diff --git a/.vscode/settings.json b/.vscode/settings.json index 109a5a7..1d9f0d4 100644 --- a/.vscode/settings.json +++ b/.vscode/settings.json @@ -5,5 +5,7 @@ "task": "Watch" } ], - "latex.linting.enabled": false + "latex.linting.enabled": false, + "latex-workshop.latex.autoBuild.run": "never", + "latex-workshop.latex.texDirs": ["${workspaceFolder}"] } \ No newline at end of file diff --git a/document.tex b/document.tex index 9b4f3a7..4ea963a 100644 --- a/document.tex +++ b/document.tex @@ -1,4 +1,4 @@ -% \documentclass{report} +\documentclass{report} \usepackage{amssymb, amsmath, hyperref} \usepackage{preamble} @@ -12,7 +12,7 @@ %\input{./src/process/index} \bibliographystyle{alpha} % We choose the "plain" reference style -\bibliography{refs} % Entries are in the refs.bib file +\bibliography{refs.bib} % Entries are in the refs.bib file \end{document} diff --git a/src/measure/measurable-maps/in-measure.tex b/src/measure/measurable-maps/in-measure.tex new file mode 100644 index 0000000..5dafcf5 --- /dev/null +++ b/src/measure/measurable-maps/in-measure.tex @@ -0,0 +1,83 @@ +\section{Convergence in Measure} +\label{section:convergence-in-measure} + +\begin{definition}[Convergence in Measure] +\label{definition:convergence-in-measure} + Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{in measure} if for every $\eps > 0$, + \[ + \lim_{n \to \infty}\mu(\bracs{d(f_n, f) > \eps}) = 0 + \] +\end{definition} + +\begin{definition}[Cauchy in Measure] +\label{definition:cauchy-in-measure} + Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is \textbf{Cauchy in measure} if for every $\eps > 0$, + \[ + \mu(\bracs{d(f_m, f_n) > \eps}) \to 0 \quad \text{as}\ m, n \to \infty + \] +\end{definition} + +\begin{lemma} +\label{lemma:ae-in-measure} + Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a 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. +\end{lemma} +\begin{proof} + Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_n(x), f(x)) < \eps$ for all $n \ge N$, so + \[ + \mu\paren{\bigcup_{N \in \natp}\bigcap_{n \ge N}\bracs{d(f_n, f) < \eps}} = \mu(X) + \] + + By continuity from above (\autoref{proposition:measure-properties}), + \[ + \limv{N}\mu{\bracs{d(f_N, f) \ge \eps}} \le \limv{N}\mu\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) \ge \eps}} = 0 + \] +\end{proof} + + +\begin{proposition}[{{\cite[Theorem 2.30]{Folland}}}] +\label{proposition:cauchy-in-measure-limit} + Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a complete metric space, and $\seq{f_n}$ be Borel measurable functions from $X \to Y$, then: + \begin{enumerate} + \item There exists a Borel measurable function $f: X \to Y$ such that $f_n \to f$ in measure. + \item There exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere. + \item For any Borel measurable function $g: X \to Y$ such that $f_n \to g$ in measure, $f = g$ almost everywhere. + \end{enumerate} +\end{proposition} +\begin{proof} + (2): 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 $j \ge k \ge K$, + \[ + \bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}} \subset \bigcup_{\ell = k}^{j - 1}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}} + \subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}} + \] + + By monotonicity and subadditivity (\autoref{proposition:measure-properties}), + \[ + \mu\paren{\bigcup_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}}} + \le \sum_{\ell \ge K}\mu\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}} + \le 2^{-K+1} + \] + + so + \[ + \mu\braks{\paren{\bigcap_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) \le 2^{-K+1}}}^c} \le 2^{-K+1} + \] + + By the \hyperref[First Borel-Cantelli Lemma]{lemma:borel-cantelli-1}, + \[ + \mu\braks{\limsup_{K \to \infty}\paren{\bigcap_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) \le 2^{-K+1}}}^c} = 0 + \] + + Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges to a Borel measurable function $f: X \to Y$. + + (1): Let $\eps, \delta > 0$, then there exists $N \in \natp$ such that $\mu(\bracs{d(f_m, f_n) > \delta/2}) < \eps/2$ for all $m, n \ge N$. Let $k \in \natp$ such that $n_k \ge N$ and $\mu(\bracs{d(f, f_{n_k}) > \delta/2}) < \eps/2$, then for any $m \ge N$, + \[ + \mu\bracs{d(f_m, f) > \delta} \le \mu\bracs{d(f_m, f_{n_k}) > \delta/2} + \mu\bracs{d(f, f_{n_k}) > \delta/2} < \eps + \] + + (3): By (2), there exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ and $f_{n_k} \to g$ almost everywhere, so $f = g$ almost everywhere. +\end{proof} + + + diff --git a/src/measure/measurable-maps/index.tex b/src/measure/measurable-maps/index.tex index 4eb8705..a56100f 100644 --- a/src/measure/measurable-maps/index.tex +++ b/src/measure/measurable-maps/index.tex @@ -6,3 +6,4 @@ \input{./real-valued.tex} \input{./simple.tex} \input{./metric.tex} +\input{./in-measure.tex} diff --git a/src/measure/measurable-maps/measurable-maps.tex b/src/measure/measurable-maps/measurable-maps.tex index fd314f3..7282062 100644 --- a/src/measure/measurable-maps/measurable-maps.tex +++ b/src/measure/measurable-maps/measurable-maps.tex @@ -6,11 +6,17 @@ Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces and $f: X \to Y$ be a mapping, then $f$ is \textbf{$(\cm, \cn)$-measurable} if $f^{-1}(E) \in \cm$ for all $E \in \cn$. \end{definition} + \begin{definition}[Borel Measurable] \label{definition:borel-measurable-function} Let $(X, \cm)$ be a measurable space, $Y$ be a topological space, and $f: X \to Y$ be a mapping, then $f$ is \textbf{Borel measurable} if $f$ is $(\cm, \cb_Y)$-measurable. \end{definition} +\begin{definition}[Convergence Almost Everywhere] +\label{definition:convergence-ae} + Let $(X, \cm, \mu)$ be a measure space, $Y$ be a topological space, $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{almost everywhere/a.e.} if there exists a $\mu$-null set $N \in \cm$ such that $f_n \to f$ pointwise on $X \setminus N$. +\end{definition} + \begin{lemma} \label{lemma:continuous-borel-measurable} Let $X, Y$ be topological spaces and $f: X \to Y$ be continuous, then $f$ is Borel measurable. diff --git a/src/measure/measure/measure.tex b/src/measure/measure/measure.tex index d3b26b9..8004c13 100644 --- a/src/measure/measure/measure.tex +++ b/src/measure/measure/measure.tex @@ -64,15 +64,19 @@ (3): Denote $E_0 = \emptyset$. For each $n \in \natp$, let $F_n = E_n \setminus E_{n - 1}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case, - \[ - \mu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigsqcup_{n \in \natp}F_n} = \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n) - \] + \begin{align*} + \mu\paren{\bigcup_{n \in \natp}E_n} &= \mu\paren{\bigsqcup_{n \in \natp}F_n} \\ + &= \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n) + \end{align*} + (4): Since $\mu(E_1) < \infty$, - \[ - \limv{n}\mu(E_n) = \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) = \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} = \mu\paren{\bigcap_{n \in \natp}E_n} - \] + \begin{align*} + \limv{n}\mu(E_n) &= \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) \\ + &= \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} \\ + &= \mu\paren{\bigcap_{n \in \natp}E_n} + \end{align*} by (3). @@ -125,5 +129,23 @@ \[ \mu(F) = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu(F) \] - \end{proof} + +\begin{lemma}[First Borel-Cantelli Lemma] +\label{lemma:borel-cantelli-1} + Let $(X, \cm, \mu)$ be a measure space and $\seq{E_n} \subset \cm$. If $\sum_{n \in \natp}\mu(E_n) < \infty$, then + \[ + \mu\paren{\limsup_{n \to \infty}E_n} = 0 + \] +\end{lemma} +\begin{proof} + For any $n \in \natp$, by monotonicity and subadditivity (\autoref{proposition:measure-properties}), + \[ + \mu\paren{\limsup_{k \to \infty}E_k} \le \mu\paren{\bigcup_{k \ge n}E_k} \le \sum_{k \ge n}\mu(E_k) + \] + As $\sum_{k \in \natp}\mu(E_k) < \infty$, + \[ + \mu\paren{\limsup_{k \to \infty}E_k} \le \inf_{n \in \natp}\sum_{k \ge n}\mu(E_k) = 0 + \] +\end{proof} + diff --git a/src/measure/sets/index.tex b/src/measure/sets/index.tex index d5b6dd9..7d2deb9 100644 --- a/src/measure/sets/index.tex +++ b/src/measure/sets/index.tex @@ -4,3 +4,4 @@ \input{./algebra.tex} \input{./lambda.tex} \input{./elementary.tex} +\input{./limits.tex} diff --git a/src/measure/sets/limits.tex b/src/measure/sets/limits.tex new file mode 100644 index 0000000..7e41fc3 --- /dev/null +++ b/src/measure/sets/limits.tex @@ -0,0 +1,35 @@ +\section{Limits of Sets} +\label{section:set-limits} + +\begin{definition}[Limit of Sets] +\label{definition:set-limits} + Let $X$ be a set and $\seq{E_n} \subset 2^X$, then the \textbf{limit superior} of $\seq{E_n}$ is + \[ + \limsup_{n \to \infty} E_n = \bigcap_{n \in \natp}\bigcup_{k \ge n}E_k + \] + and the \textbf{limit inferior} of $\seq{E_n}$ is + \[ + \liminf_{n \to \infty}E_n = \bigcup_{n \in \natp}\bigcap_{k \ge n}E_k + \] +\end{definition} + + +\begin{lemma} +\label{lemma:set-limits-characterisation} + Let $X$ be a set and $\seq{X_n} \subset 2^X$, then + \[ + \limsup_{n \to \infty} E_n = \bracs{x \in X|x \in E_n \text{for infinitely many }n} + \] + and + \[ + \liminf_{n \to \infty}E_n = \bracs{x \in X|x \in E_n \text{for all but finitely many }n} + \] +\end{lemma} +\begin{proof} + For any $x \in X$, $x \in \limsup_{n \to \infty}E_n$ if and only if for any $n \in \natp$, there exists $k \ge n$ such that $x \in E_k$, if and only if $x \in E_n$ for infinitely many $n \in \natp$. + + For any $x \in X$, $x \in \liminf_{n \to \infty}$ if and only if there exists $n \in \natp$ such that $x \in E_k$ for all $k \ge n$, if and only if $x \in E_n$ for all but finitely many $n \in \natp$. +\end{proof} + + +