avewright/proofwiki-math huggingface dataset:
Theorem: Totally Bounded Metric Space is Second-Countable

Let $M = \struct {A, d}$ be a metric space which is totally bounded.
Then $M$ is second-countable.

Proof:
Let $M = \struct {A, d}$ be totally bounded.
Let $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$
As $M$ is totally bounded, for each $\epsilon$ there exists a finite $\epsilon$-net $\CC$ for $M$.
From Net forms Basis for Metric Space, $\CC$ is a countable basis for $M$.
That is, $M$ is second-countable.
{{qed}}
Theorem: Totally Bounded Metric Space is Second-Countable

Let $M = \struct {A, d}$ be a metric space which is totally bounded.
Then $M$ is second-countable.

Proof:
Follows directly from:
: Totally Bounded Metric Space is Separable
: Separable Metric Space is Second-Countable
{{qed}}
Theorem: Heine-Borel Theorem/Metric Space

A metric space is compact {{iff}} it is both complete and totally bounded.

Proof:
=== Sufficient Condition ===
This follows directly from:
* Compact Metric Space is Complete
* Compact Metric Space is Totally Bounded
{{qed|lemma}}
Theorem: Borel Sigma-Algebra on Euclidean Space by Monotone Class

Let $\sqbrk {\R^n, \tau}$ be the $n$-dimensional Euclidean space.
Then:
:$\map \BB {\R^n, \tau} = \map {\mathfrak m} \tau$
where $\BB$ denotes Borel $\sigma$-algebra, and $\mathfrak m$ denotes generated monotone class.

Proof:
Let $U \in \tau$ be an open set, and define $C$ by:
:$C := X \setminus U$
hence $C$ is a closed set.
Further, define, for all $n \in \N$:
:$C_n := \ds \bigcup_{c \mathop \in C} \map B {c; \frac 1 n}$
where $B$ denotes open ball.
The $C_n$ are open sets, being the union of open balls.
It is clear that $C \subseteq C_n$ for all $n \in \N$.
Conversely, as $U$ is open, for any $u \in U$ (that is, $u \notin C$), find $n \in \N$ such that:
:$\map B {u; \dfrac 1 n} \subseteq U$
as is possible from the definition of open set in a metric space.
Thus, for all $c \in C = X \setminus U$, this means:
:$\map d {u, c} \ge \dfrac 1 n$
whence $u \notin C_n$.
That is, we have established that:
:$c \in C \iff \forall n \in \N: c \in C_n$
Phrased in terms of intersection, this means:
:$C = \ds \bigcap_{n \mathop \in \N} C_n$
Thus, since $C_n \in \tau \subseteq \map {\mathfrak m} \tau$:
:$C \in \map {\mathfrak m} \tau$
Now define:
:$\relcomp X \tau := \set {X \setminus U: U \in \tau}$
Then we have shown:
:$\map {\mathfrak m} {\tau \cup \relcomp X \tau} \subseteq \map {\mathfrak m} \tau$
and the reverse inclusion (and hence equality) follows from Generated Monotone Class Preserves Subset.
Now applying Generated Sigma-Algebra by Generated Monotone Class: Corollary:
:$\map \sigma \tau = \map {\mathfrak m} {\tau \cup \relcomp X \tau}$
Combining these two equalities gives the result, by definition of Borel $\sigma$-algebra.
{{qed}}
Theorem: Measure is Strongly Additive

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is strongly additive, that is:
:$\forall E, F \in \Sigma: \map \mu {E \cap F} + \map \mu {E \cup F} = \map \mu E + \map \mu F$

Proof:
Combine Measure is Finitely Additive Function with Additive Function is Strongly Additive.
{{qed}}
Theorem: Measure is Subadditive

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is subadditive, that is:
:$\forall E, F \in \Sigma: \map \mu {E \cup F} \le \map \mu E + \map \mu F$

Proof:
A measure is an additive function, and, by definition, nowhere negative.
So Additive Nowhere Negative Function is Subadditive applies.
Hence the result directly:
:$\map \mu {E \cup F} \le \map \mu E + \map \mu F$
{{qed}}
Theorem: Representation of Degree One is Irreducible

Let $\struct {G, \cdot}$ be a finite group.
Let $\rho: G \to \GL V$ be a linear representation of $G$ on $V$ of degree $1$. 
Then $\rho$ is an  irreducible linear representation.

Proof:
By the definition of degree of a linear representation, it is known that $\dim V = 1$.
Let $W$ be a proper vector subspace of $V$.
It follows from Dimension of Proper Subspace is Less Than its Superspace that:
:$\dim W < 1$
and hence $\dim W = 0$.
Now from Trivial Vector Space iff Zero Dimension, it follows that: 
:$W = \set {\mathbf 0}$
But this is not a non-trivial proper subspace of $V$.
Thus $V$ has no non-trivial proper vector subspaces.
Hence, by definition, $\rho$ is an irreducible linear representation.
{{qed}}
Category:Representation Theory
s6xt3ooq707g5vl1ipxh0n7xjkfthgr
Theorem: Irreducible Representations of Abelian Group

Let $\struct {G, \cdot}$ be a finite abelian group.
Let $V$ be a non-null vector space over an algebraically closed field $K$.
Let $\rho: G \to \GL V$ be a linear representation.
Then $\rho$ is irreducible {{iff}} $\map {\dim_K} V = 1$, where, $\dim_K$ denotes dimension.

Proof:
=== Sufficient Condition ===
Suppose that $\map {\dim_K} V = 1$.
That $\rho$ is irreducible is shown on Representation of Degree One is Irreducible.
{{qed|lemma}}
Theorem: Schur's Lemma (Representation Theory)

Let $\struct {G, \cdot}$ be a finite group.
Let $V$ and $V'$ be two irreducible $G$-modules.
Let $f: V \to V'$ be a homomorphism of $G$-modules.
Then either:
:$\map f v = 0$ for all $v \in V$
or:
:$f$ is an isomorphism.

Proof:
From Kernel is G-Module, $\map \ker f$ is a $G$-submodule of $V$.
From Image is G-Module, $\Img f$ is a $G$-submodule of $V'$.
By the definition of irreducible:
:$\map \ker f = \set 0$
or:
:$\map \ker f = V$
{{explain|Link to a result which shows this. While it does indeed follow from the definition, it would be useful to have a page directly demonstrating this.}}
If $\map \ker f = V$ then by definition:
:$\map f v = 0$ for all $v \in V$
Let $\map \ker f = \set 0$.
Then from Linear Transformation is Injective iff Kernel Contains Only Zero:
:$f$ is injective.
{{explain|Establish whether the above result (which discusses linear transformations on $R$-modules, not $G$ modules) can be directly applied. If so, amend its wording so as to make this clear.}}
It also follows that:
:$\Img f = V'$
{{Explain|Prove this}}
Thus $f$ is surjective and injective.
Thus by definition $f$ is a bijection and thence an isomorphism.
{{qed}}
{{Namedfor|Issai Schur|cat = Schur}}
Category:Representation Theory
q65wvhjw5wmyg5oiul348l1ehd5xhe4
Theorem: G-Module is Irreducible iff no Non-Trivial Proper Submodules

Let $\struct {G, \circ}$ be a finite group.
Let $\struct {V, \phi}$ be a $G$-module.
Then $V$ is an irreducible $G$-module {{iff}} $V$ has no non-trivial proper $G$-submodules.

Proof:
=== Necessary Condition ===
{{AimForCont}} that $V$ is an irreducible $G$-module, but it has a non-trivial proper $G$-submodule.
By the definition irreducible, its associated representation is irreducible.
Let this representation be denoted:
:$\tilde \phi = \rho: G \to \GL V$
In Correspondence between Linear Group Actions and Linear Representations it is defined as:
:$\map {\map \rho g} v = \map \phi {g, v}$
where $g \in G$ and $v \in V$.
Since $V$ has a proper $G$-submodule, there exists $W$ a non-trivial proper vector subspace such that:
:$\phi \sqbrk {G, W} \subseteq W$ 
and so:
$\rho \sqbrk G W \subseteq W$
Hence $W$ is invariant by every linear operators in $\set {\map \rho g: g \in G}$.
By definition, $\rho$ cannot be irreducible.
Thus we have reached a contradiction, and $V$ has then no non-trivial  proper $G$-submodules.
{{qed|lemma}}
Theorem: Kernel is G-Module

Let $\struct {G, \cdot}$ be a group.
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a homomorphism of $G$-modules.
Then its kernel $\map \ker f$ is a $G$-submodule of $V$.

Proof:
From G-Submodule Test it suffices to prove that $\phi \sqbrk {\struct {G, \map \ker f} } \subseteq \map \ker f$. 
That is, it is to be shown that, if $g \in G$ and $v \in \map \ker f$, then $\map \phi {g, v} \in \map \ker f$.
Assume that $g \in G$ and $v \in \map \ker f$.
{{begin-eqn}}
{{eqn | l = \map f {\map \phi {g, v} }
      | r = \map \mu {g, \map f v}
      | c = $f$ is a $G$-module homomorphism
}}
{{eqn | r = \map \mu {g, 0}
      | c = $v \in \map \ker f$
}}
{{eqn | r = 0
      | c = $\mu$ is a linear action
}}
{{end-eqn}}
Thus $\map \phi {g, v} \in \map \ker f$.
Hence $\map \ker f$ is a $G$-submodule of $V$.
{{qed}}
Category:Representation Theory
2woti1bqyrmd6d01v2dsexhnha80exm
Theorem: Image is G-Module

Let $\struct {G, \cdot}$ be a group.
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a $G$-Module Homomorphism.
Then $\Img f$ is a $G$-Submodule of $V'$.

Proof:
From $G$-Submodule Test it suffices to prove that $\map \mu {G, \Img f} \subseteq \Img f$.
In other words: for any $g \in G$ and $w \in \Img f$, it is to be shown that $\map \mu {g, w} \in \Img f$.
Assume that $g \in G$ and $w \in \Img f$.
Then:
:$\exists v \in V: \map f v = w$
By definition of homomorphism:
:$\map \mu {g, w} = \map \mu {g, \map f v} = \map f {\map \phi {g, v} }$
Hence:
:$\forall g \in G: \forall w \in \Img f: \map \mu {g, w} \in \Img f$
By $G$-Submodule Test, it follows that $\Img f$ is a $G$-Submodule of $V'$.
{{qed}}
Category:Representation Theory
brp65hthozfledgss2zylc6o1v3qd4z
Theorem: Set Intersection Preserves Subsets/Families of Sets

Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
:$\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \bigcap_{\alpha \mathop \in I} A_\alpha
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall \alpha \in I
      | l = x
      | o = \in
      | r = A_\alpha
      | c = {{Defof|Intersection of Family}}
}}
{{eqn | ll= \leadsto
      | q = \forall \alpha \in I
      | l = x
      | o = \in
      | r = B_\alpha
      | c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \bigcap_{\alpha \mathop \in I} B_\alpha
      | c = {{Defof|Intersection of Family}}
}}
{{end-eqn}}
By definition of subset:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
{{qed}}
Theorem: Set Intersection Preserves Subsets

Let $A, B, S, T$ be sets.
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

Proof:
Let $A \subseteq B$ and $S \subseteq T$.
Then:
{{begin-eqn}}     
{{eqn| l = x \in A
     | o = \implies
     | r = x \in B
     | c = {{Defof|Subset}}
}}
{{eqn| l = x \in S
     | o = \implies
     | r = x \in T
     | c = {{Defof|Subset}}
}}
{{end-eqn}}
Now we invoke the Praeclarum Theorema of propositional logic:
:$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$
applying it as:
:$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \land x \in S \implies x \in B \land x \in T}$
The result follows directly from the definition of set intersection:
:$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \cap S \implies x \in B \cap T}$
and from the definition of subset:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
{{qed}}
Category:Set Intersection
Category:Subsets
Category:Set Intersection Preserves Subsets
ji875znv1qa80dwb7ktw1erjpbhjpbv
Theorem: Set Intersection Preserves Subsets

Let $A, B, S, T$ be sets.
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

Proof:
Let $A \subseteq B$, and let $S$ be any set.
From Set Intersection Preserves Subsets:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From Set is Subset of Itself, $S \subseteq S$ for all sets $S$.
Hence the result:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$
{{qed}}
Theorem: Set Intersection Preserves Subsets

Let $A, B, S, T$ be sets.
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

Proof:
Recall the Factor Principles, themselves a corollary of the Praeclarum Theorema:
:$\paren {p \implies q} \vdash \paren {p \land r} \implies \paren {q \land r}$
This is applied as:
{{begin-eqn}}
{{eqn | o = 
      | r = A \subseteq B
      | c = 
}}
{{eqn | o = \leadsto
      | r = \paren {x \in A \implies x \in B}
      | c = {{Defof|Subset}}
}}
{{eqn | o = \leadsto
      | r = \paren {x \in A \land x \in S \implies x \in B \land x \in S}
      | c = Factor Principles
}}
{{eqn | o = \leadsto
      | r = \paren {x \in A \cap S \implies x \in B \cap S}
      | c = {{Defof|Set Intersection}}
}}
{{eqn | o = \leadsto
      | r = A \cap S \subseteq B \cap S
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
{{Qed}}
Theorem: Null Space Contains Zero Vector

Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
$\quad \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots  &  \vdots & \ddots &  \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n} \\
\end {bmatrix}$ 
is a matrix in the matrix space $\map {\MM_\R} {m, n}$.
Then the null space of $\mathbf A$ contains the zero vector:
:$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$
where:
:$\mathbf 0 = \mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf 0
      | r = <nowiki>\begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
 \vdots &  \vdots & \ddots &  \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n} \\
\end {bmatrix} \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}</nowiki>
}}
{{eqn | r = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}
}}
{{eqn | r = \mathbf 0
}}
{{end-eqn}}
The order is correct {{hypothesis}}.
The result follows by the definition of null space.
{{qed}}
Theorem: Null Space Contains Zero Vector

Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
$\quad \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots  &  \vdots & \ddots &  \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n} \\
\end {bmatrix}$ 
is a matrix in the matrix space $\map {\MM_\R} {m, n}$.
Then the null space of $\mathbf A$ contains the zero vector:
:$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$
where:
:$\mathbf 0 = \mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$

Proof:
From Matrix Product as Linear Transformation, $\mathbf {Ax} = \mathbf 0$ defines a linear transformation from $\R^m$ to $\R^n$.
The result then follows from Linear Transformation Maps Zero Vector to Zero Vector.
Theorem: Homogeneous System has Zero Vector as Solution

Every homogeneous system of linear equations has the zero vector as a solution.

Proof:
By the definition of null space, $\mathbf 0$ is a solution {{iff}} the null space contains the zero vector.
The result follows from Null Space Contains Zero Vector.
{{qed}}
Category:Linear Algebra
Category:Null Spaces
mp2ftq9clu23hsbzit9h6vwj15zfend
Theorem: Commutative Linear Transformation is G-Module Homomorphism

Let $\rho: G \to \GL V$ be a representation.
Let $f: V \to V$ be a linear transformation.
Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Then $f: V \to V$ is a $G$-module homomorphism.

Proof:
Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Let $v$ be a vector $v \in V$.
Then:
:$\map {\map \rho g} {\map f v} = \map f {\map {\map \rho g} v}$
Using the properties from Correspondence between Linear Group Actions and Linear Representations:
:there exists a $G$-module $\struct {V, \phi}$  associated with $\rho$ such that:
::$\map \phi {g, v} = \map {\map \rho g} v$
Applying the last formula:
:$\map {\map \rho g} {\map f v} = \map \phi {g, \map f v}$
and:
:$\map f {\map \phi {g, v} } = \map f {\map {\map \rho g} v}$
Thus our assumption is equivalent to:
:$\map f {\map \phi {g, v} } = \map \phi {g, \map f v}$
Hence, by definition of $G$-module homomorphism, $f: V \to V$ is a $G$-module homomorphism.
{{qed}}
Category:Representation Theory
j1ebqkmfcwylhefb3fyh3z59gv85ew2
Theorem: Characterization of Measures

Let $\struct {X, \Sigma}$ be a measurable space.
Denote $\overline \R_{\ge 0}$ for the set of positive extended real numbers.
A mapping $\mu: \Sigma \to \overline \R_{\ge 0}$ is a measure {{iff}}:
{{begin-itemize}}
{{item|(1):|$\map \mu \O {{=}} 0$}}
{{item|(2):|$\mu$ is finitely additive}}
{{item|(3):|For every increasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$, if $E_n \uparrow E$, then:
:$\ds \map \mu E {{=}} \lim_{n \mathop \to \infty} \map \mu {E_n}$}}
{{end-itemize}}
where $E_n \uparrow E$ denotes limit of increasing sequence of sets.
Alternatively, and equivalently, $(3)$ may be replaced by either of:
{{begin-itemize}}
{{item|(3'):|For every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map \mu {E_1}$ is finite, if $E_n \downarrow E$, then:
:$\ds \map \mu E {{=}} \lim_{n \mathop \to \infty} \map \mu {E_n}$}}
{{item|(3' '):|For every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map \mu {E_1}$ is finite, if $E_n \downarrow \O$, then:
:$\ds \lim_{n \mathop \to \infty} \map \mu {E_n} {{=}} 0$}}
{{end-itemize}}
where $E_n \downarrow E$ denotes limit of decreasing sequence of sets.

Proof:
=== Necessary Condition ===
To show is that a measure $\mu$ has the properties $(1)$, $(2)$, $(3)$, $(3')$ and $(3' ')$.
Property $(1)$ is also part of the definition of measure, and hence is immediate.
Property $(2)$ is precisely the statement of Measure is Finitely Additive Function.
Property $(3)$ is precisely the statement of Measure of Limit of Increasing Sequence of Measurable Sets.
For $(3' ')$, note that it is a special case of $(3')$.
For property $(3')$, let $\sequence {E_n}_{n \mathop \in \N} \downarrow E$ be a decreasing sequence in $\Sigma$.
Suppose that $\map \mu {E_1} < +\infty$.
By Measure is Monotone, this implies:
:$\forall n \in \N: \map \mu {E_n} < +\infty$
and also:
:$\map \mu E < +\infty$
Now define:
:$\forall n \in \N: F_n := E_1 \setminus E_n$
Then:
:$F_n \uparrow E_1 \setminus E$
{{explain|prove this claim}}
Hence, property $(3)$ can be applied as follows:
{{begin-eqn}}
{{eqn | l = \map \mu {E_1} - \map \mu E
      | r = \map \mu {E_1 \setminus E}
      | c = Measure of Set Difference with Subset
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map \mu {E_1 \setminus E_n}
      | c = by property $(3)$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\map \mu {E_1} - \map \mu {E_n} }
      | c = Measure of Set Difference with Subset
}}
{{eqn | r = \map \mu {E_1} - \lim_{n \mathop \to \infty} \map \mu {E_n}
}}
{{end-eqn}}
Here, all expressions involving subtraction are well-defined as $\mu$ takes finite values.
It follows that:
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$
as required.
{{qed|lemma}}
Theorem: Measure is Countably Subadditive

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is a countably subadditive function.

Proof:
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets in $\Sigma$.
It is required to show that:
:$\ds \map \mu {\bigcup_{n \mathop \in \N} E_n} \le \sum_{n \mathop \in \N} \map \mu {E_n}$
Now define the sequence $\sequence {F_n}_{n \mathop \in \N}$ in $\Sigma$ by:
:$F_n := \ds \bigcup_{k \mathop = 0}^n E_n$
By Subset of Union, it follows that, for all $n \in \N$, $F_n \subseteq F_{n + 1}$.
Hence, $\sequence {F_n}_{n \mathop \in \N}$ is increasing.
It is immediate that $F_n \uparrow \ds \bigcup_{n \mathop \in \N} E_n$, where $\uparrow$ signifies the limit of an increasing sequence of sets.
Now reason as follows:
{{begin-eqn}}
{{eqn | l = \map \mu {\bigcup_{n \mathop \in \N} E_n}
      | r = \lim_{n \mathop \to \infty} \map \mu {F_n}
      | c = Measure of Limit of Increasing Sequence of Measurable Sets
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map \mu {E_0 \cup \cdots \cup E_n}
      | c = Definition of $F_n$
}}
{{eqn | o = \le
      | r = \lim_{n \mathop \to \infty} \sum_{k \mathop = 0}^n \map \mu {E_k}
      | c = Measure is Subadditive: Corollary
}}
{{eqn|r = \sum_{k \mathop \in \N} \map \mu {E_k}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Dirac Measure is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Then $\delta_x$ is a measure.

Proof:
Let us verify in turn that $\delta_x$ satisfies the axioms for a measure.
Theorem: Dirac Measure is Probability Measure

Let $\struct {X, \AA}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Then $\delta_x$ is a probability measure.

Proof:
By Dirac Measure is Measure, $\delta_x$ is a measure.
Also, $\map {\delta_x} X = 1$ because $x \in X$.
Hence $\delta_x$ is a probability measure.
{{qed}}
Category:Probability Measures
gtm3ifhzjlvmmqvlih66r361v1se0cu
Theorem: Co-Countable Measure is Measure

Let $X$ be an uncountable set.
Let $\Sigma$ be the $\sigma$-algebra of countable sets on $X$.
Then the co-countable measure $\mu$ on $X$ is a measure.

Proof:
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.
Theorem: Co-Countable Measure is Probability Measure

Let $X$ be an uncountable set.
Let $\AA$ be the $\sigma$-algebra of countable sets on $X$.
Then the co-countable measure $\mu$ on $X$ is a probability measure.

Proof:
By Co-Countable Measure is Measure, $\mu$ is a measure.
By Relative Complement with Self is Empty Set, have $\relcomp X X = \O$.
As $\O$ is countable, it follows that $X$ is co-countable.
Hence $\map \mu X = 1$, and so $\mu$ is a probability measure.
{{qed}}
Category:Probability Measures
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Theorem: Counting Measure is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Then the counting measure $\size {\,\cdot\,}$ on $\struct {X, \Sigma}$ is a measure.

Proof:
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\size {\,\cdot\,}$.
Theorem: Trivial Vector Space iff Zero Dimension

Let $V$ be a vector space.
Then $V = \set \bszero$ {{iff}} $\map \dim V = 0$, where $\dim$ signifies dimension of vector space.

Proof:
=== Necessary Condition ===
Suppose $V = \set \bszero$.
We have that $V$ has no $\mathbf v \in V, \mathbf v \ne \bszero$.
Thus there exists no $\set {\mathbf v} \subseteq V$ such that $\mathbf v \ne \bszero$.
Such a $\set {\mathbf v}$ would be a linearly independent set.
Hence $\O$ is the only possible basis for $V$.
Hence $\map \dim V = \card \O = 0$.
{{qed|lemma}}
Theorem: G-Submodule Test

Let $\struct {V, \phi}$ be a $G$-module over a field $k$.
Let $W$ be a vector subspace of $V$.
Let $\phi_W: G \times W \to V$ denote the restriction of $\phi$ to $G \times W$.
Then:
:$\struct {W, \phi_W}$ is a $G$-submodule of $V$
{{iff}}:
:$\map \phi {G, W} \subseteq W$

Proof:
=== Necessary Condition ===
Let $W$ be a $G$-submodule of $V$.
Hence by definition $\phi_W: G \times W \to W$ is a linear action on $W$.
Also by definition:
:$\map {\phi_W} {G, W} = \map \phi {G, W} \subseteq W$
{{qed|lemma}}
Theorem: Composite of Continuous Mappings is Continuous/Point

Let $T_1, T_2, T_3$ be topological spaces.
Let the mapping $f : T_1 \to T_2$ be continuous at $x$.
Let the mapping $g : T_2 \to T_3$ be continuous at $\map f x$.
Then the composite mapping $g \circ f : T_1 \to T_3$ is continuous at $x$.

Proof:
Let $N$ be any neighborhood of $\map {\paren {g \circ f} } x$.
By the definition of continuity at a point:
:there exists a neighborhood $L$ of $\map f x$ such that $g \sqbrk L \subseteq N$
and
:there exists a neighborhood $M$ of $x$ such that $f \sqbrk M \subseteq L$.
Thus $\paren {g \circ f} \sqbrk M \subseteq g \sqbrk L \subseteq N$, as desired.
{{qed}}