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}
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
@@ -13,7 +13,7 @@
\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}
\begin{proof}
It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
@@ -22,7 +22,7 @@
\[
U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps')
\]
\item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathscr{M}(X; Y)$,
\item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathcal{L}^0(X; Y)$,
\[
\bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
\]
@@ -35,19 +35,19 @@
\label{definition:ky-fan}
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:
\begin{enumerate}
\item $\alpha$ is a metric on $\mathscr{M}(X; Y)$ modulo almost everywhere equality.
\item $\alpha$ induces the uniform structure of convergence in measure on $\mathscr{M}(X; Y)$.
\item $\alpha$ is a metric on $L^0(X; Y)$.
\item $\alpha$ induces the uniform structure of convergence in measure on $L^0(X; Y)$.
\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}
\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}
\item[(M)] If $\alpha(f, g) = 0$, then by \hyperref[continuity from above]{proposition:measure-properties},
\[
@@ -63,12 +63,12 @@
so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$.
\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
\[
\bracs{(f, g) \in \mathscr{M}(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset
\bracs{(f, g) \in \mathscr{M}(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps}
\bracs{(f, g) \in \mathcal{L}^0(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset
\bracs{(f, g) \in \mathcal{L}^0(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps}
\]
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.
@@ -78,7 +78,7 @@
\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
\[
U(A, \delta, \eps) = \bracs{(f, g) \in \mathscr{M}(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
U(A, \delta, \eps) = \bracs{(f, g) \in \mathcal{L}^0(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
\]
then
@@ -86,7 +86,7 @@
\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}
\begin{proof}
It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
@@ -95,7 +95,7 @@
\[
U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')
\]
\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 $f, g, h \in \mathcal{L}^0(X; Y)$,
\[
\bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
\]
@@ -149,18 +149,16 @@
\]
\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 $\seq{f_n} \subset Y^X$ be a sequence Borel measurable functions from $X \to Y$ that is Cauchy in measure, then:
\begin{theorem}
\label{theorem:cauchy-in-measure-limit}
Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a complete metric space, 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.
\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 $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete.
\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 $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
\end{theorem}
\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 $j \ge k \ge K$,
\[
@@ -168,7 +166,7 @@
\subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
\]
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}}}
\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 $f: X \to Y$.
(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.
(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)]

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@@ -197,7 +197,7 @@
\begin{proof}[Proof, {{\cite[Theorem 7.10]{Folland}}}. ]
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 $\mu(A \setminus K) < \eps/3$.