avewright/proofwiki-math huggingface dataset:
Theorem: Union of Vertical Sections is Vertical Section of Union

Let $X$ and $Y$ be sets.
Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$. 
Let $x \in X$. 
Then: 
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
where:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x$ is the $x$-vertical section of $\ds \bigcup_{\alpha \in A} E_\alpha$
:$\paren {E_\alpha}_x$ is the $x$-vertical section of $E_\alpha$.

Proof:
Note that:
:$\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
{{iff}}: 
:$y \in \paren {E_\alpha}_x$ for some $\alpha \in A$.
From the definition of the $x$-vertical section, this is equivalent to:
:$\tuple {x, y} \in E_\alpha$ for some $\alpha \in A$. 
This in turn is equivalent to: 
:$\ds \tuple {x, y} \in \bigcup_{\alpha \in A} E_\alpha$
Again applying the definition of the $x$-vertical section:
:$\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
So:
:$\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$ {{iff}} $\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
giving:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
{{qed}}
Category:Set Union
Category:Vertical Section of Sets
l8mdrbl39qfpr2xzcw3pwzqwpmtp0h2
Theorem: Tempered Distribution Space is Proper Subset of Distribution Space

Let $\map {\DD'} \R$ be the distribution space.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then $\map {\SS'} \R$ is a proper subset of $\map {\DD'} \R$:
:$\map {\SS'} \R \subsetneqq \map {\DD'} \R$

Proof:
By Convergence of Sequence of Test Functions in Test Function Space implies Convergence in Schwartz Space we have that $\map {\SS'} \R \subseteq \map {\DD'} \R$.
{{Research|how?}}
Consider the real function $\map f x = e^{x^2}$.
We have that:
:Real Power Function for Positive Integer Power is Continuous
:Exponential Function is Continuous/Real Numbers
:Composite of Continuous Mappings is Continuous
Thus, $f$ is a continuous real function.
Also:
:$\forall x \in \R : e^{x^2} < \infty$
Hence, $f$ is locally integrable.
By Locally Integrable Function defines Distribution, $T_f \in \map {\DD'} \R$.
{{AimForCont}} $T_f$ is a tempered distribution.
We have that $e^{-x^2}$ is a Schwartz test function.
Then:
{{begin-eqn}}
{{eqn | l = \map {T_f} {e^{-x^2} } 
      | r = \int_{- \infty}^\infty e^{x^2} e^{-x^2} \rd x
}}
{{eqn | r = \int_{- \infty}^\infty 1 \rd x
}}
{{eqn | r = \infty
}}
{{end-eqn}}
Hence, $\map {T_f} {e^{-x^2} } \notin \R$.
This is a contradiction.
Therefore, $T_f \notin \map {\SS'} \R$ while at the same time $T_f \in \map {\DD'} \R$.
{{qed}}
Theorem: Jordan Decomposition Theorem

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$. 
Then there exists measures $\mu^+$ and $\mu^-$ on $\struct {X, \Sigma}$ such that: 
:$\mu = \mu^+ - \mu^-$
Further, at least one of $\mu^+$ and $\mu^-$ is finite.

Proof:
From the definition of a Hahn decomposition, the set $P$ is $\mu$-positive, the set $N$ is $\mu$-negative and: 
:$X = P \cup N$ 
with $P$ and $N$ disjoint.
From Sigma-Algebra Closed under Countable Intersection, we have: 
:$A \cap P \in \Sigma$
and:
:$A \cap N \in \Sigma$
for each $A \in \Sigma$.
Now, for each $A \in \Sigma$, define: 
:$\map {\mu^+} A = \map \mu {A \cap P}$
and:
:$\map {\mu^-} A = -\map \mu {A \cap N}$
We verify that $\mu^+$ and $\mu^-$ are indeed measures by first showing that they are signed measures.
Theorem: Intersection of Positive Set and Negative Set is Null Set

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $P$ be a $\mu$-positive set. 
Let $N$ be a $\mu$-negative set.
Then: 
:$P \cap N$ is a $\mu$-null set.

Proof:
Note that, from Sigma-Algebra Closed under Countable Intersection:
:$P \cap N \in \Sigma$
We aim to show that:
:for each $E \in \Sigma$ with $E \subseteq P \cap N$ we have $\map \mu E = 0$.
Note first that from Intersection is Subset, we have: 
:$P \cap N \subseteq P$
so that: 
:$E \subseteq P$
So, since $P$ is $\mu$-positive, we have: 
:$\map \mu E \ge 0$
We also have that, from Intersection is Subset: 
:$P \cap N \subseteq N$
so that: 
:$E \subseteq N$
Since $N$ is $\mu$-negative, we have: 
:$\map \mu E \le 0$
Since: 
:$\map \mu E \ge 0$ and $\map \mu E \le 0$
we have: 
:$\map \mu E = 0$
So:
:for each $E \in \Sigma$ with $E \subseteq P \cap N$ we have $\map \mu E = 0$.
So:
:$P \cap N$ is a $\mu$-null set.
{{qed}}
Category:Positive Sets
Category:Negative Sets
pejgdad49ht4n47ovd0mhmtv8wiar1z
Theorem: Non-Negative Signed Measure is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$ such that: 
:$\map \mu A \ge 0$
for each $A \in \Sigma$. 
Then $\mu$ is a measure on $\struct {X, \Sigma}$.

Proof:
We verify each of the conditions given in the definition of a measure.
From the definition of a signed measure, $\mu$ is a function $\Sigma \to \overline \R$.
Theorem: Null Sets Closed under Countable Union/Signed Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $\sequence {N_i}_{i \mathop \in \N}$ be a sequence of $\mu$-null sets.
Then: 
:$\ds N = \bigcup_{i \mathop = 1}^\infty N_i$
is a $\mu$-null set.

Proof:
From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma$-measurable sets $\sequence {A_i}_{i \in \N}$ such that: 
:$\ds N = \bigcup_{i \mathop = 1}^\infty A_i$
We now show that if $E \in \Sigma$ has $E \subseteq N$, then $\map \mu E = 0$. 
Write: 
{{begin-eqn}}
{{eqn	| l = E
		| r = E \cap N
		| c = Intersection with Subset is Subset
}}
{{eqn	| r = \bigcup_{i \mathop = 1}^\infty \paren {E \cap A_i}
		| c = Intersection Distributes over Union of Family
}}
{{end-eqn}}
Since $\mu$ is countably additive, we have:
:$\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu {E \cap A_i}$
We have, from Intersection is Subset: 
:$E \cap A_i \subseteq A_i \subseteq N_i$
Since $N_i$ is a $\mu$-null set, we have that: 
:$\map \mu {E \cap A_i} = 0$ for each $i$. 
So:
:$\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu {E \cap A_i} = 0$
So:
:for each $E \in \Sigma$ with $E \subseteq N$, we have $\map \mu E = 0$.
So $N$ is a $\mu$-null set.
{{qed}}
Category:Null Sets Closed under Countable Union
Category:Signed Measures
kio66pjfsh482idbcwlxoax9jn07m95
Theorem: Uniqueness of Jordan Decomposition

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P_1, N_1}$ and $\tuple {P_2, N_2}$ be Hahn decompositions of $\mu$. 
Let $\tuple {\mu^+_1, \mu^-_1}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_1, N_1}$. 
Let $\tuple {\mu^+_2, \mu^-_2}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_2, N_2}$. 
Then: 
:$\mu^+_1 = \mu^+_2$
and:
:$\mu^-_1 = \mu^-_2$
So:
:the Jordan decomposition of a signed measure is unique.

Proof:
We first prove two useful identities.
Theorem: Uniqueness of Jordan Decomposition/Lemma

Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$.
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P, N}$. 
Then, for each $A \in \Sigma$, we have: 
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
and:
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$

Proof:
Since $\tuple {\mu^+, \mu^-}$ is a  Jordan decomposition of $\mu$, we have: 
:$\mu = \mu^+ - \mu^-$
with $\mu^+$ and $\mu^-$ measures.
Let $A \in \Sigma$.
We first show: 
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
{{begin-eqn}}
{{eqn	| l = \map \mu B
		| r = \map {\mu^+} B - \map {\mu^-} B
}}
{{eqn	| o = \le
		| r = \map {\mu^+} B
		| c = $\mu^+ \ge 0$ since $\mu^+$ is a measure
}}
{{eqn	| o = \le
		| r = \map {\mu^+} A
		| c = Measure is Monotone
}}
{{end-eqn}}
So:
:$\map {\mu^+} A$ is an upper bound for $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note that from Intersection is Subset, we have: 
:$A \cap P \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap P \in \Sigma$ so: 
:$\map \mu {A \cap P} \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition: 
:$\map {\mu^+} A = \map \mu {A \cap P}$
so:
:$\map {\mu^+} A \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
:$\map {\mu^+} A$ is the greatest element of $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we therefore have: 
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
We now show that: 
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
{{begin-eqn}}
{{eqn	| l = -\map \mu B
		| r = \map {\mu^-} B - \map {\mu^+} B
}}
{{eqn	| o = \le
		| r = \map {\mu^-} B
		| c = $\mu^- \ge 0$ since $\mu^-$ is a measure
}}
{{eqn	| o = \le
		| r = \map {\mu^-} A
		| c = Measure is Monotone
}}
{{end-eqn}}
So:
:$\map {\mu^-} A$ is an upper bound for $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note from Intersection is Subset, we have: 
:$A \cap N \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap N \in \Sigma$ so: 
:$-\map \mu {A \cap N} \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition: 
:$\map {\mu^-} A = -\map \mu {A \cap N}$
so:
:$\map {\mu^-} A \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
:$\map {\mu^-} A$ is the greatest element of $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we have: 
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
hence the result.
{{qed}}
Category:Uniqueness of Jordan Decomposition
ay0ykc95gm5glo2w1szme4636mjxwl7
Theorem: Radon-Nikodym Theorem

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$ such that: 
:$\nu$ is absolutely continuous with respect to $\mu$.
Then there exists a $\Sigma$-measurable function $g : X \to \hointr 0 \infty$ such that: 
:$\ds \map \nu A = \int_A g \rd \mu$
for each $A \in \Sigma$.
Further, if $g_1 : X \to \hointr 0 \infty$ and $g_2 : X \to \hointr 0 \infty$ are such that: 
:$\ds \map \nu A = \int_A g_1 \rd \mu = \int_A g_2 \rd \mu$
for each $A \in \Sigma$, then: 
:$g_1 = g_2$ $\mu$-almost everywhere.

Proof:
=== Existence ===
We first prove the case of $\mu$ and $\nu$ finite. 
Define $\FF$ to be the set of $\Sigma$-measurable functions $f : X \to \overline \R_{\ge 0}$ with: 
:$\ds \int_A f \rd \mu \le \map \nu A$
for each $A \in \Sigma$.
We show that $\FF$ is non-empty. 
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have: 
:$\ds \int_A 0 \rd \mu = 0$
for each $A \in \Sigma$, giving: 
:$\ds \int_A 0 \rd \mu \le \map \nu A$
for each $A \in \Sigma$. 
So the constant $0$ function is contained in $\FF$. 
For each $f \in F$, we have: 
:$\ds \int f \rd \mu \le \map \nu X$
Since $\nu$ is a finite measure, we have: 
:$\ds \sup \set {\int f \rd \mu : f \in \FF} \le \map \nu X < \infty$ 
We will now show that there exists $g \in \FF$ such that: 
:$\ds \int g \rd \mu = \sup \set {\int f \rd \mu : f \in \FF}$
and that this $g$ has: 
:$\ds \map \nu A = \int_A g \rd \mu$
for each $A \in \Sigma$.
We now construct a increasing sequence of functions $\sequence {g_n}_{n \mathop \in \N}$ in $\FF$ such that: 
:$\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu = \sup \set {\int f \rd \mu : f \in \FF}$
with a view to then apply the monotone convergence theorem. 
We pick $f_n \in \FF$ such that: 
:$\ds \sup \set {\int f \rd \mu : f \in \FF} - \frac 1 n < \int f_n \rd \mu \le \sup \set {\int f \rd \mu : f \in \FF}$
Theorem: Vertical Section of Measurable Set is Measurable

Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces. 
Let $E \in \Sigma_X \otimes \Sigma_Y$ where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$.
Let $x \in X$. 
Then: 
:$E_x \in \Sigma_Y$
where $E_x$ is the $x$-vertical section of $E$.

Proof:
Let:
:$\FF = \set {E \subseteq X \times Y : E_x \in \Sigma_Y}$
We will show that $\FF$ contains each $S_1 \times S_2$ with $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
We will then show that $\FF$ is a $\sigma$-algebra, at which point we will have: 
:$\map \sigma {\set {S_1 \times S_2 : S_1 \in \Sigma_X, \, S_2 \in \Sigma_Y} } \subseteq \FF$
from Sigma-Algebra Contains Generated Sigma-Algebra of Subset.
From the definition of the product $\sigma$-algebra, we will then have: 
:$\Sigma_X \otimes \Sigma_Y \subseteq \FF$
We will then have the demand.
Let $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
From Vertical Section of Cartesian Product, we have: 
:$\ds \paren {S_1 \times S_2}_x = \begin{cases}S_2 & x \in S_1 \\ \O & x \not \in S_1\end{cases}$
From the definition of a $\sigma$-algebra, we have $\O \in \Sigma_Y$, so in either case we have: 
:$\paren {S_1 \times S_2}_x \in \Sigma_Y$
That is: 
:$S_1 \times S_2 \in \FF$
It remains to show that $\FF$ is a $\sigma$-algebra.
Since $S_1 \times S_2 \in \FF$ for $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$.
Since $X \in \Sigma_X$ and $Y \in \Sigma_Y$, we obtain: 
:$X \times Y \in \FF$
We show that $\FF$ is closed under countable union.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\FF$.
We have $\paren {E_n}_x \in \Sigma_Y$ for each $n \in \N$. 
So, since $\Sigma_Y$ is a $\sigma$-algebra, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}_x \in \Sigma_Y$
From Union of Horizontal Sections is Horizontal Section of Union, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}_x = \paren {\bigcup_{n \mathop = 1}^\infty E_n}_x$ 
So we have: 
:$\ds \paren {\bigcup_{n \mathop = 1}^\infty E_n}_x \in \Sigma_Y$
That is: 
:$\ds \bigcup_{n \mathop = 1}^\infty E_n \in \FF$
We finally show that $\FF$ is closed under complementation.
Let $E \in \FF$.
We then have $E_x \in \Sigma_Y$.
Since $\Sigma_Y$ is closed under complementation, we have $Y \setminus E_x \in \Sigma_Y$. 
From Complement of Vertical Section of Set is Vertical Section of Complement, we have: 
:$Y \setminus E_x = \paren {\paren {X \times Y} \setminus E}_x$
so that: 
:$\paren {\paren {X \times Y} \setminus E}_x \in \Sigma_Y$
giving: 
:$\paren {X \times Y} \setminus E \in \FF$
So $\FF$ is a $\sigma$-algebra.
As discussed, we therefore obtain: 
:$\Sigma_X \otimes \Sigma_Y \subseteq \FF$ 
In particular, for any $E \in \Sigma_X \otimes \Sigma_Y$, we have $E \in \FF$.
That is: 
:for any $E \in \Sigma_X \otimes \Sigma_Y$ we have $E_x \in \Sigma_X$
as was the demand.
{{qed}}
Theorem: Measure of Limit of Increasing Sequence of Measurable Sets

Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $E \in \Sigma$.
Let $\sequence {E_n}_{n \mathop \in \N}$ be an increasing sequence of $\Sigma$-measurable sets such that: 
:$E_n \uparrow E$ 
where $E_n \uparrow E$ denotes limit of increasing sequence of sets.
Then: 
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$

Proof:
We define a sequence $\sequence {F_n}_{n \mathop \in \N}$ inductively.
Set $F_1 = E_1$. 
For $n > 1$, define: 
:$F_n = E_n \setminus E_{n - 1}$
From the definition of set difference we have: 
:$E_n \setminus E_{n - 1} = E_n \cap \paren {X \setminus E_{n - 1} }$
Since $\Sigma$ is closed under complementation, we have: 
:$X \setminus E_{n - 1} \in \Sigma$
and so, from Sigma-Algebra Closed under Countable Intersection:
:$F_n \in \Sigma$
for $n > 1$. 
Clearly $F_1 \in \Sigma$ also.
Since $\sequence {E_n}_{n \in \N}$ is increasing, we have that: 
:$E_i \subseteq E_j$ for $i \le j$. 
So:
:$F_n$ is disjoint to $E_1, E_2, \ldots, E_{n - 1}$.
Since $F_i \subseteq E_i$ for each $i$, we have:
:$F_n$ is disjoint to $F_1, F_2, \ldots, F_{n - 1}$.
Picking $n = \max \set {i, j}$ we obtain that: 
:$F_i \cap F_j = \O$ for $i \ne j$.
We now show that: 
:$\ds E_k = \bigcup_{i \mathop = 1}^k F_i$
We have: 
:$\ds x \in \bigcup_{i \mathop = 1}^k F_i$
{{iff}}:
:$x \in F_i$ for some $1 \le i \le k$.
This is equivalent to: 
:$x \in E_i$ for some $1 \le i \le k$.
Since $\sequence {E_n}_{n \in \N}$ increasing sequence, this is equivalent to: 
:$x \in E_k$
showing that: 
:$\ds x \in \bigcup_{i \mathop = 1}^k F_i$ {{iff}} $x \in E_k$
giving:
:$\ds E_k = \bigcup_{i \mathop = 1}^k F_i$
So, we have: 
:$\ds \bigcup_{k \mathop = 1}^\infty E_k = \bigcup_{i \mathop = 1}^\infty F_i$
That is: 
:$\ds E = \bigcup_{i \mathop = 1}^\infty F_i$
from the definition of a limit of an increasing sequence of sets.
We can now compute:
{{begin-eqn}}
{{eqn	| l = \map \mu E
		| r = \map \mu {\bigcup_{i \mathop = 1}^\infty F_i}
}}
{{eqn	| r = \sum_{i \mathop = 1}^\infty \map \mu {F_i}
		| c = since $\mu$ is countably additive
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \sum_{i \mathop = 1}^n \map \mu {F_i}
		| c = {{Defof|Infinite Series}}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \mu {\bigcup_{i \mathop = 1}^n F_i}
		| c = Measure is Finitely Additive Function
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \mu {E_n}
}}
{{end-eqn}}
{{qed}}
Category:Measures
Category:Increasing Sequences of Sets
oqr60koy9vc0l9qwdfn4drt3hsou64k
Theorem: Signed Measure of Limit of Increasing Sequence of Measurable Sets

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $E \in \Sigma$. 
Let $\sequence {E_n}_{n \mathop \in \N}$ be an increasing sequence of $\Sigma$-measurable sets such that: 
:$E_n \uparrow E$ 
where $E_n \uparrow E$ denotes the limit of an increasing sequence of sets.
Then: 
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$

Proof:
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then $\mu^+$ and $\mu^-$ are measures with: 
:$\mu = \mu^+ - \mu^-$
where at least one of $\mu^+$ and $\mu^-$ is finite.
Then we have: 
:$\map \mu E = \map {\mu^+} E - \map {\mu^-} E$
From Measure of Limit of Increasing Sequence of Measurable Sets, we have: 
:$\ds \map {\mu^+} E = \lim_{n \mathop \to \infty} \map {\mu^+} {E_n}$
and:
:$\ds \map {\mu^-} E = \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}$
with at most one of these limits infinite.
So:
{{begin-eqn}}
{{eqn	| l = \map \mu E
		| r = \lim_{n \mathop \to \infty} \map {\mu^+} {E_n} - \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \paren {\map {\mu^+} {E_n} - \map {\mu^-} {E_n} }
		| c = Difference Rule for Real Sequences
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \mu {E_n}
}}
{{end-eqn}}
{{qed}}
Category:Signed Measures
Category:Increasing Sequences of Sets
i5gyc03b8xubuq8ury33jsml6c8nl5f
Theorem: Signed Measure of Limit of Decreasing Sequence of Measurable Sets

Let $\struct {X, \Sigma}$ be a measurable space. 
Let $E \in \Sigma$. 
Let $\sequence {E_n}_{n \mathop \in \N}$ be an decreasing sequence of $\Sigma$-measurable sets such that: 
:$E_n \downarrow E$ 
where $E_n \downarrow E$ denotes limit of decreasing sequence of sets, and:
:there exists $m \in \N$ such that $\map \mu {E_m}$ is finite.
Then: 
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$

Proof:
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then $\mu^+$ and $\mu^-$ are measures with: 
:$\mu = \mu^+ - \mu^-$
where at least one of $\mu^+$ and $\mu^-$ is finite.
Then we have: 
:$\map \mu E = \map {\mu^+} E - \map {\mu^-} E$
and: 
:$\map \mu {E_m} = \map {\mu^+} {E_m} - \map {\mu^-} {E_m}$
Since $\map \mu {E_m}$ is finite, we have that: 
:$\map {\mu^+} {E_m}$ and $\map {\mu^-} {E_m}$ are finite.
From {{Corollary|Measure of Limit of Decreasing Sequence of Measurable Sets}}, we have: 
:$\ds \map {\mu^+} E = \lim_{n \mathop \to \infty} \map {\mu^+} {E_n}$
and:
:$\ds \map {\mu^-} E = \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}$
with at most one of these limits infinite.
So:
{{begin-eqn}}
{{eqn	| l = \map \mu E
		| r = \map {\mu^+} E - \map {\mu^-} E
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map {\mu^+} {E_n} - \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \paren {\map {\mu^+} {E_n} - \map {\mu^-} {E_n} }
		| c = Difference Rule for Real Sequences
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \mu {E_n}
}}
{{end-eqn}}
{{qed}}
Theorem: Union is Increasing Sequence of Sets

Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of sets.
Then: 
:the sequence $\ds \sequence {\bigcup_{k \mathop = 1}^n D_k}_{n \mathop \in \N}$ is increasing.

Proof:
We have: 
:$\ds \bigcup_{k \mathop = 1}^{n + 1} D_k = D_{n + 1} \cup \bigcup_{k \mathop = 1}^n D_k$
From Set is Subset of Union, we have: 
:$\ds \bigcup_{k \mathop = 1}^n D_k \subseteq D_{n + 1} \cup \bigcup_{k \mathop = 1}^n D_k$
so:
:$\ds \bigcup_{k \mathop = 1}^n D_k \subseteq \bigcup_{k \mathop = 1}^{n + 1} D_k$
So:
:$\ds \sequence {\bigcup_{k \mathop = 1}^n D_k}_{n \mathop \in \N}$ is increasing.
{{qed}}
Category:Set Union
Category:Increasing Sequences of Sets
hiu7ewbpzdjt424d4n5ud327az4sw94
Theorem: Vertical Section preserves Subsets

Let $X$ and $Y$ be sets.
Let $A \subseteq B$ be subsets of $X \times Y$. 
Let $x \in X$. 
Then: 
:$A_x \subseteq B_x$
where $A_x$ is the $x$-vertical section of $A$ and $B_x$ is the $x$-vertical section of $B$.

Proof:
Note that if: 
:$y \in A_x$
from the definition of $x$-vertical section, we have: 
:$\tuple {x, y} \in A$
so:
:$\tuple {x, y} \in B$
So, from the definition of $x$-vertical section, we have: 
:$y \in B_x$
So:
:if $y \in A_x$ then $y \in B_x$.
That is: 
:$A_x \subseteq B_x$
{{qed}}
Category:Vertical Section of Sets
0z9c4upcjkiutyxfxtesdgxsyd0wgo4
Theorem: Vertical Section preserves Increasing Sequences of Sets

Let $X$ and $Y$ be sets.
Let $\sequence {A_n}_{n \mathop \in \N}$ be an increasing sequence in $X \times Y$. 
Let $x \in X$. 
Then: 
:$\sequence {\paren {A_n}_x}_{n \mathop \in \N}$ is an increasing sequence.

Proof:
Since $\sequence {A_n}_{n \mathop \in \N}$ is increasing, we have: 
:$A_n \subseteq A_{n + 1}$
for each $n$. 
From Vertical Section preserves Subsets, we have: 
:$\paren {A_n}_x \subseteq \paren {A_{n + 1} }_x$
for each $n$.
So:
:$\sequence {\paren {A_n}_x}_{n \mathop \in \N}$ is an increasing sequence.
{{qed}}
Category:Vertical Section of Sets
Category:Increasing Sequences of Sets
rc0iuu26fmugr7fdrp6br54bh7a75c2
Theorem: Horizontal Section preserves Increasing Sequences of Sets

Let $X$ and $Y$ be sets.
Let $\sequence {A_n}_{n \mathop \in \N}$ be an increasing sequence in $X \times Y$. 
Let $y \in Y$. 
Then: 
:$\sequence {\paren {A_n}^y}_{n \mathop \in \N}$ is an increasing sequence.

Proof:
Since $\sequence {A_n}_{n \mathop \in \N}$ is increasing, we have: 
:$A_n \subseteq A_{n + 1}$
for each $n$. 
From Horizontal Section preserves Subsets, we have: 
:$\paren {A_n}^y \subseteq \paren {A_{n + 1} }^y$
for each $n$.
So:
:$\sequence {\paren {A_n}^y}_{n \mathop \in \N}$ is an increasing sequence.
{{qed}}
Category:Horizontal Section of Sets
Category:Increasing Sequences of Sets
au80cfr4nlapnj4b8fwn0usyun2e1ih
Theorem: Horizontal Section preserves Subsets

Let $X$ and $Y$ be sets.
Let $A \subseteq B$ be subsets of $X \times Y$. 
Let $y \in Y$. 
Then: 
:$A^y \subseteq B^y$
where $A^y$ is the $y$-horizontal section of $A$ and $B^y$ is the $y$-horizontal section of $B$.

Proof:
Note that if: 
:$x \in A^y$
from the definition of $x$-vertical section, we have: 
:$\tuple {x, y} \in A$
so:
:$\tuple {x, y} \in B$
So, from the definition of $x$-vertical section, we have: 
:$x \in B^y$
So:
:if $x \in A^y$ then $x \in B^y$.
That is: 
:$A^y \subseteq B^y$
{{qed}}
Category:Horizontal Section of Sets
bjjm3j79u37akwchevsf28oa3ajo7hr