Let $E$ be a measurable set with $m(E)<\infty$, $(f_n)$ a sequence of real-valued measurable functions on $E$ and $f$ be a real_valued measurable function on $E$. Suppose $(f_n)$ converges to $f$ pointwise a.e. in $E$. Now it is required to prove that $(f_n)$ converges to $f$ in measure. Here is my attempt.
Assume $(f_n)$ does not converge to $f$ in measure. Then there exists $\eta_0>0$ such that $\lim_{n\to\infty}m(\{x\in E:|f_n(x)-f(x)|\geq\eta_0\})\neq0$. Therefore there exists $\epsilon_0>0$ such that for all $n\in\mathbb{N},$ there exists $N>n$ such that $m(\{x\in E:|f_N(x)-f(x)|\geq\eta_0\})\geq\epsilon_0$. Then there is a subsequence $(f_{n_k})$ of $(f_n)$ such that $m(\{x\in E:|f_{n_k}(x)-f(x)|\geq\eta_0\})\geq\epsilon_0$ for each $k\in\mathbb{N}$. Now $(f_{n_k})$ is a subsequence of $(f_n)$ that does not converge to $f$ on a set of positive measure; contradiction.
Is this proof alright? The stipulation $m(E)<\infty$ worries me. Is there something wrong? Thanks.
2nd attempt-added later:
Suppose $(f_n)$ converges to $f$ pointwise a.e. in $E$. Let $\epsilon>0$. Then by Egoroff's theorem, there is a measurable set $F\subseteq E$ such that $m(F)<\epsilon$ and $(f_n)$ converges to $f$ uniformly on $E\setminus F$. Now $(|f_n-f|)$ converges to $0$ uniformly on $E\setminus F$. Thus there exists $N\in\mathbb{N}$ such that for each $n>N$ and $x\in E\setminus F$, $|f_n(x)-f(x)|<\epsilon$. Let $n>N$. Then $m(\{x\in E:|f_n(x)-f(x)|\geq \epsilon\})=m(\{x\in E\setminus F:|f_n(x)-f(x)|\geq \epsilon\})+m(\{x\in F:|f_n(x)-f(x)|\geq \epsilon\})<0+\epsilon=\epsilon.$
@Ian Is this argument alright?