avewright/proofwiki-math huggingface dataset:
Theorem: Characterization of Null Sets of Variation of Signed Measure

Let $\struct {X, \Sigma}$ be measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $\size \mu$ be the variation of $\mu$. 
Then $A \in \Sigma$ is such that $\map {\size \mu} A = 0$ {{iff}}:
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.

Proof:
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then, from the definition of variation: 
:$\size \mu = \mu^+ + \mu^-$
Suppose that $A \in \Sigma$ is such that $\map {\size \mu} A = 0$.
Then: 
:$\map {\mu^+} A + \map {\mu^-} A = 0$
Since $\mu^+ \ge 0$ and $\mu^- \ge 0$ we have: 
:$\map {\mu^+} A = 0 $ and $\map {\mu^-} A = 0$
From Null Sets Closed under Subset, we then have: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map {\mu^+} B = 0$
and:
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map {\mu^-} B = 0$
So, for each $\Sigma$-measurable set $B \subseteq A$, we have: 
:$\map {\mu^+} B - \map {\mu^-} B = 0$
From the definition of the Jordan decomposition, we therefore have: 
:$\map \mu B = 0$
for each $\Sigma$-measurable set $B \subseteq A$.
{{qed|lemma}}
Now suppose that: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$. 
From the definition of Jordan decomposition, and Uniqueness of Jordan Decomposition, we have: 
:$\map {\mu^+} A = \map \mu {P \cap A}$
and:
:$\map {\mu^+} A = -\map \mu {N \cap A}$
From Intersection is Subset, we have: 
:$P \cap A \subseteq A$
and:
:$N \cap A \subseteq A$
So, we have: 
:$\map \mu {P \cap A} = 0$
and:
:$\map \mu {N \cap A} = 0$
from the assumption.
That is: 
:$\map {\mu^+} A = 0$
and:
:$\map {\mu^-} A = 0$
So:
:$\map {\size \mu} A = \map {\mu^+} A + \map {\mu^-} A = 0$
So, if: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$
then:
:$\map {\size \mu} A = 0$
{{qed}}
Category:Variation of Signed Measure
rti89860dra95l494044bc5bjeunb31
Theorem: Signed Measure may not be Monotone

Let $\struct {X, \Sigma}$ be measurable space. 
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Then $\mu$ may not be monotone.

Proof:
Let:
:$\struct {X, \Sigma} = \struct {\R, \map \BB \R}$
where $\map \BB \R$ is the Borel $\sigma$-algebra on $\R$.
Define: 
:$\mu = \delta_1 - 2 \delta_2$
where $\delta_1$ and $\delta_2$ are the Dirac measures at $1$ and $2$ respectively.
Since $\delta_1$ and $\delta_2$ are both finite measures, we have: 
:$\mu$ is a signed measure
from Linear Combination of Signed Measures is Signed Measure.
Then, we have: 
:$\closedint 0 1 \subseteq \closedint 0 2$
with: 
{{begin-eqn}}
{{eqn	| l = \map \mu {\closedint 0 1}
		| r = \map {\delta_1} {\closedint 0 1} - 2 \map {\delta_2} {\closedint 0 1}
}}
{{eqn	| r = 1 - 0
		| c = {{Defof|Dirac Measure}}
}}
{{eqn	| r = 1
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \map \mu {\closedint 0 2}
		| r = \map {\delta_1} {\closedint 0 2} - 2 \map {\delta_2} {\closedint 0 2}
}}
{{eqn	| r = 1 - 2
		| c = {{Defof|Dirac Measure}}
}}
{{eqn	| r = -1
}}
{{end-eqn}}
So:
:$\closedint 0 1 \subseteq \closedint 0 2$ and $\map \mu {\closedint 0 2} \le \map \mu {\closedint 0 1}$
So:
:$\mu$ is not monotone.
{{qed}}
Category:Signed Measures
bbrjoe0bwu9mxl000lyh0ykl21zyu85
Theorem: Sigma-Algebra Closed under Set Difference

Let $\struct {X, \Sigma}$ be a measurable space.
Let $A, B \in \Sigma$.
Then the set difference $A \setminus B$ is contained in $\Sigma$.

Proof:
Since $\sigma$-algebras are closed under relative complement, we have: 
:$\relcomp X B \in \Sigma$
By Sigma-Algebra Closed under Finite Intersection, we have: 
:$A \cap \relcomp X B \in \Sigma$
From Set Difference as Intersection with Relative Complement, we have: 
:$A \setminus B = A \cap \relcomp X B$
so:
:$A \setminus B \in \Sigma$
{{qed}}
Category:Sigma-Algebras
Category:Set Difference
hp1lbdwohv14kvsd8e7sde67vxq4mc3
Theorem: Sigma-Algebra Closed under Symmetric Difference

Let $\struct {X, \Sigma}$ be a measurable space.
Let $A, B \in \Sigma$.
Then the symmetric difference $A \Delta B$ is contained in $\Sigma$.

Proof:
From Sigma-Algebra Closed under Set Difference, we have: 
:$A \setminus B \in \Sigma$
and:
:$B \setminus A \in \Sigma$
Since $\sigma$-algebras are closed under countable union, we have: 
:$\paren {A \setminus B} \cup \paren {B \setminus A} \in \Sigma$
From the definition of symmetric difference, we have: 
:$A \Delta B = \paren {A \setminus B} \cup \paren {B \setminus A}$
so:
:$A \Delta B \in \Sigma$
{{qed}}
Category:Sigma-Algebras
Category:Symmetric Difference
7ne94vqd8eusgq7yf98nbzkjqhfcpxl
Theorem: Characterization of Null Sets of Variation of Complex Measure

Let $\struct {X, \Sigma}$ be measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Let $\size \mu$ be the variation of $\mu$. 
Then $A \in \Sigma$ is such that $\map {\size \mu} A = 0$ {{iff}}:
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.

Proof:
Let $A \in \Sigma$.
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
From the definition of variation, we have:
:$\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
Suppose that: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
Then, for any: 
:$\set {A_1, A_2, \ldots, A_n} \in \map P A$
we have: 
:$\map \mu {A_i} = 0$ for each $i$
so:
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } = 0$
This gives: 
:$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = \set 0$
So, from the definition of supremum, we have: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = 0$
so:
:$\map {\cmod \mu} A = 0$ 
{{qed|lemma}}
Suppose that:
:$\map {\cmod \mu} A = 0$
Then: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = 0$
Note that since each element of: 
:$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$ 
is non-negative, we must have: 
:$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = \set 0$
So, for all:
:$\set {A_1, A_2, \ldots, A_n} \in \map P A$
we have: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } = 0$
Let $B \subseteq A$ be $\Sigma$-measurable.
From Sigma-Algebra Closed under Set Difference, we have: 
:$A \setminus B$ is $\Sigma$-measurable.
We also have: 
:$A = B \cup \paren {A \setminus B}$ 
so:
:$\set {B, A \setminus B}$ is a finite partition of $A$ into $\Sigma$-measurable sets.
That is: 
:$\set {B, A \setminus B} \in \map P A$
We therefore have: 
:$\cmod {\map \mu B} + \cmod {\map \mu {A \setminus B} } = 0$
Since: 
:$\cmod {\map \mu {A \setminus B} } \ge 0$
and:
:$\cmod {\map \mu B} \ge 0$ 
we must have: 
:$\cmod {\map \mu B} = \cmod {\map \mu {A \setminus B} } = 0$
and in particular: 
:$\map \mu B = 0$
So:
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
{{qed}}
Category:Variation of Complex Measure
0r6dicyx47znztrul1kdtej2g0t43k5
Theorem: Absolute Continuity of Complex Measure in terms of Jordan Decomposition

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a complex measure on $\struct {X, \Sigma}$.
Let $\tuple {\nu_1, \nu_2, \nu_3, \nu_4}$ be the Jordan decomposition of $\nu$.
Then $\nu$ is absolutely continuous with respect to $\mu$ {{iff}}:
:$\nu_1$, $\nu_2$, $\nu_3$ and $\nu_4$ are absolutely continuous with respect to $\mu$.

Proof:
Let $\cmod \nu$ be the variation of $\nu$.
Theorem: Bound for Variation of Complex Measure in terms of Jordan Decomposition

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Let $\cmod \mu$ be the variation of $\mu$. 
Let $\tuple {\mu_1, \mu_2, \mu_3, \mu_4}$ be the Jordan decomposition of $\mu$.
Then: 
:$\map {\cmod \mu} A \le \map {\mu_1} A + \map {\mu_2} A + \map {\mu_3} A + \map {\mu_4} A$
for all $A \in \Sigma$.

Proof:
Let $A \in \Sigma$.
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
From the definition of variation, we have: 
:$\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
Let: 
:$\set {A_1, A_2, \ldots, A_n} \in \map P A$ 
Consider: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} }$
We have, from the definition of Jordan decomposition: 
:$\mu = \mu_1 - \mu_2 + i \paren {\mu_3 - \mu_4}$
so that: 
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} }
		| r = \sum_{i \mathop = 1}^n \cmod {\map {\mu_1} {A_i} - \map {\mu_2} {A_i} + i \paren {\map {\mu_3} {A_i} - \map {\mu_4} {A_i} } }
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \cmod {\map {\mu_1} {A_i} } + \sum_{i \mathop = 1}^n \cmod {\map {\mu_2} {A_i} } + \sum_{i \mathop = 1}^n \cmod {\map {\mu_3} {A_i} } + \sum_{i \mathop = 1}^n \cmod {\map {\mu_4} {A_i} }
		| c = Triangle Inequality for Complex Numbers
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \map {\mu_1} {A_i} + \sum_{i \mathop = 1}^n \map {\mu_2} {A_i} + \sum_{i \mathop = 1}^n \map {\mu_3} {A_i} + \sum_{i \mathop = 1}^n \map {\mu_4} {A_i}
		| c = since $\mu_1$, $\mu_2$, $\mu_3$ and $\mu_4$ are measures
}}
{{end-eqn}}
Since: $\set {A_1, A_2, \ldots, A_n}$ is a partition of $A$, we have: 
:$\set {A_1, A_2, \ldots, A_n}$ are pairwise disjoint with: 
::$\ds A = \bigcup_{i \mathop = 1}^n A_i$
So:
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^n \map {\mu_1} {A_i} + \sum_{i \mathop = 1}^n \map {\mu_2} {A_i} + \sum_{i \mathop = 1}^n \map {\mu_3} {A_i} + \sum_{i \mathop = 1}^n \map {\mu_4} {A_i}
		| r = \map {\mu_1} {\bigcup_{i \mathop = 1}^n A_i} + \map {\mu_2} {\bigcup_{i \mathop = 1}^n A_i} + \map {\mu_3} {\bigcup_{i \mathop = 1}^n A_i} + \map {\mu_4} {\bigcup_{i \mathop = 1}^n A_i}
		| c = Measure is Finitely Additive Function
}}
{{eqn	| r = \map {\mu_1} A + \map {\mu_2} A + \map {\mu_3} A + \map {\mu_4} A
}}
{{end-eqn}}
{{qed}}
Category:Complex Measures
cqcizkxjifqkntyx1dsyi7o4jj2zlh0
Theorem: Characterization of Absolute Continuity of Signed Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a signed measure on $\struct {X, \Sigma}$.
Then $\nu$ is absolutely continuous with respect to $\mu$ {{iff}}: 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.

Proof:
Let $\tuple {\nu^+, \nu^-}$ be the Jordan decomposition of $\nu$. 
Let $\size \nu$ be the variation of $\nu$.
Theorem: Characterization of Absolute Continuity of Complex Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a complex measure on $\struct {X, \Sigma}$.
Then $\nu$ is absolutely continuous with respect to $\mu$ {{iff}}: 
:for all $A \in \Sigma$ with $\map \mu A = 0$, we have $\map \nu A = 0$.

Proof:
Let $\tuple {\nu_1, \nu_2, \nu_3, \nu_4}$ be the Jordan decomposition of $\nu$. 
Let $\size \nu$ be the variation of $\nu$.
Theorem: Characteristic Function of Disjoint Union/Corollary

Let $\set {D_1, D_2, \ldots, D_N}$ be a set of pairwise disjoint subsets of $X$. 
Let: 
:$\ds D = \bigcup_{n \mathop = 1}^N D_n$
Then: 
:$\ds \chi_D = \sum_{n \mathop = 1}^N \chi_{D_n}$
where:
:$\chi_D$ is the characteristic function of $D$
:$\chi_{D_n}$ is the characteristic function of $D_n$.

Proof:
We can extend $\set {D_1, D_2, \ldots, D_N}$ to a sequence $\sequence {D_n}_{n \mathop \in \N}$ of subsets of $X$ by setting: 
:$D_i = \O$ for $i \ge N + 1$
Clearly, from Intersection with Empty Set, we have: 
:$D_i \cap D_j = \O$ for $i \ge N + 1$ and all $j$. 
So $\sequence {D_n}_{n \mathop \in \N}$ is a sequence of pairwise disjoint subsets of $X$ with: 
:$\ds D = \bigcup_{n \mathop = 1}^\infty D_n$
We therefore have, by Characteristic Function of Disjoint Union:
:$\ds \chi_D = \sum_{n \mathop = 1}^\infty \chi_{D_n}$
We can write:
{{begin-eqn}}
{{eqn	| l = \sum_{n \mathop = 1}^\infty \chi_{D_n}
		| r = \sum_{n \mathop = 1}^N \chi_{D_n} + \sum_{n \mathop = N + 1}^\infty \chi_{D_n}
}}
{{eqn	| r = \sum_{n \mathop = 1}^N \chi_{D_n} + \sum_{n \mathop = N + 1}^\infty \chi_\O
}}
{{eqn	| r = \sum_{n \mathop = 1}^N \chi_{D_n} + \sum_{n \mathop = N + 1}^\infty 0
		| c = Characteristic Function of Empty Set
}}
{{eqn	| r = \sum_{n \mathop = 1}^N \chi_{D_n}
}}
{{end-eqn}}
{{qed}}
Category:Characteristic Function of Disjoint Union
1264rt0wvpst4jmzkay4nzfpxh27cmt
Theorem: Integral of Positive Function with respect to Counting Measure on Natural Numbers

Consider the measure space $\struct {\N, \powerset \N, \mu}$ where $\mu$ is the counting measure on $\struct {\N, \powerset \N}$.
Let $f : \N \to \R$ be a function. 
Then: 
:$\ds \int f \rd \mu = \sum_{n \mathop = 1}^\infty \map f n$

Proof:
Clearly we have: 
:$\set {x \in \N : \map f x \le \alpha} \in \powerset \N$
for each $\alpha \in \R$, so any function $f : \N \to \R$ is $\powerset \N$-measurable.
Similarly, an arbitrary subset of $\N$ is clearly $\powerset \N$-measurable.
For each $n \in \N$, define $f_n : \N \to \R$ by: 
:$\ds \map {f_n} k = \begin{cases}\map f k & k \le n \\ 0 & k > n\end{cases}$
for $k \in \N$.
We show that $\sequence {f_n}_{n \mathop \in \N}$ is increasing.
Let $k \in \N$. 
We want to show that: 
:$\map {f_n} k \le \map {f_{n + 1} } k$ 
for each $n \in \N$. 
If $k \le n$, then we have: 
:$\map {f_n} k = \map f k$
and:
:$\map {f_{n + 1} } k = \map f k$
so we have: 
:$\map {f_n} k = \map {f_{n + 1} } k$
in this case.
If $k = n + 1$, then: 
:$\map {f_n} k = 0$
and:
:$\map {f_{n + 1} } k = \map f k$
so we have: 
:$\map {f_n} k \le \map {f_{n + 1} } k$
If $k > n + 1$, then: 
:$\map {f_n} k = \map f k$
and:
:$\map {f_{n + 1} } k = \map f k$
so that: 
:$\map {f_n} k = \map {f_{n + 1} } k$
So $\sequence {f_n}_{n \mathop \in \N}$ is increasing
We show that: 
:$\ds \map f k = \lim_{n \mathop \to \infty} \map {f_n} k$
Note that for all $n \ge k$, we have: 
:$\map {f_n} k = \map f k$
so we obtain the above from Tail of Convergent Sequence.
So $\sequence {f_n}_{n \mathop \in \N}$ is a increasing sequence of $\powerset \N$-measurable functions converging to $f$. 
So, from the Monotone Convergence Theorem, we obtain: 
:$\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$ 
It remains to compute: 
:$\ds \int f_n \rd \mu$
for each $n \in \N$.
We can rewrite: 
:$\ds f_n = \sum_{i \mathop = 1}^n \map f i \chi_{\set i}$
Clearly:
:$\set {\set 1, \set 2, \ldots, \set n}$ is pairwise disjoint
so, from the definition of the $\mu$-integral of a positive simple function, we have: 
:$\ds \map {I_\mu} {f_n} = \sum_{i \mathop = 1}^n \map f i \map \mu {\set i}$
From the definition of the counting measure, we have: 
:$\map \mu {\set i} = 1$ 
for each $i$, so:
:$\ds \map {I_\mu} {f_n} = \sum_{i \mathop = 1}^n \map f i$
From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, we therefore have: 
:$\ds \int f_n \rd \mu = \sum_{i \mathop = 1}^n \map f i$
From the definition of infinite series, we then have: 
:$\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \map f k = \sum_{k \mathop = 1}^\infty \map f k$
as required.
{{qed}}
Category:Counting Measure
0evns9puxsgr24d6skayj580w22emio
Theorem: In Connected Smooth Manifold Any Two Points can be Joined by Admissible Curve

Let $M$ be a connected smooth manifold with or without a boundary.
Let $p, q \in M$ be points.
Let $\gamma : \closedint a b \to M$ be an admissible curve.
Then:
:$\forall p, q \in M : \exists \gamma \subset M : \paren {\map \gamma a = p} \land \paren {\map \gamma b = q}$

Proof:
For $p, q \in M$, we write:
:$p \sim q$
{{iff}} there exists an admissible curve $\gamma : \closedint a b \to M$ such that $\map \gamma a = p$ and $\map \gamma b = q$
Theorem: Function Measurable with respect to Power Set

Let $\struct {X, \powerset X}$ be a measurable space, where $\powerset X$ is the power set of $X$. 
Let $f : X \to \overline \R$ be a function.
Then $f$ is $\powerset X$-measurable function.

Proof:
For each $\alpha \in \R$, we have: 
:$\set {x \in X : \map f x \le \alpha} \subseteq X$
That is, from the definition of power set:
:$\set {x \in X : \map f x \le \alpha} \in \powerset X$
So for each $\alpha \in \R$:
:the set $\set {x \in X : \map f x \le \alpha}$ is $\powerset X$-measurable.
So:
:$f$ is $\powerset X$-measurable.
{{qed}}
Category:Measurable Functions
pf3wjasdppim7c084n0s9cn7wy8tes0
Theorem: Counting Measure on Natural Numbers is Sigma-Finite

Let $\mu$ be the counting measure on $\struct {\N, \powerset \N}$. 
Then $\mu$ is $\sigma$-finite.

Proof:
For each $n \in \N$, define: 
:$X_n = \N \cap \closedint 1 n = \set {1, 2, \ldots, n}$
Since $X_n \subseteq \N$ for each $n$ from Intersection is Subset, we have that:
:$X_n$ is $\powerset \N$-measurable for each $n$. 
We show that $\sequence {X_n}_{n \mathop \in \N}$ is an exhausting sequence in $\powerset \N$ and that: 
:$\map \mu {X_n} < \infty$ for each $n \in \N$. 
From the definition of counting measure, we also have: 
:$\map \mu {X_n} = n < \infty$ 
for each $n$. 
We have that: 
:$\N \cap \closedint 1 n \subseteq \N \cap \closedint 1 {n + 1}$ 
from Set Intersection Preserves Subsets, so:
:$X_n \subseteq X_{n + 1}$ 
So $\sequence {X_n}_{n \mathop \in \N}$ is an increasing sequence of $\powerset \N$-measurable sets.
It now only remains to verify: 
:$\ds \N = \bigcup_{n \mathop = 1}^\infty X_n$
We clearly have: 
:$\ds \bigcup_{n \mathop = 1}^\infty X_n \subseteq \N$
For each $k \in \N$, we have $k \in X_k$, so we also have: 
:$\ds \N \subseteq \bigcup_{n \mathop = 1}^\infty X_n$ 
so that: 
:$\ds \N = \bigcup_{n \mathop = 1}^\infty X_n$
So $\sequence {X_n}_{n \mathop \in \N}$ is exhausting sequence in $\powerset \N$ and that: 
:$\map \mu {X_n} < \infty$ for each $n \in \N$.
So:
:$\mu$ is $\sigma$-finite.
{{qed}}
Category:Counting Measure
5faqembfbqs15ng4busidpc6s229otc
Theorem: Function A.E. Equal to Measurable Function in Complete Measure Space is Measurable

Let $\struct {X, \Sigma, \mu}$ be a complete measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $g : X \to \overline \R$ be a function such that: 
:$f = g$ $\mu$-almost everywhere.
Then $g$ is $\Sigma$-measurable.

Proof:
We aim to show that: 
:$\set {x \in X : \map g x \le \alpha} \in \Sigma$
for each $\alpha \in \R$. 
Let $\alpha \in \R$. 
Since $f = g$ $\mu$-almost everywhere there exists a $\mu$-null set such that: 
:whenever $x \in X$ has $\map f x \ne \map g x$, we have $x \in N$.
We have: 
{{begin-eqn}}
{{eqn	| l = \set {x \in X : \map g x \le \alpha}
		| r = \set {x \in X : \map g x \le \alpha} \cap \paren {N \cup \paren {X \setminus N} }
		| c = Intersection with Subset is Subset, Union with Complement
}}
{{eqn	| r = \paren {\set {x \in X : \map g x \le \alpha} \cap N} \cup \paren {\set {x \in X : \map g x \le \alpha} \cap \paren {X \setminus N} }
		| c = Union Distributes over Intersection
}}
{{end-eqn}}
For $x \in X \setminus N$, we have: 
:$\map g x = \map f x$
so:
:$\set {x \in X \setminus N : \map g x \le \alpha} \cap \paren {X \setminus N} = \set {x \in X \setminus N : \map f x \le \alpha} \cap \paren {X \setminus N}$
Since $f$ is $\Sigma$-measurable, we have: 
:$\set {x \in X : \map f x \le \alpha} \in \Sigma$
Since $\sigma$-algebras are closed under complement, we have: 
:$X \setminus N \in \Sigma$ 
From Sigma-Algebra Closed under Countable Intersection, we therefore have: 
:$\set {x \in X \setminus N : \map f x \le \alpha} \cap \paren {X \setminus N} \in \Sigma$
Now, from Intersection is Subset, we have: 
:$\set {x \in X : \map g x \le \alpha} \cap N \subseteq N$
Since $N$ is $\mu$-null, we have: 
:$\set {x \in X : \map g x \le \alpha}$ is $\mu$-null
since $\struct {X, \Sigma, \mu}$ is complete. 
In particular: 
:$\set {x \in X : \map g x \le \alpha} \cap N \in \Sigma$
So, since $\sigma$-algebras are closed under countable union, we have: 
:$\paren {\set {x \in X : \map g x \le \alpha} \cap N} \cup \paren {\set {x \in X : \map g x \le \alpha} \cap \paren {X \setminus N} } \in \Sigma$
that is: 
:$\set {x \in X : \map g x \le \alpha} \in \Sigma$
Since $\alpha \in \R$ was arbitrary, we have: 
:$g$ is $\Sigma$-measurable.
{{qed}}
Category:Complete Measure Spaces
f80ocjny8kvncmosau3550a3vxwyga7
Theorem: Lower Sum of Refinement

Let $\closedint a b$ be a closed interval.
Let $P$ be a finite subdivision of $\closedint a b$. 
Let $Q$ be a refinement of $P$. 
Then: 
:$\map L {f, P} \le \map L {f, Q}$
where $\map L {f, P}$ and $\map L {f, Q}$ denotes the lower Darboux sum of $f$ with respect to $P$ and $Q$.

Proof:
Write: 
:$P = \set {x_0, x_1, \ldots, x_k}$
and:
:$Q = \set {y_0, y_1, \ldots, y_l}$
where:
:$a = x_0 < x_1 < \ldots < x_k = b$
and:
:$a = y_0 < y_1 < \ldots < y_l = b$
Since $P \subseteq Q$, we have $k \le l$ from Cardinality of Subset of Finite Set.
Set: 
:$m_i = \inf \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for each $1 \le i \le k$.
Also set: 
:${\tilde m}_j = \inf \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} }$
for each $1 \le j \le l$. 
Consider a pair of elements $\tuple {x_{i - 1}, x_i}$ in $P$. 
Since $P \subseteq Q$, there exists $a_i, b_i$ such that: 
:$\tuple {x_{i - 1}, x_i} = \tuple {y_{a_i}, y_{b_i} }$
We can see that: 
:$a_1 = 0$
and:
:$b_k = l$ 
We also clearly have: 
:$b_{i - 1} = a_i$ for each $1 \le i \le k$. 
Note that: 
:$\closedint {y_{j - 1} } {y_j} \subseteq \closedint {x_{i - 1} } {x_i}$
for all $a_i + 1 \le j \le b_i$. 
So:
:$\set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \subseteq \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$. 
So, from Infimum of Subset, we have:
:$\inf \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \ge \inf \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$.
That is: 
:$m_i \le {\tilde m}_j$
for each $\tuple {i, j}$ with $a_i + 1 \le j \le b_i$
We can then write: 
:$\ds x_i - x_{i - 1} = \sum_{j \mathop = a_i + 1}^{b_i} \paren {y_j - y_{j - 1} }$
for each $1 \le i \le k$, giving:
{{begin-eqn}}
{{eqn	| l = \map L {f, P}
		| r = \sum_{i \mathop = 1}^k m_i \paren {x_i - x_{i - 1} }
		| c = {{Defof|Lower Darboux Sum}}
}}
{{eqn	| r = \sum_{i \mathop = 1}^k m_i \paren {\sum_{j \mathop = a_i + 1}^{b_i} \paren {y_j - y_{j - 1} } }
}}
{{eqn	| r = \sum_{i \mathop = 1}^k \paren {\sum_{j \mathop = a_i + 1}^{b_i} m_i \paren {y_j - y_{j - 1} } }
}}
{{eqn	| o = \le
		| r = \sum_{i \mathop = 1}^k \paren {\sum_{j \mathop = a_i + 1}^{b_i} {\tilde m}_j \paren {y_j - y_{j - 1} } }
		| c = since $m_i \le {\tilde m}_j$ for each $\tuple {i, j}$ with $a_i + 1 \le j \le b_i$
}}
{{eqn	| r = \sum_{j \mathop = 1}^l {\tilde m}_j \paren {y_j - y_{j - 1} }
}}
{{eqn	| r = \map L {f, Q}
}}
{{end-eqn}}
{{qed}}