Is it true that $\lim_{n\to\infty}\mu(\{x\in E:|f_n(x)-f(x)|>\epsilon\})=\mu(\{x\in E: \lim_{n\to\infty}|f_n(x)-f(x)|>\epsilon\})?$ Let $(X,\mathcal{M},\mu)$ be a measure space and $E\in\mathcal{M}$. Let $f:E\to\Bbb R$ and $f_n:E\to\Bbb R$ be measurable functions for all $n\in \Bbb R$. Suppose $f_n$ converges to $f$ in measure and $\lim_{n\to\infty}|f_n(x)-f(x)|$ exists for almost all $x\in E$.   Could we conclude that for any $\epsilon>0$
$$\lim_{n\to\infty}\mu(\{x\in E:|f_n(x)-f(x)|>\epsilon\})=\mu(\{x\in E: \lim_{n\to\infty}|f_n(x)-f(x)|>\epsilon\})?$$
Attempt. For simplicity, let $g(x)=\lim_{n\to\infty}|f_n(x)-f(x)|$ for almost all $x\in E$, and $g_n(x)=|f_n(x)-f(x)|$ for all $n\in \Bbb N, x\in E$. Then $\lim_{n\to\infty}\mu(\{x\in E:|f_n(x)-f(x)|>\epsilon\})=\lim_{n\to\infty}\int_E\chi_{\{g_n>\epsilon\}}$. I'm not sure how to interchange the limit and integral here.
Thanks!
 A: With our loss of generality, assume that $X=E$. As it was pointed out by the comments, if $\mu$ is $\sigma$-finite ($\mu$ is $\sigma$ finite if there is a countable measurable partition $\{A_n:n\in\mathbb{N}\}$ of $X$ with $\mu(A_n)<\infty)$), the problem becomes trivial.
We now consider the case where $\mu$ is semi-finite ($\mu$ is semi-finite if any set $A$ of positive $\mu$ measure contains a subset of positive and finite $\mu$ measure). We claim that
$h:=\lim_n|f_n-f|=0$ $\mu$ a.s. To see this, suppose that is not the case. Then $\mu(\{h>0\})>0$ and so, there is a set $A$ of positive finite measure on which $h>0$. For any $\delta>0$,
$$\mu(A\cap\{|f_n-f|>\delta\})\leq\mu(|f_n-f|>\delta)\xrightarrow{n\rightarrow\infty}0$$
Since $0<\mu(A)<\infty$, there exists a subsequence $n_k$ such that $\lim_k|f_{n_k}(x)-f(x)|=0$ $\mu$-a.s. on $A$. This contradicts the fact that $h>0$ on $A$. 
Consequently, for any $\delta>0$, $\mu(\lim_n|f_n-f|>\delta)=0$
Putting things together, we get
$$\lim_n\mu(|f_n-f|>\delta)=0=\mu(\lim_n|f_n-f|>\delta)$$

Edit: Suppose $\mu$ is not semi-definite. For each $F\in\mathscr{F}$ define
$$\mu_0(F)=\sup\{\mu(E):E\in\mathscr{F},\,\mu(E)<\infty, \,E\subset F\}$$
It is not difficult to show that $\mu_0$ is a semi-finite measure on $\mathscr{F}$, $\mu_0\leq \mu$, and  $\mu(F)=0$ whenever $F$ is a $\mu$-atom with $\mu(F)=\infty$ ($A$ is a $\mu$-atom if for all $B\in\mathscr{F}$, $B\subset A$ implies $\mu(B)\in\{0,\mu(A)\}$). The function $\nu:\mathscr{F}\rightarrow\overline{\mathbb{R}}$ defined as
$$\nu(A)=\left\{\begin{matrix}
\infty &\text{if $A$ contains a $\mu$-atom $A'$ with $\mu(A)=\infty$}\\
0 & \text{otherwise}
\end{matrix}\right.
$$
is a measure; furthermore,
$$\mu=\mu_0+\nu$$
Since $\mu_0\leq \mu$,  the semi-finite case implies that for any $\delta>0$
$$\lim_n\mu_0\big(|f_n-f|>\delta\big)=0=\mu_0(\lim_n|f_n-f|>\delta)$$
Suppose $A$ is a $\mu$-atom with $\mu(A)=\infty$ contained in $\{x:\lim_n|f_n-f|\,\text{exists}\}$.  As $\lim_n\mu(A\cap\{|f_n-f|>\delta\})=0$ for any $\delta>0$, there is $N_\delta\in\mathbb{N}$, such that $\mu(A\cap\{|f_n-f|>\delta\})=0$ for all $n\geq N_\delta$, that is $f_n=f$ on $A$ for all $n\geq N_\delta$. This shows that
$$ \lim_n|f_n-f|=0\qquad\mu-\text{a.s.}$$
Hence, the conclusion of OP also holds for non-semi-finite measures.
