hcju/proofwikips huggingface dataset:
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 : :$\mathcal R$ 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
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 $(4)$.
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  $f$ is both a bijection and a symmetric relation.
De Morgan's Laws (Set Theory)/Relative Complement/General Case
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: :$\displaystyle \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$   $\RR$ 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  limit inferior is taken in the extended real numbers $\overline \R$.