Convergence in metric and a.e How might I show that there's no metric on the space of measurable functions on $([0,1],\mathrm{Lebesgue})$ such that a sequence of functions converges a.e. iff the sequence converges in the metric?
 A: In fact, the almost converge doesn't correspond to any topology. We use the following fact: 

Let $(X,\mathcal T)$ a topological space. A sequence $\{x_n\}$ converges to $x$ on $(X,\mathcal T)$ if and only if for all subsequence $\{x_{n_k}\}$ we can extract a converging subsequence to $x$. 

Consider a sequence $\{X_n\}$ of random variables which converges in probability but not almost surely to $X$. For each subsequence of $\{X_n\}$, we can extract an almost everywhere converging subsequence, which yield a contradiction.
But there is a metric for the convergence in probability, namely 
$$\delta(X,Y):= \int_{ \Omega}\dfrac{|X(\omega)-Y(\omega)|}{1+|X(\omega)-Y(\omega)|}d\,\mathbb P(\omega).$$
A: In a metric space, a sequence such that any subsequence has a converging subsequence must be convergent.
Considering a sequence converging in measure but not a.e. shows that this property is not true for a.e. convergence.
A: Just to add to the other answers (since it was not explicitly stated): there is a sequence in $L_1[0,1]$ that converges in measure but not pointwise a.e. 
For example: 
$f_1(x)=1$, 
$f_2(x)= \chi_{[0,1/2]}$
$f_3(x)=\chi_{[1/2,1]}$
$f_4(x)=\chi_{[0,1/4]}  $
$f_5(x)=\chi_{[1/4,1/2]}  $
$f_6(x)=\chi_{[1/2,3/4]}  $
$f_7(x)=\chi_{[3/4,1]}  $
$f_8(x)=\chi_{[0,1/8]}  $
$\phantom{f_8(x)}\ \ \vdots$
Where $\chi_A$ is the indicator function on $A$.
$\{f_n\}$ converges to measure to 0 but does not converge pointwise a.e.
A: There is no topology on $L^1([0,1])$ which describes the notion of "convergence almost everywhere".
Well, I just noticed that this has been answered a few seconds ago. Anyway, I'd like to point out the nice note "Convergence Almost Everywhere is Not Topological" which can be read here...
