Updated formulation of convergence in measure.
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Bokuan Li
2026-06-22 13:06:15 -04:00
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@@ -5,7 +5,7 @@
\label{definition:in-measure} \label{definition:in-measure}
Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$, let Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$, let
\[ \[
U(\delta, \eps) = \bracs{(f, g) \in \mathscr{M}(X; Y)| \mu\bracs{d(f, g) > \delta} < \eps} U(\delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu\bracs{d(f, g) > \delta} < \eps}
\] \]
then then
@@ -13,7 +13,7 @@
\fB = \bracs{U(\delta, \eps)|\eps, \delta > 0} \fB = \bracs{U(\delta, \eps)|\eps, \delta > 0}
\] \]
forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of convergence in measure} on $\mathscr{M}(X; Y)$. forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of convergence in measure} on $\mathcal{L}^0(X; Y)$.
\end{definition} \end{definition}
\begin{proof} \begin{proof}
It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}: It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
@@ -22,7 +22,7 @@
\[ \[
U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps') U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps')
\] \]
\item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathscr{M}(X; Y)$, \item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathcal{L}^0(X; Y)$,
\[ \[
\bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta} \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
\] \]
@@ -35,19 +35,19 @@
\label{definition:ky-fan} \label{definition:ky-fan}
Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and
\[ \[
\alpha: \mathscr{M}(X; Y)^2 \to [0, \infty) \quad (f, g) \mapsto \inf\bracs{\eps > 0| \mu\bracs{d(f, g) > \eps} \le \eps} \wedge 1 \alpha: L^0(X; Y)^2 \to [0, \infty) \quad (f, g) \mapsto \inf\bracs{\eps > 0| \mu\bracs{d(f, g) > \eps} \le \eps} \wedge 1
\] \]
then: then:
\begin{enumerate} \begin{enumerate}
\item $\alpha$ is a metric on $\mathscr{M}(X; Y)$ modulo almost everywhere equality. \item $\alpha$ is a metric on $L^0(X; Y)$.
\item $\alpha$ induces the uniform structure of convergence in measure on $\mathscr{M}(X; Y)$. \item $\alpha$ induces the uniform structure of convergence in measure on $L^0(X; Y)$.
\end{enumerate} \end{enumerate}
The mapping $\alpha$ is the \textbf{Ky Fan metric} on $\mathscr{M}(X; Y)$. The mapping $\alpha$ is the \textbf{Ky Fan metric} on $L^0(X; Y)$.
\end{definition} \end{definition}
\begin{proof} \begin{proof}
(1): Let $f, g, h \in \mathscr{M}(X; Y)$, then (1): Let $f, g, h \in \mathcal{L}^0(X; Y)$, then
\begin{enumerate} \begin{enumerate}
\item[(M)] If $\alpha(f, g) = 0$, then by \hyperref[continuity from above]{proposition:measure-properties}, \item[(M)] If $\alpha(f, g) = 0$, then by \hyperref[continuity from above]{proposition:measure-properties},
\[ \[
@@ -63,12 +63,12 @@
so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$. so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$.
\end{enumerate} \end{enumerate}
so $\alpha$ is a metric on $\mathscr{M}(X; Y)$, modulo almost everywhere equality. so $\alpha$ is a metric on $\mathcal{L}^0(X; Y)$, modulo almost everywhere equality.
(2): Let $f, g \in \mathscr{M}(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r} \le r$. Thus (2): Let $f, g \in \mathcal{L}^0(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r} \le r$. Thus
\[ \[
\bracs{(f, g) \in \mathscr{M}(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset \bracs{(f, g) \in \mathcal{L}^0(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset
\bracs{(f, g) \in \mathscr{M}(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps} \bracs{(f, g) \in \mathcal{L}^0(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps}
\] \]
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. 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.
@@ -78,7 +78,7 @@
\label{definition:locally-in-measure} \label{definition:locally-in-measure}
Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$ and $A \in \cm$ with $\mu(A) < \infty$, let Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$ and $A \in \cm$ with $\mu(A) < \infty$, let
\[ \[
U(A, \delta, \eps) = \bracs{(f, g) \in \mathscr{M}(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps} U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
\] \]
then then
@@ -86,7 +86,7 @@
\fB = \bracs{U(A, \delta, \eps)|\eps, \delta > 0, A \in \cm, \mu(A) < \infty} \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 $\mathscr{M}(X; Y)$. 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} \end{definition}
\begin{proof} \begin{proof}
It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}: It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
@@ -95,7 +95,7 @@
\[ \[
U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps') U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')
\] \]
\item[(UB3)] For each $\eps, \delta > 0$, $A \in \cm$ with $\mu(A) < \infty$, and $f, g, h \in \mathscr{M}(X; Y)$, \item[(UB3)] For each $\eps, \delta > 0$, $A \in \cm$ with $\mu(A) < \infty$, and $f, g, h \in \mathcal{L}^0(X; Y)$,
\[ \[
\bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta} \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
\] \]
@@ -149,18 +149,16 @@
\] \]
\end{proof} \end{proof}
\begin{theorem}
\begin{proposition}[{{\cite[Theorem 2.30]{Folland}}}] \label{theorem:cauchy-in-measure-limit}
\label{proposition:cauchy-in-measure-limit} Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a complete metric space, then:
Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a complete metric space, and $\seq{f_n} \subset Y^X$ be a sequence Borel measurable functions from $X \to Y$ that is Cauchy in measure, then:
\begin{enumerate} \begin{enumerate}
\item There exists a Borel measurable function $f: X \to Y$ such that $f_n \to f$ in measure. \item For any $\seq{f_n} \subset L^0(X; Y)$ that is Cauchy in measure, there exists $f \in L^0(X; Y)$ and a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere.
\item There exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere. \item $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete.
\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{enumerate}
\end{proposition} \end{theorem}
\begin{proof} \begin{proof}[Proof, [{{\cite[Theorem 2.30]{Folland}}}]. ]
(2): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$. (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 $j \ge k \ge K$, In this case, for any $K \in \natp$ and $j \ge k \ge K$,
\[ \[
@@ -168,7 +166,7 @@
\subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}} \subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
\] \]
By monotonicity and subadditivity (\autoref{proposition:measure-properties}), By \hyperref[monotonicity and subadditivity]{proposition:measure-properties},
\[ \[
\mu\paren{\bigcup_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}}} \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 \sum_{\ell \ge K}\mu\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
@@ -187,12 +185,7 @@
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 $f: X \to Y$. 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 $m \ge N$, (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}.
\[
\mu\bracs{d(f_m, f) > \delta} \le \mu\bracs{d(f_m, f_{n_k}) > \delta/2} + \mu\bracs{d(f, f_{n_k}) > \delta/2} < \eps
\]
(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} \end{proof}
\begin{theorem}[Monotone Convergence Theorem (in Measure)] \begin{theorem}[Monotone Convergence Theorem (in Measure)]

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@@ -197,7 +197,7 @@
\begin{proof}[Proof, {{\cite[Theorem 7.10]{Folland}}}. ] \begin{proof}[Proof, {{\cite[Theorem 7.10]{Folland}}}. ]
First assume that $f$ is bounded. First assume that $f$ is bounded.
(1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \autoref{theorem:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere.
By \hyperref[Egoroff's Theorem]{theorem:egoroff}, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_n \to f$ uniformly and $\mu(\bracs{f \ne 0} \setminus A) < \eps/3$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$. By \hyperref[Egoroff's Theorem]{theorem:egoroff}, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_n \to f$ uniformly and $\mu(\bracs{f \ne 0} \setminus A) < \eps/3$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$.