $B$-algebra.  Then: :$\forall x \in X: 0 \circ \paren {0 \circ x} = x$


B-Algebra Induces Group

Let $\struct {X, \circ}$ be a $B$-algebra. Let $*$ be the binary operation on $X$ defined as: :$\forall a, b \in X: a * b := a \circ \paren {0 \circ b}$ Then the algebraic structure $\struct {X, *}$ is a group such that: :$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$. That is: :$\forall a, b \in X: a * b^{-1} := a \circ b$


Convolution of Measurable Functions is Bilinear

Let $\BB^n$ be the Borel $\sigma$-algebra on $\R^n$. Let $f, f', g, g': \R^n \to \R$ be $\BB^n$-measurable functions. Then for all $\lambda \in \R$: :$\paren {\lambda f + f'} * g = \lambda \paren {f * g} + f' * g$ :$f * \paren {\lambda g + g'} = \lambda \paren {f * g} + f * g'$ provided the convolutions in these expressions exist. That is, convolution $*$ is a bilinear operation.


Convolution of Measurable Function and Measure is Bilinear

Let $\mu$ and $\nu$ be measures on the Borel $\sigma$-algebra $\mathcal B^n$ on $\R^n$. Let $f, f': \R^n \to \R$ be $\mathcal B^n$-measurable functions. Then for all $\lambda \in \R$: :$\left({\lambda f + f'}\right) * \mu = \lambda \left({f * \mu}\right) + f' * \mu$ :$f * \left({\lambda \mu + \nu}\right) = \lambda \left({f * \mu}\right) + f * \nu$ provided the convolutions in these expressions exist. That is, convolution $*$ is a bilinear operation.


Convolution of Measures is Bilinear

Let $\mu, \mu', \nu$ and $\nu'$ be measures on the Borel $\sigma$-algebra $\mathcal B^n$ on $\R^n$. Then for all $\lambda \in \R$: :$\left({\lambda \mu + \mu'}\right) * \nu = \lambda \left({\mu * \nu}\right) + \mu' * \nu$ :$\mu * \left({\lambda \nu + \nu'}\right) = \lambda \left({\mu * \nu}\right) + \mu * \nu'$ where $*$ denotes convolution. That is, convolution is a bilinear operation.


Convolution of Measures as Pushforward Measure

Let $\mu$ and $\nu$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$. Let $\alpha: \R^n \times \R^n \to \R^n, \ \map \alpha {\mathbf x, \mathbf y} = \mathbf x + \mathbf y$ be vector addition on $\R^n$. Then we have the following equality of measures on $\BB^n$: :$\mu * \nu = \map {\alpha_*} {\mu \times \nu}$ where $\mu * \nu$ is the convolution of $\mu$ and $\nu$, and $\map {\alpha_*} {\mu \times \nu}$ is the pushforward of the product measure $\mu \times \nu$ under $\alpha$.


Continuous Functions with Compact Support Dense in Lebesgue P-Space

Let $C_c \left({\R^n}\right)$ be the space of continuous functions with compact support on $\R^n$. Let $p \in \R$, $p \ge 1$, and let $\mathcal L^p \left({\lambda^n}\right)$ be Lebesgue $p$-space for Lebesgue measure $\lambda^n$. Then $C_c \left({\R^n}\right)$ is everywhere dense in $\mathcal L^p \left({\lambda^n}\right)$ with respect to the $p$-seminorm $\left\Vert{\, \cdot \,}\right\Vert_p$.


Convolution of Integrable Function with Bounded Function

Let $f: \R^n \to \R$ be a Lebesgue integrable function. Let $g: \R^n \to \R$ be an essentially bounded function under Lebesgue measure $\lambda^n$. Then the convolution $f * g$ of $f$ and $g$ is bounded and continuous. In particular, $f * g$ is again essentially bounded.


F-Sigma Sets Closed under Union

Let $T = \struct {S, \tau}$ be a topological space. Let $F, F'$ be $F_\sigma$ sets of $T$. Then their union $F \cup F'$ is also an $F_\sigma$ set of $T$.


F-Sigma Sets Closed under Intersection

Let $T = \left({S, \tau}\right)$ be a topological space. Let $F, F'$ be $F_\sigma$ sets of $T$. Then their intersection $F \cap F'$ is also a $F_\sigma$ set of $T$.


G-Delta Sets Closed under Union

Let $T = \left({S, \tau}\right)$ be a topological space. Let $G, G'$ be $G_\delta$ sets of $T$. Then their union $G \cup G'$ is also a $G_\delta$ set of $T$.


G-Delta Sets Closed under Intersection

Let $T = \struct {S, \tau}$ be a topological space. Let $G, G'$ be $G_\delta$ sets of $T$. Then their intersection $G \cap G'$ is also a $G_\delta$ set of $T$.


Limit Point is Limit of Convergent Sequence

Let $M = \left({X, d}\right)$ be a metric space. Let $E \subseteq X$ be a subset of $X$. Let $p$ be a limit point of $E$. Then there exists a sequence $\left\langle{x_n}\right\rangle \subseteq E$ which converges to a limit: :$\displaystyle \lim_{n \mathop \to \infty} x_n = p$ where $\displaystyle \lim_{n \mathop \to \infty} x_n$ is the limit of the sequence $\left\langle{x_n}\right\rangle$.


F-Sigma Sets form Lattice

Let $T = \left({S, \tau}\right)$ be a topological space. Let $\mathcal F$ be the collection of all $F_\sigma$ sets of $T$. Then $\left({\mathcal F, \subseteq}\right)$ is a lattice, where $\subseteq$ denotes the subset relation.


G-Delta Sets form Lattice

Let $T = \left({S, \tau}\right)$ be a topological space. Let $\mathcal G$ be the collection of all $G_\delta$ sets of $T$. Then $\left({\mathcal G, \subseteq}\right)$ is a lattice, where $\subseteq$ denotes the subset relation.


Bounded Sequence in Euclidean Space has Convergent Subsequence

Let $\left\langle{x_i}\right\rangle_{i \in \N}$ be a bounded sequence in the Euclidean space $\R^n$. Then some subsequence of $\left\langle{x_i}\right\rangle_{i \in \N}$ converges to a limit.


Exponential Growth Equation/Special Case

All solutions of the differential equation $y' = y$ take the form $y = C e^x$.


Existence-Uniqueness Theorem for First-Order Differential Equation

Let $P$ and $Q$ be continuous functions on some open interval $I \subseteq \R$. Let $a \in I$. Let $b \in \R$. There is a unique function $f(x)=y$ on $I$ that satisfies the differential equation  :$y' + P(x)y = Q(x)$ along with the initial condition  :$f(a)=b$ This function is :$\displaystyle f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$ where $\displaystyle A(x) = \int_a^x P(t)dt$


Proper Well-Ordering determines Smallest Elements

Let $S$ be a class. Let $\preceq$ be a proper well-ordering on $S$. Let $B$ be a nonempty subclass of $S$. Then $B$ has a $\preceq$-smallest element.


Subset of Ordinals has Minimal Element

Let $A$ be an ordinal (we shall allow $A$ to be a proper class). Let $B$ be a nonempty subset of $A$. Then $B$ has an $\Epsilon$-minimal element. {{explain|$\Epsilon$}} That is: : $\exists x \in B: B \cap x = \varnothing$


Union of Subset of Ordinals is Ordinal

{{explain|If $A$ is a proper class, then its union may be the class of all ordinals, which we do not consider an ordinal.}} Let $A$ be a class of ordinals. That is, $A \subseteq \On$, where $\On$ denotes the ordinal class. Then $\bigcup A$ is an ordinal.


Union of Ordinals is Least Upper Bound

Let $A \subset \operatorname{On}$. That is, let $A$ be a class of ordinals (every member of $A$ is an ordinal). Then $\bigcup A$, the union of $A$, is the least upper bound of $A$: :$\displaystyle \forall x \in A: x \le A$ :$\displaystyle \forall y \in A: y \le x \implies \bigcup A \le x$


Ordinal is Less than Successor

Let $x$ be an ordinal. Let $x^+$ denote the successor of $x$. Then: :$x \in x^+$ :$x \subset x^+$


No Natural Number between Number and Successor

Let $\N$ be the natural numbers. Let $n \in \N$. Then no natural number $m$ exists strictly between $n$ and its successor: :$\neg \exists m \in \N: \paren {n < m < n^+}$ That is: :If $m \le n \le m^+$, then $m = n$ or $m = n^+$.


No Largest Ordinal

Let $a$ be a set of ordinals. Then: :$\forall x \in a: x \prec \left({\bigcup a}\right)^+$


Minimal Infinite Successor Set is Ordinal

Let $\omega$ denote the minimal infinite successor set. Then $\omega$ is an ordinal.


No Infinitely Descending Membership Chains

Let $\omega$ denote the minimal infinite successor set. Let $F$ be a function whose domain is $\omega$. Then: :$\exists n \in \omega: \map F {n^+} \notin \map F n$


Transfinite Recursion/Uniqueness of Transfinite Recursion

Let $f$ be a mapping with a domain $y$ where $y$ is an ordinal. Let $f$ satisfy the condition that: : $\forall x \in y: f \left({x}\right) = G \left({f \restriction x}\right)$ where $f \restriction x$ denotes the restriction of $f$ to $x$. {{explain|What is $G$?}} Let $g$ be a mapping with a domain $z$ where $z$ is an ordinal. Let $g$ satisfy the condition that: : $\forall x \in z: g \left({x}\right) = G \left({g \restriction x}\right)$ Let $y \subseteq z$. Then: : $\forall x \in y: f \left({x}\right) = g \left({x}\right)$


Transfinite Induction/Schema 1

Let $P \left({x}\right)$ be a property  Suppose that: : If $P \left({x}\right)$ holds for all ordinals $x$ less than $y$, then $P \left({y}\right)$ also holds. Then $P \left({x}\right)$ holds for all ordinals $x$.