Prove if $f_n\rightarrow f$ in almost every point in E, then $f_n\rightarrow f$ in measure. Let $f_n:E\rightarrow\mathbb{\bar{R}}$, with $m(E)<\infty$. Prove if $f_n\rightarrow f$ in almost every point in E, then $f_n\rightarrow f$ in measure.
My attempt:
Let $\epsilon, \delta >0$.
As $f_n\rightarrow f$ in almost every point then exists $N\in\mathbb{N}$ such that  $m\{x:|f_n(x)-f(x)|>\epsilon\}=0$
As $\delta>0$ then $m\{x:|f_n(x)-f(x)|>\epsilon\}<\delta$ and this implies $f_n\rightarrow f$ in measure by definition.
Is correct this?
 A: Your argument is not correct as ponied out above. 
One way of proving this is to use DCT. DCT is probably an overkill but it makes the proof very simple. If $f_n \to f$ a.e. in a finite measure space then $\int \min \{1,|f_n-f|\} d\mu \to 0$ . But $\min \{1,|f_n-f|\} >\epsilon$ whenever $|f_n-f| >\epsilon$  (and $0<\epsilon <1$) from which it follows that $\mu (|f_n-f| >\epsilon) \to 0$.  
A: Kavi Rama Murthy already gave a very elegant proof with aid of integral. So let me try to provide a less elegant but elementary proof. First, define
$$ Z_{n}(\epsilon) = \{ x \in E : |f_n(x) - x| \geq \epsilon \}. $$
Then for each $\epsilon > 0$, we easily find that
\begin{align*}
\{ x \in E : f_n(x) \not\to f(x) \}
&\supseteq \{x \in E : |f_n(x) - f(x)| \geq \epsilon \text{ for infinitely many $n$'s}\} \\
&= \cap_{N\geq 1}\cup_{n\geq N} Z_n(\epsilon).
\end{align*}
So the almost-everywhere convergence assumption asserts that $m\left( \cap_{N\geq 1}\cup_{n\geq N} Z_n(\epsilon) \right) = 0$. Then, since $E$ has finite measure and $N \mapsto \cup_{n\geq N} Z_n(\epsilon)$ is decreasing, we have
$$ \lim_{N\to\infty} m\left(\cup_{n\geq N} Z_n(\epsilon)\right) = 0. $$
Then from the inequality $ m(Z_N(\epsilon)) \leq m\left(\cup_{n\geq N} Z_n(\epsilon)\right) $, we find that $m(Z_N(\epsilon)) \to 0$ as $N\to\infty$. This implies the desired convergence in measure.
