\section{Convergence in Measure}
\label{section:convergence-in-measure}

\begin{definition}[In Measure]
\label{definition:in-measure}
    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, then the uniform
\end{definition}




\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 $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, and $f$ be a $(\cm, \cb_Y)$-measurable function, then $\fF \to f$ \textbf{in measure} if for each $\eps > 0$, 
    \[
    \lim_{g, \fF}\mu(\bracs{d(f, g) > \eps}) = 0
    \]

    and $\fF \to f$ \textbf{locally in measure} if $\fF \to f$ in measure on every set of finite measure. 

    Alternatively, if $\net{f}$ is a net of $(\cm, \cb_Y)$-measurable functions, then $f_\alpha \to f$ \textbf{in measure} if for each $\eps > 0$, 
    \[
    \lim_{\alpha \in A}\mu(\bracs{d(f, f_\alpha) > \eps}) = 0
    \]

    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 $f, g \in A$. 

    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.
\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 $\seq{f_n} \subset Y^X$ be a sequence Borel measurable functions from $X \to Y$ that is Cauchy in measure, 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}

\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}

