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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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src/measure/index.tex
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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}