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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?

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But there is always a metric such that convergence in measure iff convergence with respect to the metric. –  Ashok Dec 1 '11 at 9:01
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See also this MO thread and this blog post. –  t.b. Dec 1 '11 at 10:55
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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 $$\delta(X,Y):= \int_{ \Omega}\dfrac{|X(\omega)-Y(\omega)|}{1+|X(\omega)-Y(\omega)|}d\,\mathbb P(\omega).$$

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+1! "Consider a sequence {Xn} of random variables which converges in probability but not almost surely to X." Why "for each subsequence of {Xn}, we can extract an almost everywhere converging subsequence, which yield a contradiction"? –  Tim Dec 29 '12 at 5:31
    
Is the problem about how to prove the claim or why it is useful? –  Davide Giraudo Dec 29 '12 at 14:23
    
Yes, it is. @Davide. –  Tim Dec 29 '12 at 14:27
    
I found a theorem.Thanks! –  Tim Dec 29 '12 at 22:43
    
"A sequence {xn} converges to x on (X,T) if and only if for all subsequence {xnk} we can extract a converging subsequence to x." Isn't it that "if and only if every subsequence of {xn} converges"? Why do you mention subsequence of subsequence, instead of subsequence of the original sequence? –  Tim Dec 29 '12 at 22:54
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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.

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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...

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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.

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