avewright/proofwiki-math huggingface dataset:
Theorem: Fixed Point of g-Tower is Greatest Element

Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x = \map g x$.
Then $x$ is the greatest element (under the subset relation) of $M$.

Proof:
Let $x$ be an element of $M$ such that $\map g x = x$.
We will demonstrate by the Principle of Superinduction on $y$ that:
:$\forall y \in M: y \subseteq x$
The proof proceeds by superinduction.
For all $y \in M$, let $\map P y$ be the proposition:
:$y \subseteq x$
Theorem: Restriction of Union of Mappings which Agree Equals Mapping

Let $Y$ be a set.
Let $\family {A_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\family {f_i: A_i \to Y}$ be a family of mappings indexed by $I$.
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : X \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.
Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $A_i \cap A_j$.
Let $f \restriction_{A_i} : A_i \to Y$ denote the restriction of $f$ to $A_i$.
Then:
:$\forall i \in I: f \restriction_{A_i} = f_i: A_i \to Y$.

Proof:
From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $X$ to $Y$.
Let $i \in I$ be an arbitrary index of $I$.
Consider the restriction of $f$ to $A_i$:
{{begin-eqn}}
{{eqn | l = f \restriction _ { A_i }
      | r = f \cap \paren {A_i \times Y}
      | c = {{Defof|Restriction of Mapping}}
}}
{{eqn | r = \paren{\bigcup_{j \in I} f_j} \cap \paren{A_i \times Y}
      | c = Definition of $f$
}}
{{eqn | r = \bigcup_{j \in I} \paren{ f_j \cap \paren{A_i \times Y} }
      | c = Intersection Distributes over Union
}}
{{end-eqn}}
Let $j$ be an arbitrary index of $I$.
Theorem: Metric on Shift of Finite Type is Non-Archimedean

Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $\theta \in \openint 0 1$.
Then the metric $d_\theta$ on $X_\mathbf A$ is non-archimedean metric.
That is, $\struct {X_\mathbf A, d _\theta} $ is an ultrametric space.

Proof:
Let $x, y, z \in X_\mathbf A$.
Let:
:$N_1 := \sup \set {n \in \N \cup \set \infty : x_i = y_i \text { for all } i \in \openint {-n} n}$
and:
:$N_2 := \sup \set {n \in \N \cup \set \infty : y_i = z_i \text { for all } i \in \openint {-n} n}$
so that:
:$\map {d_\theta} {x, y} = \theta^{N_1}$
and:
:$\map {d_\theta} {y, z} = \theta^{N_2}$
Let $M := \min \set {N_1, N_2}$.
Then for all $i \in \openint {-M} M$:
:$x_i = y_i = z_i$
Hence:
:$\sup \set {n \in \N \cup \set \infty : x_i = z_i \text { for all } i \in \openint {-n} n} \ge M$
so that:
:$\map {d_\theta} {x, z} \le \theta ^M$
On the other hand:
:$\theta^M = \theta^{\min \set {N_1, N_2} } = \max \set {\theta^{N_1}, \theta^{N_2} }$
Therefore:
:$\map {d_\theta} {x, z} \le \max \set {\theta^{N_1}, \theta^{N_2} } = \max \set {\map {d_\theta} {x, y}, \map {d_\theta} {y, z} }$
{{qed}}
Category:Metric Spaces
oe4xiq5w7uauk8do7cd91f6wh3ue690
Theorem: Completely Metrizable Space is Metrizable Space

Let $\struct {S, \tau}$ be a completely metrizable topological space.
Then, $\struct {S, \tau}$ is a metrizable space.

Proof:
Since $\struct {S, \tau}$ is completely metrizable, $\tau$ is induced by a (complete) metric $d$.
In particular, $\struct {S, d}$ is a metric space.
As $\struct {S, \tau}$ is homeomorphic to itself by the identity mapping, it is metrizable.
{{qed}}
Category:Completely Metrizable Spaces
Category:Metrizable Spaces
pcwkc8mbzqaqdwj9h8oknqmlbvbfoah
Theorem: Pasting Lemma/Continuous Mappings on Closed Sets

Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be a finite indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a finite family of closed sets of $T$.
Let $\family {f_i : C_i \to Y}_{i \mathop \in I}$ be a finite family of continuous mappings.
Let $C = \ds \bigcup_{i \mathop \in I} C_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : C \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.
Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $C_i \cap C_j$.
Then $f$ is a continuous mapping on $C = \ds \bigcup_{i \mathop \in I} C_i$.

Proof:
From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $C$ to $Y$.
From Restriction of Union of Mappings which Agree Equals Mapping:
:$\forall i \in I : f \restriction_{C_i} = f_i$
Hence:
:$\forall i \in I : f \restriction_{C_i}$ is continuous
From Pasting Lemma for Finite Union of Closed Sets
:$f$ is continuous
{{qed}}
Theorem: Pasting Lemma/Pair of Continuous Mappings on Open Sets

Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be open in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
Then the mapping $f \cup g : A \cup B \to Y$ is continuous.

Proof:
Follows directly from Pasting Lemma for Continuous Mappings on Open Sets.
{{qed}}
Theorem: Pasting Lemma/Pair of Continuous Mappings on Open Sets

Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be open in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
Then the mapping $f \cup g : A \cup B \to Y$ is continuous.

Proof:
From Union of Mappings which Agree is Mapping
:$f \cup g$ is well-defined.
By {{Defof|Continuous Mapping}}:
:$f \cup g$ is continuous {{iff}} $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$ for every open $U$ in $Y$.
Let $U$ be an arbitrary open subset in $Y$.
From Preimage of Union Mapping is Union of Preimages:
:$\paren {f \cup g}^{-1} \sqbrk U = f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$
By {{Defof|Continuous Mapping}}:
:$f^{-1} \sqbrk U$ is open in $A$
:$g^{-1} \sqbrk U$ is open in $B$.
From Open Set in Open Subspace:
:$f^{-1} \sqbrk U$ and $g^{-1} \sqbrk U$ are open in $X$.
By {{Open-set-axiom|1}}:
:$A \cup B$ is open in $X$
:$f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$ is open in $X$
From Open Set in Open Subspace:
:$f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$ is open in $A \cup B$
Hence:
:$\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$
Since $U$ was an arbitrary open subset of $Y$:
:for all open subsets $U$ of $Y$, $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$
By {{Defof|Continuous Mapping}}:
:$f \cup g$ is continuous.
{{qed}}
Theorem: Pasting Lemma/Pair of Continuous Mappings on Closed Sets

Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be closed in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
Then the mapping $f \cup g : A \cup B \to Y$ is continuous.

Proof:
Follows directly from Pasting Lemma for Continuous Mappings on Closed Sets
{{qed}}
Theorem: Shift of Finite Type is Compact

Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Then $X_\mathbf A $ is compact.

Proof:
{{ProofWanted}}
Category:Compact Topological Spaces
Category:Topology
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Theorem: Union of g-Tower is Greatest Element and Unique Fixed Point

Let $M$ be a set.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then:
:$\ds \bigcup M \in M$
:$\ds \bigcup M$ is the greatest element (under the subset relation) of $M$
:$\ds \bigcup M$ is the unique fixed point of $M$.

Proof:
Let the hypothesis be assumed.
By G-Tower is Nest, $M$ is a nest. By assumption, $M$ is a set.
Hence by definition $M$ is a chain.
By $g$-Tower is Closed under Chain Unions:
:$\ds \bigcup M \in M$
It follows by Draft:Union of Set of Sets is Greatest Element under Subset Relation that $\ds \bigcup M$ is the greatest element (under the subset relation) of $M$.
Hence $\ds \bigcup M$ cannot be a proper subset of $\ds \map g {\bigcup M}$.
Thus:
:$\ds \bigcup M = \map g {\bigcup M}$
and the result follows.
{{qed}}
Theorem: G-Tower is Closed under Chain Unions

Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is closed under chain unions.

Proof:
By definition:
:a $g$-tower is a class which is minimally superinductive under $g$
:a class which is minimally superinductive under $g$ is superinductive under $g$
:a superinductive class is closed under chain unions.
Hence the result.
{{qed}}
Category:G-Towers
Category:Closure under Chain Unions
reymphwfm5emep94xq63afo0dcz37rw
Theorem: Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping

Let $S$ be a set of sets.
Let $g$ be a mapping such that:
:for every $x \in S$, $x$ is closed under $g$.
Then the intersection $\bigcap S$ of $S$ is also closed under $g$.

Proof:
The domain of $g$ is not made clear, but the assumption is that:
:$\forall x \in S: \forall y \in x: y \in \Dom g$
First we note that by definition of intersection of $S$:
:$\bigcap S := \set {y: \forall x \in S: y \in x}$
Then:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = \bigcap S
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall x \in S
      | l = y
      | o = \in
      | r = x
      | c = {{Defof|Intersection of Set of Sets}}
}}
{{eqn | ll= \leadsto
      | q = \forall x \in S
      | l = \map g y
      | o = \in
      | r = x
      | c = as $x$ is closed under $g$
}}
{{eqn | ll= \leadsto
      | l = \map g y
      | o = \in
      | r = \bigcap S
      | c = {{Defof|Intersection of Set of Sets}}
}}
{{end-eqn}}
Hence the result by definition of closed under $g$.
{{qed}}
Theorem: Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions

Let $S$ be a set of sets.
Let $S$ be such that:
:$\forall x \in S: x$ is closed under chain unions.
Then the intersection $\ds \bigcap S$ of $S$ is also closed under chain unions.

Proof:
First we note that by definition of intersection of $S$:
:$\ds \bigcap S := \set {y: \forall x \in S: y \in x}$
Recall the definition of closed under chain unions:
$S$ is '''closed under chain unions''' {{iff}}:
:for every chain $C$ of elements of $S$, $\ds \bigcup C$ is also in $S$.
Let $C_\cap$ be a chain in $\ds \bigcap S$.
Then by {{Defof|Intersection of Set of Sets}}:
:$\forall x \in S: C_\cap$ is a chain in $x$.
Hence as $x$ is closed under chain unions for all $x \in S$:
:$\forall x \in S: \ds \bigcup C_\cap$ is in $x$
Hence by {{Defof|Intersection of Set of Sets}}:
:$\ds \bigcup C_\cap$ is in $S$
Hence the result by definition of closed under chain unions.
{{qed}}
Theorem: Inverse of Composition of Subsets of Topological Group

Let $\struct {G, \odot, \tau}$ be a topological group.
Let $A, B \subseteq G$.
Then: 
:$\paren {A \odot B}^{-1} = B^{-1} \odot A^{-1}$

Proof:
The proof is identical to Inverse of Product of Subsets of Group.
{{qed}}
Theorem: Set which is Superinductive under Progressing Mapping has Fixed Point

Let $S$ be a set.
Let $g: S \to S$ be a progressing mapping on $S$.
Let $S$ be superinductive under $g$.
Then there exists $x \in S$ such that $x$ is a fixed point of $g$.

Proof:
Let us assume the hypothesis.
Let $M$ be the intersection of all subsets of $S$ that are superinductive under $g$.
From:
:Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
:Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions
it follows by definition that $M$ is minimally superinductive under $g$.
Hence by definition $M$ is a $g$-Tower.
By Intersection of Non-Empty Class is Set, $M$ is a set.
Thus by Union of $g$-Tower is Greatest Element and Unique Fixed Point:
:$\ds \bigcup M$ is a fixed point of $g$
:$\ds \bigcup M \in M$
Because $M \subseteq S$ it follows that:
:$\ds \bigcup M \in S$
and so $\ds \bigcup M$ is the $x$ that is being proved to exist.
{{qed}}
Theorem: Equivalence of Definitions of Strictly Progressing Mappings

Let $g$ be a mapping.
{{TFAE|def = Strictly Progressing Mapping}}

Proof:
=== $(1)$ implies $(2)$ ===
Let $g$ be a strictly progressing mapping by definition $1$.
Then by definition:
:$\forall x \in \Dom g: x \subsetneqq \map g x$
Hence {{afortiori}}:
:$\forall x \in \Dom g: x \subseteq \map g x$
and so $g$ is a progressing mapping.
Also {{afortiori}}:
:$\forall x \in \Dom g: x \ne \map g x$
and so $g$ has no fixed points.
Thus $g$ is a strictly progressing mapping by definition $2$.
{{qed|lemma}}
Theorem: Superinductive Class under Strictly Progressing Mapping is Proper Class

Let $A$ be a class.
Let $g: A \to A$ be a strictly progressing mapping on $A$.
Let $A$ be superinductive under $g$.
Then $A$ cannot be a set, and thus is a proper class.

Proof:
{{AimForCont}} $A$ is a set.
Then from Set which is Superinductive under Progressing Mapping has Fixed Point, $A$ has a fixed point.
However, we have that $g$ is a strictly progressing mapping on $A$.
Hence {{afortiori}} $g$ has no fixed point in $A$.
Hence the result by Proof by Contradiction.
{{Qed}}
Theorem: Modus Ponendo Ponens for Semantic Consequence in Predicate Logic

Let $\mathrm{PL}$ be the formal semantics of structures for predicate logic.
Denote with $\models_{\mathrm{PL}}$ $\mathrm{PL}$-semantic consequence.
Let $\mathbf A$ and $\mathbf B$ be sentences.
Let $\FF$ be a set of sentences.
Suppose that:
:$\FF \models_{\mathrm{PL}} \mathbf A$
:$\FF \models_{\mathrm{PL}} \mathbf A \implies \mathbf B$
Then:
:$\FF \models_{\mathrm{PL}} \mathbf B$
establishing Modus Ponendo Ponens in $\mathrm{PL}$.

Proof:
To establish $\FF \models_{\mathrm{PL}} \mathbf B$, we need to show that for all structures $\AA$:
:$\AA \models_{\mathrm{PL}} \FF$ implies $\AA \models_{\mathrm{PL}} \mathbf B$
where $\models_{\mathrm{PL}}$ denotes the models relation.
Then since $\AA \models_{\mathrm{PL}} \FF$, it follows that:
:$\AA \models_{\mathrm{PL}} \mathbf A$
:$\AA \models_{\mathrm{PL}} \mathbf A \implies \mathbf B$
By definition then, the latter means:
:$\map {f^\to} {\map {\operatorname{val}_\AA} {\mathbf A}, \map {\operatorname{val}_\AA} {\mathbf B} } = \T$
where $f^\to$ is the truth function of $\implies$ and $\map {\operatorname{val}_\AA} {\mathbf A}$ is the value of $\mathbf A$ in $\AA$.
Since $\map {\operatorname{val}_\AA} {\mathbf A} = \T$, it then follows from the definition of $f^\to$ that necessarily:
:$\map {\operatorname{val}_\AA} {\mathbf B} = \T$
which is to say:
:$\AA \models_{\mathrm{PL}} \mathbf B$
Since $\AA$ was arbitrary such that $\AA \models_{\mathrm{PL}} \FF$, it follows that:
:$\FF \models_{\mathrm{PL}} \mathbf B$
{{qed}}
Category:Model Theory for Predicate Logic
Category:Modus Ponendo Ponens
fi93oq8jmwp3xxmoypora4bhe2q8265
Theorem: Short Exact Sequence Condition of Noetherian Modules

Let $A$ be a commutative ring with unity.
Let:
:$0 \longrightarrow M' \stackrel {\alpha} {\longrightarrow} M \stackrel {\beta} {\longrightarrow} M' ' \longrightarrow 0$
be a short exact sequence of $A$-modules.
Then:
:$M$ is Noetherian
{{iff}}:
:$M'$ and $M' '$ are Noetherian

Proof:
=== Necessary condition ===
Assume $M$ is Noetherian.
Theorem: Subspace of Euclidean Space is Closed

Let $\R^n$ be a Euclidean space.
Every subspace of $\R^n$ is closed in $\R^n$.

Proof:
First we note that from Euclidean Space is Normed Vector Space, $\R^n$ has a norm $\norm {\, \cdot \,}$.
Let $V$ be a subspace of $\R^n$.
Let $\set {\mathbf v_1, \ldots, \mathbf v_k}$ be a basis for $V$.
Extend this to basis $\set {\mathbf v_1, \ldots, \mathbf v_k, \mathbf v_{k + 1}, \ldots, \mathbf v_n}$ for $\R^n$.
By using Gram-Schmidt procedure, there exists a set of orthonormal vectors $\set {\mathbf u_1, \ldots, \mathbf u_n}$ such that:
:$\forall k \in \N_{> 0} : k \le n : \map \span {\mathbf v_1, \ldots, \mathbf v_k} = \map \span {\mathbf u_1, \ldots, \mathbf u_k}$
where $\span$ stands for the linear span.
Let $A \in \R^{\paren {n - k} \times n}$ be a matrix as follows:
:$A = \begin {bmatrix} {\mathbf u_{k + 1} }^\intercal \\ \vdots \\ {\mathbf u_n}^\intercal \end {bmatrix}$
where $\mathbf u^\intercal$ denotes the transpose of $\mathbf u$ when expressed as a column matrix.
Theorem: Characterization of Lipschitz Continuity on Shift of Finite Type by Variations

Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $f : X_\mathbf A \to \C$ be a continuous mapping.
Let $\theta \in \openint 0 1$.
Let $C > 0$.
Then:
:$\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$
{{iff}}:
:$\forall n \in \N : \map {\mathrm {var}_n} f \le C \theta ^n$
where:
:$d_\theta$ denotes the metric
:$\map {\mathrm {var}_n} f$ denotes the $n$th variation of $f$.
That is:
:$\ds \size f_\theta = \sup \set { \dfrac {\cmod {\map f x - \map f y} }{\map {d_\theta} {x,y} } : x,y \in X_\mathbf A, x \ne y}$
where $\size \cdot_\theta $ is the Lipschitz seminorm.
{{improve|Maybe a separate page for this "That is"}}
{{improve|... and we still need to reference "Lipschitz continuity" somehow}}

Proof:
=== Sufficient Condition ===
Suppose:
:$\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$
Let $n\in\N$.
By definition of the metric:
:$\forall x, y \in X_\mathbf A:$
::$\forall i \in \openint {-n} n : x_i = y_i \implies \map {d _\theta} {x, y} \le \theta ^n$
Thus, by the assumption:
:$\forall x, y \in X_\mathbf A:$
::$\forall i \in \openint {-n} n :x_i = y_i \implies \size {\map f x - \map f y} \le C \map {d_\theta} {x, y} \le C \theta ^n$
Therefore:
:$\map {\mathrm {var}_n} f \le C \theta ^n$
{{qed|lemma}}
Theorem: G-Tower is Closed under Mapping

Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is closed under $g$:
:$\forall x: \paren {x \in M \implies \map g x \in M}$

Proof:
By definition:
:a $g$-tower is a class which is minimally superinductive under $g$
:a class which is minimally superinductive under $g$ is superinductive under $g$
:a superinductive class is inductive under $g$
:an inductive class under $g$ is closed under $g$.
Hence the result.
{{qed}}
Category:G-Towers
Category:Closedness under Mappings
jwpnuelosjzwkvegxpgl9gdmfvaid8l
Theorem: Strict Lower Closure of G-Tower is Set of Elements which are Proper Subsets

Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x \in M$.
Then the strict lower closure of $x$ is the set of all elements of $M$ that are proper subsets of $M$.

Proof:
Follows directly from the definitions.
{{qed}}
Theorem: P-Seminorm is Seminorm

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \hointr 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space on $\struct {X, \Sigma, \mu}$. 
Let $\norm {\, \cdot \,}_p$ be the $p$-seminorm on $\map {\LL^p} {X, \Sigma, \mu}$.
Then $\norm {\, \cdot \,}_p$ is a seminorm on $\map {\LL^p} {X, \Sigma, \mu}$.

Proof:
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
From the construction of the integral of a positive $\Sigma$-measurable function, we have: 
:$\ds \paren {\int \size f^p \rd \mu}^{1/p} \ge 0$
so:
:$\norm f_p \ge 0$ 
We also have: 
:$\ds \int \size f^p \rd \mu < \infty$
So $\norm {\, \cdot \,}$ maps from $\map {\LL^p} {X, \Sigma, \mu}$ to the non-negative reals.
We now verify {{SeminormAxiom|2}} and {{SeminormAxiom|3}}.
Theorem: Lp Space is Subset of Space of Real-Valued Measurable Functions Identified by A.E. Equality

Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \closedint 1 \infty$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$ be the space of real-valued measurable functions identified by $\mu$-A.E. equality.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ space of $\struct {X, \Sigma, \mu}$.
Then: 
:$\map {L^p} {X, \Sigma, \mu} \subseteq \map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$
In particular:
:if $f \in \map {\LL^p} {X, \Sigma, \mu}$ and $f \sim_\mu F$ we have $F \in \map {\LL^p} {X, \Sigma, \mu}$

Proof:
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
Denote by $\eqclass f {\sim_\mu}$ the $\sim_\mu$-equivalence class of $f$ in $\map {\LL^p} {X, \Sigma, \mu}$.
Let $\eqclass f {\sim_\mu}^\ast$ be the $\sim_\mu$-equivalence class of $f$ in $\map {\mathcal M} {X, \Sigma, \R}$.
We aim to show that: 
:$\eqclass f {\sim_\mu} = \eqclass f {\sim_\mu}^\ast$
Since $\map {\LL^p} {X, \Sigma, \mu} \subseteq \map {\mathcal M} {X, \Sigma, \R}$, we have:
:$\eqclass f {\sim_\mu} \subseteq \eqclass  f {\sim_\mu}^\ast$
Now let $F \in \eqclass f {\sim_\mu}^\ast$.
Then: 
:$F \sim_\mu f$ 
From Pointwise Exponentiation preserves A.E. Equality, we have: 
:$\size F^p \sim_\mu \size f^p$
Then, from {{Corollary|A.E. Equal Positive Measurable Functions have Equal Integrals|1}}: 
:$\ds \int \size F^p \rd \mu = \int \size f^p \rd \mu$
so that: 
:$\ds \paren {\int \size F^p \rd \mu}^{1/p} = \paren {\int \size f^p \rd \mu}^{1/p}$
so:
:$\norm F_p = \norm f_p < \infty$
We therefore finally have: 
:$F \in \map {\LL^p} {X, \Sigma, \mu}$
giving:
:$\eqclass f {\sim_\mu} = \eqclass f {\sim_\mu}^\ast$
{{qed}}
Category:Lp Spaces
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
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Theorem: Space of Real-Valued Measurable Functions Identified by A.E. Equality is Vector Space

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map \MM {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$. 
Let $\sim_\mu$ be the almost-everywhere equality relation on $\map \MM {X, \Sigma, \R}$ with respect to $\mu$.
Let $+$ denote pointwise addition on $\map \MM {X, \Sigma, \R} / \sim_\mu$.
Let $\cdot$ be pointwise scalar multiplication on $\map \MM {X, \Sigma, \R} / \sim_\mu$.
Then: 
:$\struct {\map \MM {X, \Sigma, \R} / \sim_\mu, +, \cdot}_\R$
forms a vector space.

Proof:
We verify each of the vector space axioms.
Theorem: G-Tower is Well-Ordered under Subset Relation/Empty Set

$\O$ is the smallest element of $M$.

Proof:
Follows directly from $g$-Tower is Well-Ordered under Subset Relation.
{{qed}}
Theorem: G-Tower is Well-Ordered under Subset Relation/Successor of Non-Greatest Element

Let $x \in M$ such that $x$ is not the greatest element of $M$.
Then the immediate successor of $x$ is $\map g x$.

Proof:
We have that $g$-Tower is Well-Ordered under Subset Relation.
Let $x \in M$ such that $x$ is not the greatest element of $M$.
Then from Fixed Point of $g$-Tower is Greatest Element:
:$x \ne \map g x$
Hence:
:$x \subsetneqq \map g x$
Hence by the Sandwich Principle for $g$-Towers, there is no $y \in M$ such that:
:$x \subsetneqq y \subsetneqq \map g x$
Hence $\map g x$ is the immediate successor of $x$.
{{qed}}
Theorem: G-Tower is Well-Ordered under Subset Relation/Union of Limit Elements

Let $x \in M$ be a limit element of $M$.
Then:
:$x = \bigcup x^\subset$
where $\bigcup x^\subset$ denotes the union of the lower section of $x$.

Proof:
We have that $g$-Tower is Well-Ordered under Subset Relation.
Let $x \in M$ be a limit element of $M$.
Then by definition $x$ is not an immediate successor element.
Hence by $g$-Tower is Well-Ordered under Subset Relation: Successor of Non-Greatest Element:
:there exists no $y \in M$ such that $x = \map g y$
otherwise $x$ would be the immediate successor of $y$.
Hence there is no $y \in x^\subset$ such that $x = \map g y$.
Thus condition $(\text C 1)$ of $g$-Tower is Well-Ordered under Subset Relation: Corollary does not hold.
Hence condition $(\text C 2)$ of $g$-Tower is Well-Ordered under Subset Relation: Corollary ''does'' hold.
That is:
:$\bigcup x^\subset = x$
{{qed}}