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Added facts about vector measures.

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Bokuan Li
2026-03-15 19:30:28 -04:00
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"prefix": "cor",
"body": [
"\\begin{corollary}",
"\\label{corollary:$1}",
" $2",
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"Proof Block": {
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\section{Bornologic Spaces}
\label{section:bornologic}
\begin{definition}[Bornologic Space]
\label{definition:bornologic-space}
Let $E$ be a locally convex space, then the following are equivalent:
\begin{enumerate}
\item For any $U \subset E$ convex and balanced, if $U$ absorbs every bounded set of $E$, then $U \in \cn_E(0)$.
\item For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous.
\end{enumerate}
If the above holds, then $E$ is a \textbf{bornologic space}.
\end{definition}
\begin{proof}
(1) $\Rightarrow$ (2): Let $B \subset E$ be bounded, then there exists $R > 0$ such that $\rho(B) \subset [0, R)$. In which case,
\[
B \subset \bracs{\rho < R} = R\bracs{\rho < 1}
\]
By assumption, $\bracs{\rho < 1} \in \cn_E(0)$, so $\rho$ is continuous by \autoref{lemma:continuous-seminorm}.
(2) $\Rightarrow$ (1): Let $\rho$ be the \hyperref[gauge]{definition:gauge} of $U$, then for any $B \subset E$ bounded, there exists $R > 0$ such that $B \subset RU$. In which case, $\rho(B) \subset [0, R]$.
\end{proof}
\begin{proposition}
\label{proposition:metrisable-bornologic}
Let $E$ be a metrisable locally convex space, then $E$ is bornologic.
\end{proposition}
\begin{proof}
Let $U \subset E$ be convex and balanced such that $U$ absorbs every bounded set of $E$. Let $\seq{U_n} \subset \cn^o(0)$ be a decreasing countable fundamental system of neighbourhoods at $0$. If $U_n \setminus nA \ne \emptyset$ for all $n \in \natp$, then there exists $\seq{x_n}$ such that $x_n \in U_n \setminus nA$ for all $n \in \natp$. In which case, $x_n \to 0$ as $n \to \infty$, so $\seq{x_n}$ is bounded. By assumption, there exists $n \in\natp$ such that $nA \supset \seq{x_n}$, which contradicts the fact that $\seq{x_n} \cap A = \emptyset$.
\end{proof}
\begin{proposition}
\label{proposition:bornologic-bounded}
Let $E$ be a bornologic space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
\begin{enumerate}
\item $T$ is continuous.
\item $T$ is bounded.
\end{enumerate}
\end{proposition}
\begin{proof}
(1) $\Rightarrow$ (2): By \autoref{proposition:continuous-bounded}.
(2) $\Rightarrow$ (1): Let $\rho: F \to [0, \infty)$ be a continuous seminorm, then $\rho \circ T$ is a seminorm on $E$ that is bounded on bounded sets. Since $E$ is bornologic, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
\end{proof}
\begin{proposition}
\label{proposition:bornologic-continuous-complete}
Let $E$ be a bornologic space and $F$ be a complete Hausdorff locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
\end{proposition}
\begin{proof}
By \autoref{proposition:bornologic-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
\end{proof}

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@@ -99,10 +99,16 @@
\item $[\cdot]$ is uniformly continuous.
\item $[\cdot]$ is continuous.
\item $[\cdot]$ is continuous at $0$.
\item $\bracs{x \in E| [x] < 1} \in \cn_E(0)$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \autoref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
$(4) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
\[
x - y \in \bracs{x \in E|[x] < r} = r\bracs{x \in E|[x] < 1} \in \cn_E(0)
\]
then $[x - y] < r$.
\end{proof}

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@@ -4,6 +4,7 @@
\input{./convex.tex}
\input{./continuous.tex}
\input{./bornologic.tex}
\input{./quotient.tex}
\input{./projective.tex}
\input{./inductive.tex}

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Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges absolutely} if $\sum_{n \in \natp}\norm{x_n}_E < \infty$.
\end{definition}
\begin{lemma}
\label{lemma:banach-absolute}
Let $E$ be a normed space, then the following are equivalent:
\begin{enumerate}
\item $E$ is a Banach space.
\item For any absolutely convergent series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$, there exists $x \in E$ such that $x = \sum_{n = 1}^\infty x_n$.
\end{enumerate}
\end{lemma}
\begin{proof}
(2) $\Rightarrow$ (1): Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence ${n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_k}} \le 2^{-k}$ for all $k \in \natp$.
For each $k \in \natp$, let $y_k = x_{n_{k+1}} - x_{n_k}$, then $\sum_{n = 1}^\infty y_k$ is absolutely convergent, and there exists $y \in E$ such that $y = \sum_{n = 1}^\infty y_n$. Let $x = x_{n_1} + y$, then $x_{n_k} \to x$ as $k \to \infty$. Since $\seq{x_n}$ is Cauchy, $x_n \to x$ as well by \autoref{lemma:cauchy-subnet}.
\end{proof}
\begin{definition}[Unconditional Convergence]
\label{definition:unconditional-convergence}
Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges unconditionally} if for any bijection $\sigma: \natp \to \natp$,

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@@ -31,6 +31,7 @@
is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
\end{proof}
\begin{theorem}[Successive Approximation]
\label{theorem:successive-approximation}
Let $E, F$ be normed spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:

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@@ -39,9 +39,53 @@
for all $x \in S$.
\end{proof}
\begin{definition}[Space of Bounded Functions]
\label{definition:bounded-function-space}
Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
\begin{enumerate}
\item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
\item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication.
Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
(2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
\[
f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
\]
so $f \in B(T; E)$.
If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
\end{proof}
\begin{definition}[Space of Bounded Continuous Functions]
\label{definition:bounded-continuous-function-space}
Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
\begin{enumerate}
\item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
\item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
(2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace.
If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete.
\end{proof}
\begin{definition}[Space of Bounded Linear Maps]
\label{definition:bounded-linear-map-space}
Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $B_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\mathfrak{S}}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$.
\end{definition}
\begin{proposition}
@@ -127,3 +171,35 @@
The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
\end{definition}
\begin{proposition}
\label{proposition:operator-space-completeness}
Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
\begin{enumerate}
\item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
\item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): For each $x, y \in E$ and $\lambda \in K$, the mappings
\[
\phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
\]
and
\[
\psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - Tx
\]
are continuous with respect to the product topology. Since
\[
\hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
\]
and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$.
(2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
\end{proof}

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\input{./sets/index.tex}
\input{./measure/index.tex}
\input{./vector/index.tex}
\input{./measurable-maps/index.tex}
\input{./lebesgue-integral/index.tex}
\input{./bochner-integral/index.tex}

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\section{Absolutely Continuous}
\label{section:absolutely-continuous-measure}
\begin{definition}[Absolutely Continuous]
\label{definition:absolutely-continuous}
Let $(X, \cm)$ be a measurable space and $\mu, \nu$ be signed/vector measures on $X$, then $\nu$ is \textbf{absolutely continuous} with respect to $\mu$, denoted $\nu \ll \mu$, if every $\mu$-null set is $\nu$-null.
\end{definition}

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\section{Signed and Complex Measures}
\label{section:signed-complex-measure}
\begin{definition}[Signed Measure]
\label{definition:signed-measure}
Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$, then $\mu$ is a \textbf{signed measure} if
\begin{enumerate}
\item[(M1)] $\mu(\emptyset) = 0$.
\item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$ where the sum converges absolutely.
\end{enumerate}
By \hyperref[Riemann's Rearrangement Theorem]{theorem:riemann-rearrangement}, (M2) implies that $\mu$ can only take at most one value in $\bracs{-\infty, \infty}$.
\end{definition}
\begin{definition}[Complex Measure]
\label{definition:complex-measure}
Let $(X, \cm)$ be a measurable space and $\mu: \cm \to \complex$, then $\mu$ is a \textbf{complex measure} if
\begin{enumerate}
\item[(M1)] $\mu(\emptyset) = 0$.
\item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$ where the sum converges absolutely.
\end{enumerate}
\end{definition}
\begin{definition}[Positive/Negative/Null Sets]
\label{definition:positive-negative-sets}
Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, and $A \in \cm$, then $A$ is...
\begin{enumerate}
\item \textbf{positive} if $\mu(B) \ge 0$ for all $B \in \cm$ with $B \subset A$.
\item \textbf{negative} if $\mu(B) \le 0$ for all $B \in \cm$ with $B \subset A$.
\item \textbf{null} if $\mu(B) = 0$ for all $B \in \cm$ with $B \subset A$.
\end{enumerate}
\end{definition}
\begin{proposition}
\label{proposition:signed-measure-properties}
Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then:
\begin{enumerate}
\item For any $\seq{E_n} \subset \cm$ and $E \in \cm$ with $E_n \upto E$, $\mu\paren{\bigcup_{n \in \natp}E_n} = \limv{n}\mu(E_n)$.
\item For any $\seq{E_n} \subset \cm$ and $E \in \cm$ with $E_n \downto E$ and $|\mu(E_1)| < \infty$, $\mu\paren{\bigcap_{n \in \natp}E_n} = \limv{n}\mu(E_n)$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): For each $N \in \natp$, let $F_N = E_N \setminus \bigcup_{n = 1}^{N-1}E_n$, then $E_N = \bigsqcup_{n = 1}^N F_n$. By (M2),
\[
\mu(E) = \limv{N}\sum_{n = 1}^N \mu(F_N) = \limv{N}\mu(E_N)
\]
(2): By (1),
\[
\mu(E) = \mu(E_1) - \mu(E_1 \setminus E) = \limv{n}[\mu(E_1) - \mu(E_1 \setminus E_n)] = \limv{n}\mu(E_n)
\]
\end{proof}
\begin{lemma}
\label{lemma:positive-sets-properties}
Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then
\begin{enumerate}
\item For any $A \in \cm$ positive and $B \in \cm$ with $B \subset A$, $B$ is positive.
\item For any $\seq{A_n} \subset \cm$ positive, $\bigcup_{n \in \natp}A_n$ is positive.
\end{enumerate}
\end{lemma}
\begin{proof}
(1): For any $C \in \cm$ with $C \subset B$, $C \subset A$, so $\mu(C) \ge 0$. Hence $B$ is positive.
(2): For each $N \in \natp$, let $B_N = A_N \setminus \sum_{n = 1}^{N-1} A_n$. By (1), $B_N$ is positive with $\bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}B_n$. For any $C \subset \bigcup_{n \in \natp}B_n$,
\[
\mu(C) = \sum_{n \in \natp}\mu(C \cap B_n) \ge 0
\]
\end{proof}
\begin{theorem}[Hahn Decomposition, {{\cite[Theorem 3.3]{Folland}}}]
\label{theorem:hahn-decomposition}
Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then there exists $P, N \in \cm$ such that:
\begin{enumerate}
\item $P$ is positive.
\item $N$ is negative.
\item $X = P \sqcup N$.
\item[(U)] For any $P', N' \in \cm$ satisfying (1)-(3), $P \Delta P'$ and $N \Delta N'$ are null.
\end{enumerate}
The disjoint union $X = P \sqcup N$ is the \textbf{Hahn decomposition} of $\mu$.
\end{theorem}
\begin{proof}
By flipping the sign of $\mu$, assume without loss of generality that $\mu < \infty$.
(1): Let $M = \sup\bracs{\mu(P)|P \in \cm \text{ positive}}$, then there exists positive sets $\seq{P_n} \subset \cm$ such that $\mu(P_n) \upto M$. Let $P = \bigcup_{n \in \natp}P_n$, then $P$ is positive by \autoref{lemma:positive-sets-properties}, and $\sup_{n \in \natp}\mu(P_n) \le \mu(P) \le M$.
(3): Let $N = X \setminus P$, then $P$ is positive and $X = P \sqcup N$.
(2): Suppose for contradiction that $N$ is not negative, then
\[
m(N) = \sup\bracs{\mu(A)|A \in \cm, A \subset N} > 0
\]
Let $A_1 \in \cm$ such that $A_1 \subset N$ and $\mu(A_1) > 0$. Since $A_{1} \cap P = \emptyset$ and $\mu(P) = M$, $A_{1}$ cannot be positive, so $M_1 = m(A_1) - \mu(A_1) > 0$. Let $n \in \natp$ and suppose inductively that $\bracs{A_n}_1^n$ has been constructed such that
\begin{enumerate}
\item[(i)] For each $1 \le k \le n$, $M_k = m(A_k) - \mu(A_k) > 0$.
\item[(ii)] For each $1 \le k \le n - 1$, $A_{k+1} \subset A_k$ with $\mu(A_{k+1}) - \mu(A_k) > M_{k}/2$.
\end{enumerate}
Let $A_n \in \cm$ with $A_{n+1} \subset A_n$ such that $\mu(A_{n+1}) - \mu(A_n) > M_n/2$. Since $A_{n+1} \cap P = \emptyset$ and $\mu(P) = M$, $A_{n+1}$ cannot be positive, so
\[
M_{n+1} = m(A_{n+1}) - \mu(A_{n+1}) > 0
\]
Let $A = \bigcap_{n \in \natp}A_n$, then since $0 < \mu(A_2) < \infty$, $\mu(A) = \limv{n}\mu(A_n)$ by \autoref{proposition:signed-measure-properties}, so
\[
\frac{1}{2}\sum_{n = 1}^\infty M_n \le \sum_{n = 1}^{\infty}\mu(A_{n+1}) - \mu(A_n) = \mu(A) - \mu(A_1) < \infty
\]
and $M_n \to 0$ as $n \to \infty$.
Since $A \cap P = \emptyset$ and $\mu(P) = M$, $A$ cannot be positive. Thus there exists $B \in \cm$ with $B \subset A$ with $\mu(B) - \mu(A) > 0$. As $M_n \to 0$ as $n \to \infty$, there exists $n \in \natp$ such that
\[
\mu(B) - \mu(A_n) > \mu(B) - \mu(A) > M_n = m(A_n) - \mu(A_n)
\]
As $\mu(B) - \mu(A_n) > \mu(B) - \mu(A)$, this contradicts the definition of $m(A_n)$.
Therefore $N$ must be negative, and $X = P \sqcup N$ is the desired decomposition.
(U): Since $X = P \sqcup N = P' \sqcup N'$,
\[
P \Delta P' = P \setminus P' \sqcup P' \setminus P = P \cap N' \sqcup P' \cap N
\]
is a union of two null sets, which is null. Likewise,
\[
N \Delta N' = N \cap P' \sqcup N' \cap P
\]
is also a null set.
\end{proof}
\begin{corollary}
\label{corollary:measure-bounded}
Let $(X, \cm)$ be a measurable space, then
\begin{enumerate}
\item For any signed measure $\mu: \cm \to (-\infty, \infty)$, $\mu$ is bounded.
\item For any complex measure $\mu: \cm \to \complex$, $\mu$ is bounded.
\end{enumerate}
\end{corollary}
\begin{proof}
(1): Let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$, then for any $E \in \cm$, $\mu(E) \in [\mu(N), \mu(P)]$.
(2): Since $\text{Re}(\mu)$ and $\text{Im}(\mu)$ are both signed measures taking values in $(-\infty, \infty)$, they are bounded by (1). Thus $\mu$ is also bounded.
\end{proof}
\begin{theorem}[Jordan Decomposition]
\label{theorem:jordan-decomposition}
Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$ be a signed measure on $X$, then there exists positive measures $\mu^+, \mu^-: \cm \to [0, \infty]$ such that
\begin{enumerate}
\item $\mu^+ \perp \mu^-$.
\item $\mu = \mu^+ - \mu^-$.
\item[(U)] For any other pair $(\nu^+, \nu^-)$ satisfying (1) and (2), $\mu^+ = \nu^+$ and $\mu^- = \nu^-$.
\end{enumerate}
\end{theorem}
\begin{proof}
(1), (2): Let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$. For each $E \in \cm$, let
\[
\mu^+(E) = \mu(E \cap P) \quad \mu^-(E) = -\mu(E \cap N)
\]
then $(\mu^+, \mu^-)$ satisfies (1) and (2).
(U): Let $X = P' \sqcup N'$ such that $P'$ is $\nu^-$-null and $N'$ is $\nu^+$-null, then $X = P' \sqcup N'$ is a Hahn decomposition of $\mu$. Since $P' \Delta P = 0$ and $N' \Delta N = 0$, for any $E \in \cm$,
\[
\mu^+(E) = \mu(E \cap P) = \mu(E \cap P') = \nu^+(E)
\]
Likewise,
\[
\mu^-(E) = -\mu(E \cap N) = -\mu(E \cap N') = \nu^-(E)
\]
\end{proof}

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\chapter{Signed, Complex, and Vector Measures}
\label{chap:vector-measures}
\input{./complex.tex}
\input{./vector.tex}
\input{./ac.tex}
\input{./ms.tex}
\input{./variation.tex}

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\section{Mutually Singular}
\label{section:mutually-singular}
\begin{definition}[Mutually Singular]
\label{definition:mutually-singular}
Let $(X, \cm)$ be a measurable space and $\mu, \nu$ be signed/vector measures, then $\mu$ and $\nu$ are \textbf{mutually singular}, denoted $\mu \perp \nu$, if there exists $U, V \in \cm$ such that:
\begin{enumerate}
\item $U$ is $\nu$-null.
\item $V$ is $\mu$-null.
\item $X = U \sqcup V$.
\end{enumerate}
\end{definition}

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\section{Total Variation}
\label{section:total-variation-measure}
\begin{definition}[Total Variation of Vector Measures]
\label{definition:total-variation-vector}
Let $(X, \cm)$ be a measure space, $\mu$ be a vector measure taking values in a normed space $E$, and
\[
|\mu|(A) = \sup\bracs{\sum_{j = 1}^n \norm{\mu(A_j)}_E \bigg | \seqf{A_j} \subset \cm, A = \bigsqcup_{j = 1}^n A_j}
\]
then:
\begin{enumerate}
\item For any $A \in \cm$,
\[
|\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}\norm{\mu(A_i)} \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}
\]
\item $|\mu|$ is a measure on $(X, \cm)$.
\end{enumerate}
and the measure $|\mu|$ is the \textbf{total variation} of $\mu$.
\end{definition}
\begin{proof}
(1): Since all finite partitions are partitions, $|\mu| \le \nu$. On the other hand, let $\seqi{A} \subset \cm$ and $A = \bigsqcup_{i \in I}A_i$, then for any $J \subset I$ finite,
\[
|\mu|(A) \ge \norm{\mu\paren{A \setminus \bigsqcup_{j \in J}A_j}}_E + \sum_{j \in J}\norm{\mu(A_j)}_E \ge \sum_{j \in J}\norm{\mu(A_j)}_E
\]
so
\[
|\mu|(A) \ge \sum_{i \in I}\norm{\mu(A_i)}_E
\]
As this holds for all $\seqi{A}$ such that $A = \bigsqcup_{i \in I}A_i$, $|\mu|(A) \ge \nu(A)$.
(2): Let $A \in \cm$ and $\seq{A_n}$ such that $A = \bigsqcup_{n \in \natp}A_n$. For each $n \in \natp$, let $\seq{B_{n, k}} \subset \cm$ such that $A_n = \bigsqcup_{k \in \natp}B_{n, k}$, then for any $N \in \natp$,
\[
|\mu|(A) \ge \sum_{n, k \in \natp}\norm{\mu(B_{n, k})}_E \ge \sum_{n = 1}^N \sum_{k \in \natp}\norm{\mu(B_{n, k})}_E
\]
As the above inequality is independent of the choice of $\bracsn{B_{n, k}|n, k \in \natp}$, $|\mu|(A) \ge \sum_{n = 1}^N |\mu|(A_n)$ for all $N \in \natp$. Thus $|\mu|(A) \ge \sum_{n \in \natp}|\mu|(A_n)$.
On the other hand, let $\bracs{B_k}_1^K \subset \cm$ such that $A = \bigsqcup_{k = 1}^K B_k$, then for each $N \in \natp$,
\[
\sum_{n = 1}^N |\mu|(A_n) \ge \sum_{k = 1}^K \sum_{n = 1}^N \norm{\mu(A_n \cap B_k)}_E
\]
so
\[
\sum_{n \in \natp}|\mu|(A_n) \ge \sum_{k = 1}^K \sum_{n \in \natp} \norm{\mu(A_n \cap B_k)}_E \ge \sum_{k = 1}^K \normn{\mu(B_k)}_E
\]
As the above holds for all choices of $\bracs{B_k}_1^K$, $\sum_{n \in \natp}|\mu|(A_n) \ge |\mu|(A)$.
\end{proof}
\begin{definition}[Total Variation of Signed Measures]
\label{definition:total-variation-signed}
Let $(X, \cm)$ be a measure space, $\mu$ be a signed measure on $(X, \cm)$, and
\[
|\mu|(A) = \sup\bracs{\sum_{j = 1}^n |\mu(A_j)| \bigg | \seqf{A_j} \subset \cm, A = \bigsqcup_{j = 1}^n A_j}
\]
then:
\begin{enumerate}
\item For any $A \in \cm$,
\[
|\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}|\mu(A_i)| \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}
\]
\item $|\mu|$ is a measure on $(X, \cm)$.
\item Let $\mu = \mu^+ - \mu^-$ be the \hyperref[Jordan decomposition]{theorem:jordan-decomposition} of $\mu$, then $|\mu| = \mu^+ + \mu^-$.
\end{enumerate}
and the measure $|\mu|$ is the \textbf{total variation} of $\mu$.
\end{definition}
\begin{proof}
(1) and (2): Same as \autoref{definition:total-variation}.
(3): For any $A \in \cm$, $|\mu(A)| \le \mu^+(A) + \mu^-(A)$, so $(\mu^+ + \mu^-) \ge |\mu|$. On the other hand, let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$, then
\[
|\mu|(A) \ge \mu(A \cap P) + \mu(A \cap N) = \mu^+(A) + \mu^-(A)
\]
so $(\mu^+ + \mu^-) \le |\mu|$.
\end{proof}
\begin{definition}[Finite Measure]
\label{definition:vector-measure-finite}
Let $(X, \cm)$ be a measurable space and $\mu$ be a signed/vector measure, then $\mu$ is \textbf{finite} if $|\mu|$ is finite.
\end{definition}
\begin{definition}[Space of Finite Measures]
\label{definition:vector-measure-finite-space}
Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $M(X, \cm; E)$ be the set of all finite $E$-valued vector measures on $(X, \cm)$. For each $\mu \in M(X, \cm; E)$, let $\norm{\mu}_{\text{var}} = |\mu|(X)$, then:
\begin{enumerate}
\item $M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
\item If $E$ is a Banach space, then so is $M(X, \cm; E)$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j} \subset \cm$ such that $X = \bigsqcup_{j = 1}^n A_j$,
\[
|\mu|(X) + |\nu|(X) \ge \sum_{j = 1}^n \norm{\mu(A_j)}_E + \norm{\nu(A_j)}_E \ge \sum_{j = 1}^n \norm{(\mu + \nu)(A_j)}_E
\]
As this holds for all choices of $\seqf{A_j}$, $\norm{\mu + \nu}_{\text{var}} \le \norm{\mu}_{\text{var}} + \norm{\nu}_{\text{var}}$.
(2): Let $\seq{\mu_n} \subset M(X, \cm; E)$ such that $\sum_{n \in \natp}\norm{\mu_n}_{\text{var}} < \infty$. For each $A \in \cm$, since $E$ is complete, let
\[
\mu(A) = \sum_{n = 1}^\infty \mu_n(A)
\]
then for any $\seq{A_k} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{k \in \natp}A_k$,
\[
\mu(A) = \sum_{n = 1}^\infty \mu_n(A) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \mu_n(A_k) = \sum_{k = 1}^\infty \mu(A_k)
\]
by Fubini's theorem.
% TODO: Actually link Fubini once it's there.
\end{proof}

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\section{Vector Measures}
\label{section:vector-measures}
\begin{definition}[Vector Measure]
\label{definition:vector-measure}
Let $(X, \cm)$ be a measurable space, $E$ be a normed space over $K \in \RC$, and $\mu: \cm \to E$, then $\mu$ is a \textbf{vector measure} if:
\begin{enumerate}
\item[(M1)] $\mu(\emptyset) = 0$.
\item[(M2)] For any $\seq{A_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in \natp} \mu(A_n)$ where the sum converges absolutely.
\end{enumerate}
\end{definition}
\begin{proposition}
\label{proposition:vector-measure-bounded}
Let $(X, \cm)$ be a measurable space, $E$ be a normed space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measure, then
\[
\sup_{A \in \cm}\norm{\mu(A)}_E < \infty
\]
\end{proposition}
\begin{proof}
For each $\phi \in E^*$, the mapping
\[
\mu_\phi: \cm \to \complex \quad A \mapsto \dpn{\mu(A), \phi}{E}
\]
is a complex measure. For each $A \in \cm$, let
\[
x_A: E^* \to \complex \quad \phi \mapsto \dpn{\mu(A), \phi}{E}
\]
then for each $\phi \in E^*$,
\[
\sup_{A \in \cm}|\dpn{\phi, x_A}{E^{**}}| = \sup_{A \in \cm}|\dpn{\mu(A),\phi}{E}| < \infty
\]
By \autoref{proposition:bornologic-continuous-complete} and \autoref{proposition:metrisable-bornologic}, $E^*$ is a Banach space. The \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness} implies that
\[
\sup_{A \in \cm}\norm{\mu(A)}_{E} = \sup_{A \in \cm}\norm{x_A}_{E^{**}} < \infty
\]
\end{proof}

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\end{remark}
\begin{definition}[Space of Bounded Functions]
\label{definition:bounded-function-space}
Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
\begin{enumerate}
\item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
\item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication.
Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
(2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
\[
f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
\]
so $f \in B(T; E)$.
If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
\end{proof}
\begin{proposition}
\label{proposition:uniform-limit-continuous}
Let $X$ be a topological space and $Y$ be a uniform space, then:
@@ -72,20 +49,3 @@
If $Y$ is complete, then $Y^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $C(T; X)$ and $UC(T; X)$ are both complete subspaces by \autoref{proposition:complete-closed}.
\end{proof}
\begin{definition}[Space of Bounded Continuous Functions]
\label{definition:bounded-continuous-function-space}
Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
\begin{enumerate}
\item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
\item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
(2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace.
If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete.
\end{proof}

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Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
\end{definition}
\begin{definition}[Cauchy Net]
\label{definition:cauchy-net}
Let $(X, \fU)$ be a uniform space and $\net{x} \subset X$ be a net, then $\net{x}$ is \textbf{Cauchy} if for any $V \in \fU$, there exists $\alpha_0 \in A$ such that $(x_\alpha, x_\beta) \in V$ for all $\alpha, \beta \ge \alpha_0$.
\end{definition}
\begin{lemma}
\label{lemma:cauchy-subnet}
Let $(X, \fU)$ be a uniform space and $\net{x} \subset X$ be a Cauchy net. If there exists a subnet $\angles{x_\beta}_B \subset \net{x}$ and $x \in X$ such that $x_\beta \to x$, then $x_\alpha \to x$.
\end{lemma}
\begin{proof}
Let $U \in \fU$ and $V \in \fU$ such that $V \circ V \subset U$, then there exists $\alpha_0 \in A$ such that $(x_\alpha, x_{\alpha'}) \in V$ for all $\alpha, \alpha' \ge \alpha_0$, and $\beta \in B$ with $\beta \ge \alpha_0$ such that $(x_\beta, x) \in V$. Thus, $(x_\alpha, x) \in V \circ V \subset U$ for all $\alpha \ge \alpha_0$, so $x_\alpha \to x$.
\end{proof}
\begin{proposition}[Cauchy Criterion]
\label{proposition:cauchycriterion}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy.