Added a local to global characterisation of convergence in measure.

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Bokuan Li
2026-06-20 20:45:36 -04:00
parent 1026ff782e
commit 10c6215427

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@@ -18,6 +18,7 @@
and $f_\alpha \to f$ \textbf{locally in measure} if $f_\alpha \to f$ in measure on every set of finite measure.
\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 $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $A \in \fF$ such that $\mu(\bracs{d(f, g) > \delta}) < \eps$ for all $f, g \in A$.
@@ -25,6 +26,35 @@
Alternatively, if $\net{f}$ is a net of $(\cm, \cb_Y)$-measurable functions, then $\net{f}$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $\alpha_0 \in A$ such that for each $\alpha, \beta \in A$ with $\alpha, \beta \ge \alpha_0$, $\mu(\bracs{d(f_\alpha, f_\beta) > \delta}) < \eps$.
\end{definition}
\begin{proposition}
\label{proposition:convergence-in-measure}
Let $(X, \cm, \mu)$ be a semifinite measure space, $(Y, d)$ be a 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 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.