diff --git a/.vscode/project.code-snippets b/.vscode/project.code-snippets index a726f38..ad0c0e0 100644 --- a/.vscode/project.code-snippets +++ b/.vscode/project.code-snippets @@ -176,5 +176,10 @@ "scope": "latex", "prefix": "cproof", "body": ["[Proof, {{\\cite[$1]{$2}}}. ]$0"] + }, + "Scaffold": { + "scope": "latex", + "prefix": "scaf", + "body": ["\\hyperref[scaffolded]{definition:measure-scaffold}$0"] } } diff --git a/src/measure/measurable-maps/in-measure.tex b/src/measure/measurable-maps/in-measure.tex index 9dbed05..688b685 100644 --- a/src/measure/measurable-maps/in-measure.tex +++ b/src/measure/measurable-maps/in-measure.tex @@ -74,65 +74,6 @@ On the other hand, if $\mu\bracs{d(f, g) > \eps} \le \eps$, then $d(f, g) \le \eps$. Therefore $\alpha$ induces the uniform structure of convergence in measure. \end{proof} -\begin{definition}[Locally In Measure] -\label{definition:locally-in-measure} - Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$ and $A \in \cm$ with $\mu(A) < \infty$, let - \[ - U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps} - \] - - then - \[ - \fB = \bracs{U(A, \delta, \eps)|\eps, \delta > 0, A \in \cm, \mu(A) < \infty} - \] - - forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of local convergence in measure} on $\mathcal{L}^0(X; Y)$. -\end{definition} -\begin{proof} - It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}: - \begin{enumerate} - \item[(FB1)] For each $\eps, \eps', \delta, \delta' > 0$ and $A, A' \in \cm$ with $\mu(A), \mu(A') < \infty$, - \[ - U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps') - \] - \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cm$ with $\mu(A) < \infty$, and $f, g, h \in \mathcal{L}^0(X; Y)$, - \[ - \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta} - \] - - so $U(A, \delta/2, \eps/2) \circ U(A, \delta/2, \eps/2) \subset U(A, \delta, \eps)$. - \end{enumerate} -\end{proof} - -\begin{proposition} -\label{proposition:convergence-in-measure} - Let $(X, \cm, \mu)$ be a semifinite measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if: - \begin{enumerate} - \item[(L)] $\fF$ is locally Cauchy in measure. - \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that - \[ - \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps - \] - \end{enumerate} -\end{proposition} -\begin{proof} - (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_1 \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that - \[ - \sup_{f, g \in F_1}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps - \] - - By (L), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that - \[ - \sup_{f, g \in F_2}\mu(A \cap \bracs{d(f, g) > \delta}) < \eps - \] - - Therefore - \[ - \sup_{f, g \in F_2}\mu\bracs{d(f, g) > \delta} < 2\eps - \] -\end{proof} - - \begin{lemma} \label{lemma:ae-in-measure} Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a separable metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then $f_n \to f$ in measure. @@ -157,7 +98,7 @@ \item $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete. \end{enumerate} \end{theorem} -\begin{proof}[Proof, [{{\cite[Theorem 2.30]{Folland}}}]. ] +\begin{proof}[Proof, {{\cite[Theorem 2.30]{Folland}}}. ] (1): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$. In this case, for any $K \in \natp$ and $j \ge k \ge K$, @@ -196,48 +137,3 @@ (2): Since the uniform structure of convergence in measure on $L^0(X; Y)$ is defined by the \hyperref[Ky Fan metric]{definition:ky-fan}, completeness follows from (1) and \autoref{proposition:complete-metric-space}. \end{proof} -\begin{theorem}[Monotone Convergence Theorem (in Measure)] -\label{theorem:mct-measure} - Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that - \begin{enumerate}[label=(\alph*)] - \item For each $x \in X$, $f_\alpha(x) \upto f(x)$. - \item $f_\alpha \to f$ locally in measure. - \end{enumerate} - - then - \[ - \lim_{\alpha \in A}\int f_\alpha d\mu = \int f d\mu - \] -\end{theorem} -\begin{proof} - By \autoref{definition:lebesgue-non-negative}, $\int f_\alpha d\mu \le \int f d\mu$ for each $\alpha \in A$. - - On the other hand, using \autoref{lemma:lebesgue-non-negative-strict}, it is sufficient to show that - \[ - \lim_{\alpha \in A}\int f_\alpha d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu - \] - - for any $\phi \in \Sigma^+(X, \cm)$ satisfying: - \begin{enumerate}[label=(\roman*)] - \item There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$. - \item $\phi \in L^1(X, \cm)$. - \end{enumerate} - - To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0} < \infty$, by (b), there exists $\alpha \in A$ such that - \[ - \mu\bracs{\phi > 0, f_\alpha + \delta < \phi} \le \mu\bracs{\phi > 0, f_\alpha + \delta < f} < \frac{\eps}{\norm{\phi}_u} - \] - - In which case, by \hyperref[linearity]{proposition:lebesgue-simple-properties}, - \begin{align*} - \int \phi d\mu &= \int_{\bracs{\phi > 0}}\phi d\mu = \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} \phi d\mu + \int_{\bracs{\phi > 0, f_\alpha + \delta < \phi}} \phi d\mu \\ - &\le \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} f_\alpha d\mu +\norm{\phi}_u\mu \bracs{\phi > 0, f_\alpha + \delta < \phi} \\ - &\le \int f_\alpha d\mu + \norm{\phi}_u \frac{\eps}{\norm{\phi}_u} = \int f_\alpha d\mu + \eps - \end{align*} - - As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu$. Therefore - \[ - \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu = \lim_{\alpha \in A}\int f_\alpha d\mu - \] -\end{proof} - diff --git a/src/measure/measurable-maps/index.tex b/src/measure/measurable-maps/index.tex index 1ef3fd3..4024fc6 100644 --- a/src/measure/measurable-maps/index.tex +++ b/src/measure/measurable-maps/index.tex @@ -8,3 +8,4 @@ \input{./metric.tex} \input{./approx.tex} \input{./in-measure.tex} +\input{./locally-in-measure.tex} diff --git a/src/measure/measurable-maps/local-in-measure.tex b/src/measure/measurable-maps/local-in-measure.tex new file mode 100644 index 0000000..c48ae6d --- /dev/null +++ b/src/measure/measurable-maps/local-in-measure.tex @@ -0,0 +1,116 @@ +\section{Local Convergence in Measure} +\label{section:locally-in-measure} + +\begin{definition}[Locally In Measure*] +\label{definition:locally-in-measure} + Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$ and $A \in \cf$, let + \[ + U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps} + \] + + then + \[ + \fB = \bracs{U(A, \delta, \eps)|\eps, \delta > 0, A \in \cm, \mu(A) < \infty} + \] + + forms a fundamental system of entourages for a uniformity. + + The uniformity defined by $\fB$ is the \textbf{uniform structure of local convergence in measure}, and $\mathcal{L}_\cf^0(X; Y)$ denotes $\mathcal{L}^0(X; Y)$ equipped with this uniformity. +\end{definition} +\begin{proof} + It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}: + \begin{enumerate} + \item[(FB1)] For each $\eps, \eps', \delta, \delta' > 0$ and $A, A' \in \cm$ with $\mu(A), \mu(A') < \infty$, + \[ + U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps') + \] + \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cf$, and $f, g, h \in \mathcal{L}^0(X; Y)$, + \[ + \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta} + \] + + so $U(A, \delta/2, \eps/2) \circ U(A, \delta/2, \eps/2) \subset U(A, \delta, \eps)$. + \end{enumerate} +\end{proof} + +\begin{proposition} +\label{proposition:convergence-in-measure} + Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if: + \begin{enumerate} + \item[(L)] $\fF$ is \hyperref[definition:locally-in-measure]{definition:locally-in-measure}. + \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cf$ such that + \[ + \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps + \] + \end{enumerate} +\end{proposition} +\begin{proof} + (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_1 \in \fF$ and $A \in \cf$ such that + \[ + \sup_{f, g \in F_1}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps + \] + + By (L), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that + \[ + \sup_{f, g \in F_2}\mu(A \cap \bracs{d(f, g) > \delta}) < \eps + \] + + Therefore + \[ + \sup_{f, g \in F_2}\mu\bracs{d(f, g) > \delta} < 2\eps + \] +\end{proof} + +\begin{theorem} +\label{theorem:locally-in-measure-complete} + Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space and $(Y, d)$ be a Polish space, then \hyperref[$\mathcal{L}^0_\cf(X; Y)$]{definition:locally-in-measure} is complete. +\end{theorem} +\begin{proof} + Let $\fF \subset \mathcal{L}^0_\cf(X; Y)$ be a Cauchy filter. By \autoref{theorem:cauchy-in-measure-limit}, for each $A \in \cf$, there exists an almost everywhere unique $f_A \in \mathcal{L}^0(A; Y)$ such that $\fF$ converges to $f_A$ when restricted to $A$. Thus for any $A, B \in \cf$, $f_{A \cup B}|_{A \cap B} = f_A|_{A \cap B} = f_B|_{B \cap A}$ almost everywhere. By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists $f \in \mathcal{L}^0(X; Y)$ such that $f|_A = f_A$ for all $A \in \cf$. Thus $\fF \to f$ locally in measure, and $\mathcal{L}^0(X; Y)$ is complete. +\end{proof} + +\begin{theorem}[Monotone Convergence Theorem (in Measure)] +\label{theorem:mct-measure} + Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that + \begin{enumerate}[label=(\alph*)] + \item For each $x \in X$, $f_\alpha(x) \upto f(x)$. + \item $f_\alpha \to f$ locally in measure. + \end{enumerate} + + then + \[ + \lim_{\alpha \in A}\int f_\alpha d\mu = \int f d\mu + \] +\end{theorem} +\begin{proof} + By \autoref{definition:lebesgue-non-negative}, $\int f_\alpha d\mu \le \int f d\mu$ for each $\alpha \in A$. + + On the other hand, using \autoref{lemma:lebesgue-non-negative-strict}, it is sufficient to show that + \[ + \lim_{\alpha \in A}\int f_\alpha d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu + \] + + for any $\phi \in \Sigma^+(X, \cm)$ satisfying: + \begin{enumerate}[label=(\roman*)] + \item There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$. + \item $\phi \in L^1(X, \cm)$. + \end{enumerate} + + To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0} < \infty$, by (b), there exists $\alpha \in A$ such that + \[ + \mu\bracs{\phi > 0, f_\alpha + \delta < \phi} \le \mu\bracs{\phi > 0, f_\alpha + \delta < f} < \frac{\eps}{\norm{\phi}_u} + \] + + In which case, by \hyperref[linearity]{proposition:lebesgue-simple-properties}, + \begin{align*} + \int \phi d\mu &= \int_{\bracs{\phi > 0}}\phi d\mu = \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} \phi d\mu + \int_{\bracs{\phi > 0, f_\alpha + \delta < \phi}} \phi d\mu \\ + &\le \int_{\bracs{\phi > 0, f_\alpha + \delta \ge \phi}} f_\alpha d\mu +\norm{\phi}_u\mu \bracs{\phi > 0, f_\alpha + \delta < \phi} \\ + &\le \int f_\alpha d\mu + \norm{\phi}_u \frac{\eps}{\norm{\phi}_u} = \int f_\alpha d\mu + \eps + \end{align*} + + As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu$. Therefore + \[ + \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu = \lim_{\alpha \in A}\int f_\alpha d\mu + \] +\end{proof} + diff --git a/src/measure/measure/localisable.tex b/src/measure/measure/localisable.tex index 8515b62..f6f07c7 100644 --- a/src/measure/measure/localisable.tex +++ b/src/measure/measure/localisable.tex @@ -184,9 +184,9 @@ \item For each $A \in \cf$, $f_n|_A = f_{A, n}$ almost everywhere. \end{enumerate} - Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, $f_{A, n} \to f_n$ pointwise by (i). For each$n \in \natp$, $f_{A, n} = f_n|_A$ almost everywhere on $A$ by (2). Thus + 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 \begin{align*} - \bracs{\limv{n}f_n \text{ exists}} \cap A &\subset \bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_A} \\ + \bracs{\limv{n}f_n \text{ exists}} \cap A &\supset \bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_A} \\ \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) \\ \mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} &= 0 \end{align*} diff --git a/src/measure/measure/semifinite.tex b/src/measure/measure/semifinite.tex index d08b5bb..ba9972e 100644 --- a/src/measure/measure/semifinite.tex +++ b/src/measure/measure/semifinite.tex @@ -63,23 +63,15 @@ \begin{definition}[Scaffold*] \label{definition:measure-scaffold} - Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$, then $\cf$ is a \textbf{scaffold} for $\mu$ if for all $E \in \cm$, + Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, then $\cf$ is a \textbf{scaffold} for $\mu$ if for all $E \in \cm$, \[ \mu(E) = \sup\bracs{\mu(E \cap A)|A \in \cf} \] and the quadruple $(X, \cm, \cf, \mu)$ is a \textbf{scaffolded measure space}. -\end{definition} -\begin{lemma} -\label{lemma:measure-scaffold-semifinite} - Let $(X, \cm, \mu)$ be a measure space: - \begin{enumerate} - \item If $\cf \subset \cm$ is a \hyperref[scaffold]{definition:measure-scaffold} for $\mu$, then $\mu$ is semifinite. - \item If $\mu$ is semifinite, then $\bracs{A \in \cm|\mu(A) < \infty}$ is a scaffold for $\mu$. - \end{enumerate} -\end{lemma} -% Omitted + For any semifinite measure space $(X, \cm, \mu)$, $\cf = \bracs{A \in \cm|\mu(A) < \infty}$ is the \textbf{canonical scaffold} for $\mu$, and $(X, \cm, \mu)$ will be equipped with this scaffold unless specified otherwise. +\end{definition} \begin{example} Let $X$ be a LCH space, $\mu$ be a Radon measure, and $\mathcal{K}$ be the collection of compact subsets of $X$, then $\mathcal{K}$ is a scaffold for $\mu$.