Convergence in measure If a sequence $\{f_n\}$ of measurable functions converges to $f$ almost everywhere, then it converges to $f$ in measure.
(This is from Introductory of real analysis, Kolmogorov-Fomin)
I have read a prove that says:
"Assume that $\mu(E) < \infty$, then if $f_n \rightarrow f$ almost everywhere then $f_n \rightarrow f$ in measure". 
I don't quite understand why we need $\mu(E) < \infty$. The proof given in there does indeed use $\mu(E) < \infty$. 
However I understand that if the measure space is finite then we have the equivalence of convergences.
Could someone explain me this differences please.
 A: The condition $\mu(E)<\infty$ is definitely needed. For example, if $E=\mathbb{R}$ with Lebesgue measure and $f_n=1_{[n,\infty)}$ then $f_n\to 0$ almost everywhere but $f_n\not\to 0$ in measure. Here $1_A$ is the indicator function of the set $A$.
Also, your statement that the two modes of convergence are equivalent when $\mu(E)<\infty$ is incorrect. If $\mu(E)<\infty$ then almost everywhere convergence implies convergence in measure, but not conversely. One example is when $E=[0,1]$ with Lebesgue measure and $\{f_n\}$ is the sequence defined by
$$ f_1=1_{[0,1]},\; f_2=1_{[0,\frac{1}{2}]},\;f_3=1_{[\frac{1}{2},1]},\; f_4=1_{[0,\frac{1}{4}]},\;f_5=1_{[\frac{1}{4},\frac{1}{2}]},\dots$$
In this case $f_n\to 0$ in measure, but
$$ \limsup_{n\to\infty}f_n(x)=1$$
and 
$$ \liminf_{n\to\infty}f_n(x)=0$$
for all $x\in[0,1]$.
The sequence $\{f_n\}$ is sometimes called the "typewriter sequence" or the "floating, shrinking interval".
A: Convergence in measure is the condition that $\mu( \{ | f - f_n| \geq \epsilon \} ) \to 0$ for each $\epsilon >0$.
Let $f = 1$, and $f_n = 1 - x/n$, on $[0, \infty)$. Then for all $\epsilon$, $\mu( \{ | f - f_n| \geq \epsilon \} = \infty$. However, $f_n \to f$ pointwise.
