avewright/proofwiki-math huggingface dataset:
Theorem: Characteristic Function of Preimage

Let $X$ and $Y$ be sets.
Let $f : X \to Y$ be a function.
Let $A \subseteq Y$.
Then: 
:$\chi_{f^{-1} \sqbrk A} = \chi_A \circ f$
where $\chi_{f^{-1} \sqbrk A}$ denotes the characteristic function of $f^{-1} \sqbrk A$.

Proof:
We show that if $x \in f^{-1} \sqbrk A$, then: 
:$\map {\paren {\chi_A \circ f} } x = 1$
and if $x \in X \setminus f^{-1} \sqbrk A$, then: 
:$\map {\paren {\chi_A \circ f} } x = 0$
Let $x \in f^{-1} \sqbrk A$.
Then $\map f x \in A$, and so: 
:$\map {\paren {\chi_A \circ f} } x = \map {\chi_A} {\map f x} = 1$
Now let $x \in X \setminus f^{-1} \sqbrk A$. 
Then $\map f x \in Y \setminus A$, and so: 
:$\map {\paren {\chi_A \circ f} } x = \map {\chi_A} {\map f x} = 0$
{{qed}}
Category:Characteristic Functions
6fsxsny6dyxnv2iliosf3v02idr06mv
Theorem: Cauchy Sequence of Subring iff Cauchy Sequence of Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {S, \norm {\, \cdot \,}_S }$ be a normed division subring of $\struct {R, \norm {\, \cdot \,} }$.
Let $\sequence{x_n}$ be a sequence in $S$.
Then:
:$\sequence{x_n}$ is a Cauchy sequence in $\struct {S, \norm {\, \cdot \,}_S }$
{{iff}}:
:$\sequence{x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$

Proof:
The result follows immediately from:
:{{Defof|Cauchy Sequence (Normed Division Ring)}}
:{{Defof|Normed Division Subring}}
{{qed}}
Category:Normed Division Rings
kkccrrvgnse5m9bs91foaz42j7hkpxo
Theorem: Convergent Sequence of Continuous Real Functions is Integrable Termwise

Let $I = \closedint a b \subseteq \R$ be a closed real interval.
Let $\struct {\map CI, \norm {\, \cdot \,}_\infty}$ be the normed vector space of real-valued functions continuous on $I$ equipped with the supremum norm.
Let $\sequence {f_k}_{k \mathop \in \N_{>0} } \in C \closedint a b$ be a sequence.
Suppose $\ds \sum_{k \mathop = 1}^\infty f_k$ converges to $f$ in $\struct {\map C I, \norm {\, \cdot \,}_\infty}$.
Then:
:$\ds \sum_{k \mathop = 1}^\infty \int_a^b f_k \rd t = \int_a^b \map f t \rd t$

Proof:
{{begin-eqn}}
{{eqn | l = \int_a^b \map f t \rd t
      | r = T f
      | c = {{Defof|Riemann Integral Operator}}
}}
{{eqn | r = T \lim_{n \mathop \to \infty} s_n
      | c = {{Defof|Convergent Series/Normed Vector Space/Definition 2|Convergent Series in Normed Vector Space}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} T s_n
      | c = Riemann Integral Operator is Continuous Linear Transformation, Continuous Mappings preserve Convergent Sequences
}}
{{eqn | r = \lim_{n \mathop \to \infty} \int_a^b \map {s_n} t \rd t
      | c = {{Defof|Riemann Integral Operator}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \int_a^b \map {f_k} t \rd t
      | c = {{Defof|Sequence of Partial Sums}}
}}
{{eqn | r = \sum_{k \mathop = 1}^\infty \int_a^b \map {f_k} t \rd t
}}
{{end-eqn}}
{{qed}}
Theorem: Tukey's Lemma implies Zorn's Lemma

Let Tukey's Lemma be accepted as true.
Then Zorn's Lemma holds.

Proof:
Recall Tukey's Lemma:
{{:Tukey's Lemma/Formulation 2}}{{qed|lemma}}
Recall Zorn's Lemma:
{{:Zorn's Lemma/Formulation 2}}{{qed|lemma}}
So, let us assume Tukey's Lemma.
Let $S$ be a non-empty ordered set, with $T$ as defined.
From Property of being Totally Ordered is of Finite Character:
:$T$ is of finite character.
From Ordering on Singleton is Total Ordering it follows trivially that $T$ is non-empty.
Then by Tukey's Lemma:
:every element of $T$ is a subset of a maximal element of $T$ under the subset relation.
Thus it is seen that Zorn's Lemma likewise holds.
{{qed}}
Theorem: Property of being Totally Ordered is of Finite Character

Let $P$ be the property of sets defined as:
:$\forall x: \map P x$ denotes that $x$ is totally ordered under a relation $\RR$.
Then $P$ is of finite character.
That is:
:$x$ is totally ordered under $\RR$
{{iff}}:
:every finite subset of $x$ is totally ordered under $\RR$.

Proof:
=== Sufficient Condition ===
Let $x$ be totally ordered under $\RR$.
Let $y \subseteq x$.
From Restriction of Total Ordering is Total Ordering it follows that $y$ is also a totally ordered under $\RR$.
This holds in particular if $y$ is a finite set.
Hence every finite subset of $x$ is totally ordered under $\RR$.
{{qed|lemma}}
Theorem: Midpoint-Convex Function is Rational Convex

Let $I$ be a non-empty real interval.
Let $f: I \to \R$ be a real function.
If $f$ is midpoint-convex, then $f$ is rational-convex.

Proof:
It suffices to show that for each $n \in \N$ and for any choice of $n$ elements $x_1, \dots, x_n \in I$:
{{begin-eqn}}
{{eqn | l = \map f {\frac {x_1 + \dots + x_n} n}
      | o = \leq
      | r = \frac {\map f {x_1} + \dots + \map f {x_n} } n
}}
{{end-eqn}}
via Forward-Backward Induction.
Theorem: Condition for Existence of Expectation of Real-Valued Measurable Function composed with Absolutely Continuous Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be an absolutely continuous random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $P_X$ be the probability distribution of $X$.
Let $\map \BB \R$ be the Borel $\sigma$-algebra of $\R$.
Let $h : \R \to \R$ be a $\map \BB \R$-measurable function.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$.
Then $\map h X$ is integrable {{iff}} $h$ is $\lambda$-integrable function.
In this case: 
:$\ds \expect {\map h X} = \int_\R \map h x \rd \map {P_X} x$
where $\expect {\map h X}$ denotes the expectation of $\map h X$.

Proof:
From Composition of Measurable Mappings is Measurable, we have: 
:$\map h X$ is $\Sigma$-measurable.
So:
:$\map h X$ is a real-valued random variable.
From the definition of the probability distribution, we have: 
:$P_X = X_\ast \Pr$
where $X_\ast \Pr$ denotes the pushforward measure of $\Pr$ under $X$.
Since $X$ is an absolutely continuous random variable, it is $\Sigma$-measurable.
So, from Integral with respect to Pushforward Measure: Corollary:
:$\map h X$ is integrable {{iff}} $h$ is $\lambda$-integrable function.
If $\map h X$ is integrable, Integral with respect to Pushforward Measure: Corollary gives: 
:$\ds \int \map h X \rd \Pr = \int_\R \map h x \rd \map {P_X} x$
that is: 
:$\ds \expect {\map h X} = \int_\R \map h x \rd \map {P_X} x$
{{qed}}
Theorem: Expectation of Real-Valued Measurable Function composed with Absolutely Continuous Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be an absolutely continuous random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $P_X$ be the probability distribution of $X$.
Let $\map \BB \R$ be the Borel $\sigma$-algebra of $\R$.
Let $h : \R \to \R$ be a $\map \BB \R$-measurable function.
Let $f_X$ be a probability density function for $X$.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$.
Then $\map h X$ is integrable {{iff}}:
:$\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$
and in this case: 
:$\ds \expect {\map h X} = \int_\R \map h x \map {f_X} x \rd \map \lambda x$

Proof:
From Composition of Measurable Mappings is Measurable:
:$\map h X$ is $\Sigma$-measurable.
So:
:$\map h X$ is a real-valued random variable.
From Characterization of Integrable Functions, we have that:
:$\map h X$ is integrable {{iff}} $\size {\map h X}$ is integrable.
We have:
{{begin-eqn}}
{{eqn	| l = \int \size {\map h X} \rd \Pr
		| r = \int \size h \circ X \rd \Pr
}}
{{eqn	| r = \int_\R \size {\map h x} \rd \map {P_X} x
		| c = Integral with respect to Pushforward Measure, {{Defof|Probability Distribution}}
}}
{{eqn	| r = \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x
		| c = Change of Measures Formula for Integrals, {{Defof|Probability Density Function}}
}}
{{end-eqn}}
Then: 
:$\ds \int \size {\map h X} \rd \Pr$ {{iff}} $\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$
so:
:$\size {\map h X}$ is integrable {{iff}} $\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$.
So:
:$\map h X$ is integrable {{iff}} $\ds \int_\R \size {\map h x} \map {f_X} x \rd \map \lambda x < \infty$.
If $\map h X$ is integrable, then we have:
{{begin-eqn}}
{{eqn	| l = \expect {\map h X}
		| r = \int \map h X \rd \Pr
		| c = {{Defof|Expectation}}
}}
{{eqn	| r = \int_\R \map h x \rd \map {P_X} x
		| c = Integral with respect to Pushforward Measure, {{Defof|Probability Distribution}}
}}
{{eqn	| r = \int_\R \map h x \map {f_X} x \rd \map \lambda x
		| c = Change of Measures Formula for Integrals: Corollary
}}
{{end-eqn}}
{{qed}}
Theorem: Axiom of Choice implies Hausdorff's Maximal Principle

Let the Axiom of Choice be accepted.
Then Hausdorff's Maximal Principle holds.

Proof:
Let $S$ be the set of all chains of $\PP$.
$S \ne \O$ since the empty set is an element of $S$. 
From Subset Relation is Ordering, we have that $\struct {S, \subseteq}$ is partially ordered by inclusion.
Let $C$ be a totally ordered subset of $\struct {S, \subseteq}$.
Then $\bigcup C$ is a chain in $C$ by Set of Chains is Closed under Chain Unions under Subset Relation.
This shows that $S$, ordered by set inclusion, is an inductive ordered subset.
By applying Zorn's Lemma, the result follows.
{{qed}}
Theorem: Axiom of Choice implies Hausdorff's Maximal Principle

Let the Axiom of Choice be accepted.
Then Hausdorff's Maximal Principle holds.

Proof:
Let $\preceq$ be an ordering on the set $\PP$.
Let $X$ be a chain in $\struct {\PP, \preceq}$.
By definition, a maximal chain in $\PP$ that includes $X$ is a chain $Y$ in $\PP$ such that $X \subseteq Y$ and there is no chain $Z$ in $\PP$ with $X \subseteq Z$ and $Y \subsetneq Z$.
Let us define $\CC$ as:
:$\CC = \leftset {Y : Y}$ is a chain in $\PP$ and $\rightset {X \subseteq Y}$
Then:
:a maximal chain in $\PP$ that includes $X$
and:
:a maximal element of $\CC$ under the partial order induced on $\CC$ by inclusion
are one and the same.
It thus suffices to show that $\CC$ contains such a maximal element.
According to Zorn's Lemma, $\CC$ contains a maximal element if each chain in $\CC$ has an upper bound in $\CC$.
Let $W$ be an arbitrary chain in $\CC$.
Let $Z = \bigcup W$.
It will be shown that $Z$ is an upper bound for $W$ in $\CC$.
Let $a, b \in Z$.
Then:
:$\exists A, B \in W: a \in A, b \in B$
Since $W$ is a chain in $\CC$, one of $A$ and $B$ includes the other.
{{WLOG}}, suppose $A \subseteq B$.
Then $a, b \in B$.
Since $B$ is a chain in $\PP$, either $a \preceq b$ or $b \preceq a$.
Therefore $Z$ is a chain in $\PP$.
By definition of $Z$, we have that:
:$\forall A \in W: A \subseteq Z$
Thus $Z$ is an upper bound for $W$ under inclusion.
Finally we have that:
:$\forall A \in W: X \subseteq A$
and so $X \subseteq Z$.
Thus:
:$Z \in \CC$
We have shown that for an arbitrary chain $W$ in $\CC$, $W$ has an upper bound $Z$ in $\CC$.
The result follows.
{{qed}}
Theorem: Axiom of Choice implies Hausdorff's Maximal Principle

Let the Axiom of Choice be accepted.
Then Hausdorff's Maximal Principle holds.

Proof:
Let $\struct {\CC, \subseteq}$ be the set of all chains in $P$ ordered by inclusion. 
By Set of Chains is Closed under Chain Unions under Subset Relation, $\CC$ is a chain complete ordered set.
Now define $f: \CC \to \CC$ as follows:
:If $C$ is a maximal chain then $\map f C = C$.
:Otherwise $f$ chooses arbitrarily, using the axiom of choice, some chain $D$ which strictly contains $C$.
By construction, $\map f C \supseteq C$.
By the above, $\CC$ is chain complete.
Therefore the Bourbaki-Witt Fixed Point Theorem applies and $f$ has a fixed point $\map F M = M$.
But by the above construction, the only fixed points of $f$ are maximal chains.
Therefore $M$ is a maximal chain.
{{qed}}
Theorem: Hausdorff's Maximal Principle implies Kuratowski's Lemma

Let Hausdorff's Maximal Principle be accepted as true.
Then Kuratowski's Lemma holds.

Proof:
Recall Hausdorff's Maximal Principle:
{{:Hausdorff's Maximal Principle/Formulation 2}}{{qed|lemma}}
Recall Kuratowski's Lemma:
{{:Kuratowski's Lemma/Formulation 2}}{{qed|lemma}}
So, let us assume Hausdorff's Maximal Principle.
Let $S$ be a non-empty set of sets which is closed under chain unions.
$\set b$ is trivially a chain.
Hence by Hausdorff's Maximal Principle $\set b$ is a subset of a maximal chain of elements of $S$.
Hence $b$ is an element of a maximal chain $C$ of elements of $S$.
We have {{hypothesis}} that $S$ is closed under chain unions.
Hence $\bigcup C \in S$.
Then $b \subseteq \bigcup C$.
If $\bigcup C$ were a proper subset of some element $x \in S$, $C$ would be a proper subset of the chain $C \cup \set x$.
But $C$ is maximal so that cannot happen.
Hence $\bigcup C$ is maximal .
Thus it is seen that Kuratowski's Lemma holds.
{{qed}}
Theorem: Hausdorff's Maximal Principle implies Kuratowski's Lemma

Let Hausdorff's Maximal Principle be accepted as true.
Then Kuratowski's Lemma holds.

Proof:
We have:
* Hausdorff's Maximal Principle implies Axiom of Choice
* Axiom of Choice implies Kuratowski's Lemma
{{qed}}
Theorem: Zorn's Lemma implies Hausdorff's Maximal Principle

Let Zorn's Lemma be accepted as true.
Then Hausdorff's Maximal Principle holds.

Proof:
Recall Zorn's Lemma:
{{:Zorn's Lemma/Formulation 2}}{{qed|lemma}}
Recall Hausdorff's Maximal Principle:
{{:Hausdorff's Maximal Principle/Formulation 2}}{{qed|lemma}}
It is seen directly that Hausdorff's Maximal Principle is a special case of Zorn's Lemma where the ordering is the subset relation.
{{qed}}
Theorem: Trace Sigma-Algebra of Measurable Set

Let $\struct {X, \Sigma}$ be a measurable space.
Let $A \in \Sigma$. 
Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.
Then: 
:$\Sigma_A = \set {B \in \Sigma : B \subseteq A}$
That is, the elements of $\Sigma_A$ are precisely the $\Sigma$-measurable sets that are subsets of $A$.

Proof:
Let: 
:$\Sigma' = \set {B \in \Sigma : B \subseteq A}$
We first show that:
:$\Sigma_A \subseteq \Sigma'$.
Let $B \in \Sigma_A$, then there exists $S \in \Sigma$ such that: 
:$B = A \cap S$
From Sigma-Algebra Closed under Finite Intersection, we have: 
:$B \in \Sigma$
while, from Intersection is Subset, we have: 
:$A \cap S \subseteq A$
so:
:$B \subseteq A$
giving: 
:$B \in \Sigma'$
So we have:
:$\Sigma_A \subseteq \Sigma'$
by the definition of set inclusion.
Now let $B \in \Sigma'$.
From Intersection with Subset is Subset, we have: 
:$B = B \cap A$ 
while $B \in \Sigma$, so: 
:$B \in \Sigma_A$
from the definition of trace $\sigma$-algebra.
So we have:
:$\Sigma' \subseteq \Sigma_A$
by the definition of set inclusion.
So, we have: 
:$\Sigma_A = \Sigma' = \set {B \in \Sigma : B \subseteq A}$
{{qed}}
Category:Trace Sigma-Algebras
6r51jvyqcw68r8iyw0ia5t2d79i0lsi
Theorem: Maximal Principles are Equivalent

The Maximal Principles are equivalent:
* Kuratowski's Lemma
* Tukey's Lemma
* Zorn's Lemma 
* Hausdorff's Maximal Principle

Proof:
We have:
* Kuratowski's Lemma implies Tukey's Lemma
* Tukey's Lemma implies Zorn's Lemma
* Zorn's Lemma implies Hausdorff's Maximal Principle
* Hausdorff's Maximal Principle implies Kuratowski's Lemma
{{qed}}
Theorem: Axiom of Choice implies Maximal Principles

Let the Axiom of Choice be accepted.
Then the Maximal Principles hold.

Proof:
From Maximal Principles are Equivalent, it is sufficient to demonstrate that any one of them is implied by the Axiom of Choice.
Indeed, we have several such theorems:
* Axiom of Choice implies Kuratowski's Lemma
* Axiom of Choice implies Tukey's Lemma
* Axiom of Choice implies Zorn's Lemma
* Axiom of Choice implies Hausdorff's Maximal Principle
{{qed}}
Theorem: Hausdorff's Maximal Principle implies Axiom of Choice/Lemma

$\Sigma$ is closed under chain unions.

Proof:
Let $f$ and $g$ be choice functions.
Then:
:$f \subseteq g$
{{iff}}:
:$\Dom f \subseteq \Dom g$
:$\forall x \in \Dom f: \map f x = \map g x$
{{ProofWanted|etc.}}
Theorem: Integral of Probability Density Function over the Reals is Equal to One

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an absolutely continuous random variable. 
Let $f_X$ be a probability density function for $X$. 
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$. 
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$. 
Then: 
:$\ds \int_\R f_X \rd \lambda = 1$
where $\ds \int_\R f_X \rd \lambda$ denotes the Lebesgue integral of $f_X$.

Proof:
Let $P_X$ be the probability distribution of $X$. 
From the definition of a probability density function, $f_X$ is a Radon-Nikodym derivative of $P_X$ with respect to $\lambda$.
We then have:
{{begin-eqn}}
{{eqn	| l = 1
		| r = \map {P_X} \R
		| c = {{Defof|Probability Measure}}, Probability Distribution is Probability Measure
}}
{{eqn	| r = \int_\R f_X \rd \lambda
		| c = {{Defof|Radon-Nikodym Derivative}}
}}
{{end-eqn}}
{{qed}}
Category:Probability Density Functions
1xnpc15ik9t122wr0g0vj0cx1rkfv03
Theorem: Equivalence of Definitions of Bounded Variation for Real Function on Closed Bounded Interval

Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a real function. 
{{TFAE|def = Bounded Variation (Closed Bounded Interval)}}
=== Definition 1 ===
{{:Definition:Bounded Variation/Closed Bounded Interval/Definition 1}} 
=== Definition 2 ===
{{:Definition:Bounded Variation/Closed Bounded Interval/Definition 2}}

Proof:
=== Definition 1 implies Definition 2 ===
Suppose that there exists a $M \in \R$ such that:
:$\map {V_f} {P ; \closedint a b} \le M$
for all finite subdivisions $P$.
Let $\SS$ be a finite non-empty subset of $\closedint a b$. 
Define: 
:$\SS^\ast = \SS \cup \set {a, b}$
and write: 
:$\SS^\ast = \set {x_0, x_1, \ldots, x_n}$
with: 
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then $\SS^\ast$ is a finite subdivision of $\closedint a b$ and:
{{begin-eqn}}
{{eqn	| l = \map {V_f^\ast} {\SS; \closedint a b}
		| r = \sum_{i \mathop = 2}^{n - 1} \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| o = \le
		| r = \size {\map f {x_1} - \map f {x_0} } + \sum_{i \mathop = 2}^{n - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } + \size {\map f {x_n} - \map f {x_{n - 1} } }
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn	| r = \map {V_f} {\SS^\ast ; \closedint a b}
}}
{{eqn	| o = \le
		| r = M
}}
{{end-eqn}}
so: 
:$\map {V_f^\ast} {\SS; \closedint a b} \le M$ 
for all finite non-empty subset of $\closedint a b$.
{{qed|lemma}}
Theorem: Principle E is Equivalent to Kuratowski's Lemma

Kuratowski's Lemma is equivalent to Principle E.

Proof:
Recall Kuratowski's Lemma:
{{:Kuratowski's Lemma/Formulation 2}}{{qed|lemma}}
Recall Principle E:
{{:Principle E}}{{qed|lemma}}
Theorem: Zorn's Lemma implies Kuratowski's Lemma

Let Zorn's Lemma be accepted as true.
Then Kuratowski's Lemma holds.

Proof:
Recall Zorn's Lemma:
{{:Zorn's Lemma/Formulation 1}}{{qed|lemma}}
Recall Kuratowski's Lemma:
{{:Kuratowski's Lemma/Formulation 1}}{{qed|lemma}}
Let $\struct {S, \preceq}$ be a non-empty ordered set.
Let $C$ be a chain in $\struct {S, \preceq}$.
Let $T$ be the set of all chains in $\struct {S, \preceq}$ that are supersets of $C$.
Let $\CC$ be a chain in the power set $\powerset T$ of $T$ under the ordering that is the subset relation.
Define $C' = \bigcup \CC$ be the union of $\CC$.
We have that the elements of $T$ are chains on $\struct {S, \preceq}$.
We have that $\CC$ is a subset of $T$.
Hence the elements of $\CC$ are also chains in $\struct {S, \preceq}$.
Thus $\bigcup \CC$ contains elements in $\struct {S, \preceq}$.
Hence:
:$C' \subseteq S$
We have that $C'$ is a chain in $\struct {S, \preceq}$.
Let $x, y \in C'$.
Then $x \in X$ and $y \in Y$ for some $X, Y \in \CC$.
As $\CC$ is a chain in $\powerset T$:
:either $X \subseteq Y$ or $Y \subseteq X$.
So $x$ and $y$ belong to the same chain in $\struct {S, \preceq}$.
Thus either $x \preceq y$ or $y \preceq x$.
Thus $C'$ is a chain on $\struct {S, \preceq}$. 
Now let $x \in C$.
Then:
:$\forall A \in T: x \in A$
Then because $\CC \subseteq T$:
:$\forall A \in \CC: x \in A$
So:
:$x \in \bigcup \CC$
and so:;
:$C \subseteq C'$
Thus:
:$C' \in T$
We now show that $C'$ is an upper bound on $\CC$.
Indeed, consider $x \in D \in \CC$.
This means:
:$x \in \bigcup \CC = C'$
so:
:$D\subseteq C'$
Thus $C'$ is an upper bound on $\CC$.
The chain in $T$ was arbitrary.
Hence every chain in $T$ has an upper bound.
Thus, by Zorn's Lemma, $T$ has a maximal element.
This must be a maximal chain containing $C$.
{{qed}}
Theorem: Bolzano-Weierstrass Theorem/Lemma 2

Let $S$ be a non-empty subset of the real numbers such that its infimum $\map \inf s$ exists.
Let $\map \inf s \notin S$.
Then $\map \inf s$ is a limit point of $S$.

Proof:
The proof follows exactly the same lines as Lemma $1$.
{{qed}}
Theorem: Restriction of Measure to Trace Sigma-Algebra of Measurable Set is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$. 
Let $A \in \Sigma$. 
Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.
Let $\mu \restriction_{\Sigma_A}$ be the restriction of $\mu$ to $\Sigma_A$.
Then $\mu \restriction_{\Sigma_A}$ is a measure on $\struct {A, \Sigma_A}$.

Proof:
We verify the three conditions required of a measure for $\mu \restriction_{\Sigma_A}$.
Note that from Trace Sigma-Algebra of Measurable Set, we have $\Sigma_A \subseteq \Sigma$.
Theorem: Closure under Chain Unions with Choice Function implies Elements with no Immediate Extension

Let $S$ be a set of sets which:
:is closed under chain unions
:has a choice function $C$ for its union $\bigcup S$.
Let $b \in S$.
Then $b$ is the subset of an element of $S$ which has no immediate extension in $S$.

Proof:
Let the hypothesis be assumed.
Let $A = \bigcup S$
Let $x \in S$ be arbitrary.
Let $\map E x$ be the set of all elements $a \in A$ such that $x \cup \set a$ is an immediate extension of $x$.
Note that:
:$\map E x \subseteq \bigcup S$
Suppose $x$ has an immediate extension in $S$.
Then:
:$\map E x$ is non-empty
:because $\map E x \subseteq \bigcup S$, $\map E x$ is in the domain of the choice function $C$.
Let us define a progressing mapping $g: S \to S$ as follows:
:<nowiki>$\forall x \in S: \map g x = \begin {cases}
x & : \text {$x$ has no immediate extension in $S$} \\ \\ 
x \cup \set {\map C {\map E x} } & : \text {$x$ does have an immediate extension in $S$}
\end{cases}$</nowiki>
Thus $g$ is a progressing mapping whose fixed points are those elements of $S$ that have no immediate extensions in $S$.
Hence the criteria hold for {{Corollary|Set which is Superinductive under Progressing Mapping has Fixed Point}} to be applied.
That tells us that an arbitrary $b \in S$ is the subset of a fixed point $x$ of $g$.
Such an element, by definition of $g$, has no immediate extension in $S$.
{{qed}}
Theorem: Swelled Set which is Closed under Chain Unions with Choice Function is Type M

Let $S$ be a set of sets which:
:is closed under chain unions
:has a choice function $C$ for its union $\ds \bigcup S$.
Then:
:$S$ is swelled
{{iff}}:
:$S$ is of type $M$.

Proof:
=== Sufficient Condition ===
Let $S$ be swelled.
Let $b \in S$ be arbitrary.
From Closure under Chain Unions with Choice Function implies Elements with no Immediate Extension:
:$b$ is the subset of an element of $S$ which has no immediate extension in $S$.
Let $x \in S$ have no immediate extension in $S$.
Then from Element of Swelled Set with no Immediate Extension is Maximal, $x$ is a maximal element under the subset relation on $S$.
That is:
:$b$ is the subset of a maximal element of $S$ under the subset relation.
As $b$ is arbitrary:
:every element of $S$ is a subset of a maximal element of $S$ under the subset relation.
Hence, by definition, $S$ is of type $M$.
{{qed|lemma}}
Theorem: Element of Swelled Set with no Immediate Extension is Maximal

Let $S$ be a swelled set of sets.
Let $x \in S$ have no immediate extension.
Then $x$ is a maximal element of $S$ {{WRT}} the subset relation.

Proof:
{{ProofWanted|S&F merely say "We recall that ..."}}
Theorem: Equivalence of Definitions of Concentration of Signed Measure on Measurable Set

Let $\struct {X, \Sigma}$ be a measurable space.
{{TFAE|def = Concentration of Signed Measure on Measurable Set}}

Proof:
From Characterization of Null Sets of Variation of Signed Measure, we have that: 
:$\map {\size \mu} {E^c} = 0$ {{iff}}:
::for each $\Sigma$-measurable set $A \subseteq E^c$, we have $\map \mu A = 0$.
Hence the desired equivalence. 
{{qed}}
Category:Concentration of Signed Measure on Measurable Set
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