201 lines
11 KiB
TeX
201 lines
11 KiB
TeX
\section{Localisable Measures}
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\label{section:localisable-measure}
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\begin{definition}[Essential Supremum]
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\label{definition:esssup-measure-space}
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Let $(X, \cm, \mu)$ be a measure space, $\ce \subset \cm$, and $S \in \cm$, then $S$ is an \textbf{essential upper bound} of $\ce$ if for any $E \in \ce$, $\mu(E \setminus S) = 0$. If in addition, for any essential upper bound $T$ of $\ce$, $\mu(S \setminus T) = 0$, then $S$ is an \textbf{essential supremum} of $\ce$.
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\end{definition}
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\begin{definition}[Localisable]
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\label{definition:localisable-measure}
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Let $(X, \cm, \mu)$ be a measure space, then $X$ is \textbf{localisable} if:
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\begin{enumerate}
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\item $\mu$ is semifinite.
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\item For every $\ce \subset \cm$, there exists an essential supremum $A$ of $\ce$.
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\end{enumerate}
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\end{definition}
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\begin{definition}[Decomposable]
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\label{definition:decomposable-measure}
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Let $(X, \cm, \mu)$ be a measure space and $\seqi{A} \subset \cm$, then $\seqi{A}$ is a \textbf{decomposition} of $(X, \cm, \mu)$ if:
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\begin{enumerate}
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\item For each $i \in I$, $\mu(A_i) < \infty$.
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\item $X = \bigsqcup_{i \in I}X_i$.
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\item $\cm = \bracs{E \subset X|E \cap A_i \in \cm \forall i \in I}$.
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\item For each $E \in \cm$, $\mu(E) = \sum_{i \in I}\mu(E \cap A_i)$.
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\end{enumerate}
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If $(X, \cm, \mu)$ admits a decomposition, then it is \textbf{decomposable}.
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\end{definition}
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\begin{lemma}
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\label{lemma:essential-upperbound-finite}
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Let $(X, \cm, \mu)$ be a finite measure space, $\ce \subset \cm$, and $\mathcal{S}$ be the set of all essential upper bounds of $\ce$, then:
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\begin{enumerate}
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\item $\mathcal{S} \ne \emptyset$.
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\item For any $S \in \cm$, $S \in \mathcal{S}$ if and only if $\mu(S \cap E) = \mu(E)$ for all $E \in \ce$.
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\item For any $\seq{S_n} \subset \mathcal{S}$, $\bigcap_{n \in \natp}S_n \in \mathcal{S}$.
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\item There exists $S \in \mathcal{S}$ such that $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$.
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\item For any $S \in \mathcal{S}$ with $\mu(S) = \inf\bracs{\mu(T)|T \in \mathcal{S}}$, $S$ is an essential supremum of $\ce$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1): $X \in \mathcal{S}$.
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(2): Since $\mu$ is finite, for any $E \in \ce$, $\mu(S \cap E) = \mu(E)$ if and only if $\mu(E \setminus S) = 0$.
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(3): Firstly, for any $S, T \in \mathcal{S}$ and $E \in \ce$,
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\[
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\mu(E \setminus (S \cap T)) \le \mu(E \setminus S) + \mu(E \setminus T) = 0
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\]
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so $S \cap T \in \mathcal{S}$. Now let $\seq{S_n} \subset X$ and $E \in \ce$, then since $\mu$ is finite,
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\[
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\mu(E) = \limv{N}\mu\paren{E \cap \bigcap_{n = 1}^N S_n} = \mu\paren{E \cap \bigcap_{n \in \natp}S_n}
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\]
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by \hyperref[continuity from above]{proposition:measure-properties}. By (2), $\bigcap_{n \in \natp}S_n \in \mathcal{S}$.
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(4): Let $M = \inf\bracs{\mu(T)|T \in \mathcal{S}}$ and $\seq{S_n} \subset \mathcal{S}$ such that $\limv{n}\mu(S_n) = M$, then by continuity from above,
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\[
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M \le \mu\paren{\bigcap_{n \in \natp}S_n} \le \limv{n}\mu(S_n) = M
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\]
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By (3), $\bigcap_{n \in \natp}S_n \in \mathcal{S}$, therefore the minimum is achieved.
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(5): Let $R \in \mathcal{S}$. By (3), $S \cap R \in \mathcal{S}$ with $\mu(S \cap R) = \inf\bracs{\mu(T)|T \in \mathcal{S}} = \mu(S)$, so
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\[
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\mu(S \setminus R) = \mu(S \setminus (S \cap R)) = \mu(S) - \mu(S \cap R) = 0
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\]
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and $S$ is the essential supremum of $\ce$.
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\end{proof}
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\begin{proposition}
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\label{proposition:decomposable-localisable}
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Let $(X, \cm, \mu)$ be a decomposable measure space, then $X$ is localisable.
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\end{proposition}
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\begin{proof}
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Let $\ce \subset \cm$ and $\seqi{A} \subset \cm$ be a decomposition of $X$.
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For each $i \in I$, let $\ce_i = \bracs{E \cap A_i|E \in \ce}$. By \autoref{lemma:essential-upperbound-finite}, there exists an essential upper bound $S_i \in \cm$ of $\ce_i$ contained in $A_i$ with respect to the restricted measure $\mu|_{A_i}$. In other words,
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\begin{enumerate}[label=(\roman*)]
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\item $S_i$ is an essential upper bound of $\ce_i$.
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\item For any essential upper bound $T \in \cm$ of $\ce_i$ with $T \subset A_i$, $\mu(S_i \setminus T) = 0$.
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\end{enumerate}
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Now, let $S = \bigsqcup_{i \in I}S_i$, then since $X = \bigsqcup_{i \in I}A_i$ and $S_i \subset A_i$ for all $i \in I$, $S \cap A_i = S_i \in \cm$ for all $i \in I$, so $S \in \cm$. For any $E \in \ce$, $E \cap A_i \in \ce_i$. Passing through the decomposition, (i) implies that,
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\begin{align*}
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\mu(E \setminus S) &= \sum_{i \in I}\mu((E \setminus S) \cap A_i) = \sum_{i \in I}\mu((E \cap A_i) \setminus (S \cap A_i)) \\
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&= \sum_{i \in I}\mu((E \cap A_i) \setminus S_i) = \sum_{i \in I}0 = 0
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\end{align*}
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and $S$ is an essential upper bound of $\ce$.
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Finally, let $T \in \cm$ be an essential upper bound of $\ce$, then $T \cap A_i$ is an essential upper bound of $\ce_i$ for all $i \in I$. The decomposition and (ii) then shows that
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\[
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\mu(S \setminus T) = \sum_{i \in I}\mu((S \cap A_i) \setminus (T \cap A_i)) = \sum_{i \in I}\mu(S_i \setminus (T \cap A_i)) = 0
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\]
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therefore $S$ is an essential supremum of $\ce$.
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\end{proof}
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\begin{lemma}
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\label{lemma:gluing-measurable-sets}
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Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space and $\bracs{(E_A, F_A)}_{A \in \cf}$ be pairs of measurable sets such that:
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\begin{enumerate}[label=(\alph*)]
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\item For each $A \in \cf$, $E_A, F_A \in \cm$, $E_A, F_A \subset A$, and $E_A \cap F_A = \emptyset$.
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\item For each $A, B \in \cf$, $\mu((E_A \cap B) \Delta (E_B \cap A)) = 0$ and $\mu((F_A \cap B) \Delta (F_B \cap A)) = 0$.
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\end{enumerate}
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Let $E$ and $F$ be essential suprema of $\bracsn{E_A}_{A \in \cf}$ and $\bracsn{F_A}_{A \in \cf}$, respectively, then
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\begin{enumerate}
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\item For each $B \in \cf$, $\mu(E \cap F_B) = 0$.
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\item $\mu(E \cap F) = 0$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1): Let $A, B \in \cf$, then
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\begin{align*}
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\mu(E_A \setminus F_B^c) &= \mu(E_A \cap F_B) = \mu(E_A \cap F_B \cap A \cap B) \\
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&\le \mu(E_A \cap F_A) + \mu((F_A \cap B) \Delta (F_B \cap A)) = 0
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\end{align*}
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so $F_B^c$ is an essential upper bound of $\bracs{E_A}_{A \in \cf}$. Since $E$ is an essential supremum of $\bracs{E_A}_{A \in \cf}$, $\mu(E \setminus F_B^c) = \mu(E \cap F_B) = 0$.
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(2): For any $B \in \cf$, $\mu(F_B \setminus E^c) = \mu(F_B \cap E) = \mu(E \cap F_B) =0$. Thus $E^c$ is an essential upper bound of $\bracs{F_B}_{B \in \cf}$. Given that $F$ is an essential supremum of $\bracsn{F_B}_{B \in \cf}$, $\mu(F \cap E) = \mu(F \setminus E^c) = 0$.
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\end{proof}
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\begin{lemma}[Gluing Lemma for Measurable Functions]
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\label{lemma:gluing-measurable}
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Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space, $Y$ be a Polish space, and $\bracsn{f_A: A \to Y|A \in \cf}$ such that:
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\begin{enumerate}[label=(\alph*)]
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\item For each $A \in \cf$, $f_A \in \mathcal{L}^0(A; Y)$.
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\item For each $A, B \in \cf$, $f_A|_{A \cap B} = f_B|_{A \cap B}$ almost everywhere.
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\end{enumerate}
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then there exists $f: X \to Y$ such that:
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\begin{enumerate}
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\item $f \in \mathcal{L}^0(X; Y)$.
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\item For each $A \in \cf$, $f|_A = f_A$ almost everywhere.
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\item[(U)] For any $g: X \to Y$ satisfying (1) and (2), $f = g$ almost everywhere.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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First suppose that $Y$ is finite. For each $y \in Y$, let $P(y)$ be an essential supremum of $\bracs{f_A^{-1}(y)|A \in \cf}$. By \autoref{lemma:gluing-measurable-sets}, for any $x, y \in Y$ with $x \ne y$, $\mu(P(x) \cap P(y)) = 0$. After modification by null sets, assume without loss of generality that $X = \bigsqcup_{y \in Y}P(y)$.
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For each $x \in X$, let $f(x) \in Y$ be the unique element of $Y$ such that $x \in P(f(x))$, then:
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\begin{enumerate}
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\item For each $y \in Y$, $f^{-1}(y) = P(y)$, so $f \in \mathcal{L}^0(X; Y)$ is measurable.
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\item Let $A \in \cf$, then for each $y \in Y$, $\mu(f_A^{-1}(y) \setminus P(y)) = 0$. On the other hand,
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\begin{align*}
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\mu\braks{(P(y) \cap A) \setminus f_A^{-1}(y)} &\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap f_A^{-1}(z) ) \\
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&\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap P(z)) \\
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&+ \sum_{z \in Y \setminus \bracs{y}}\mu(f_A^{-1}(z) \setminus P(z)) \\
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&= 0
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\end{align*}
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so $f|_A = f_A$ almost everywhere on $A$.
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\item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere.
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\end{enumerate}
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Therefore $f$ is the desired function.
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Now suppose that $Y$ is an arbitrary separable metrisable space. By \autoref{lemma:separable-metric-space-approx-identity}, there exists $\seq{I_n} \subset Y^Y$ such that:
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\begin{enumerate}[label=(\roman*)]
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\item $I_n \to \text{Id}$ pointwise as $n \to \infty$.
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\item For each $n \in \natp$, $I_n(Y)$ is finite and Borel measurable.
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\end{enumerate}
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For each $n \in \natp$, let $f_{A, n} = I_n \circ f_A$, then
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\begin{enumerate}[label=(\alph*)]
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\item For each $A \in \cf$, $f_{A, n} \in \mathcal{L}^0(A; Y)$.
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\item For each $A, B \in \cf$, $f_{A, n}|_{A \cap B} = f_{B, n}|_{A \cap B}$ almost everywhere.
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\item For each $A \in \cf$, $f_{A, n}(A) \subset I_n(Y)$.
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\end{enumerate}
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By the finite case, there exists $f_n: X \to Y$ such that:
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\begin{enumerate}
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\item $f_n \in \mathcal{L}^0(X; Y)$.
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\item For each $A \in \cf$, $f_n|_A = f_{A, n}$ almost everywhere.
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\end{enumerate}
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Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, $f_{A, n} \to f_A$ pointwise by (i). For each$n \in \natp$, $f_{A, n} = f_n|_A$ almost everywhere on $A$ by (2). Thus
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\begin{align*}
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\bracs{\limv{n}f_n \text{ exists}} \cap A &\supset \bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_A} \\
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\mu\paren{\bracs{\limv{n}f_n \text{ exists}} \cap A} &= \mu\paren{\bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_{A}}} = \mu(A) \\
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\mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} &= 0
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\end{align*}
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As $\cf$ is a scaffold for $\mu$, $\mu\bracsn{\limv{n}f_n \text{ does not exist}} = 0$. By (1) and \autoref{proposition:metric-measurable-limit}, there exists $f \in \mathcal{L}^0(X; Y)$ such that $f = \limv{n}f_n$ almost everywhere. In which case,
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\begin{enumerate}
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\item $f \in \mathcal{L}^0(X; Y)$.
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\item For each $A \in \cf$, $f|_A = \limv{n}f_n|_A = \limv{n}f_{A, n} = f_A$ almost everywhere.
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\item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere.
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\end{enumerate}
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\end{proof}
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