Almost everywhere convergence and $L^1$ convergence

Question

In class we have seen how, given the solution for the heat problem (the one done by convoluting with $\frac{1}{(4 \pi t)^{n/2}} e^{-|x|^2 /4t}$), even if the initial condition was only $L^1$, we still had a.e. convergence. For that we introduced some stuff like Hardy-Littlewood maximal function. However, the teacher said that in general if I have a sequence $\{f_n\}$ converging to $f$ in $L^1$ then I can conclude that there is a subsequence converging almost everywhere. Then he said that there is this general result:

"Given (X,d) a metric space and a succession $\{x_n\}$, then if for every subsequence there is a sub-subsequence converging to a certain $x$, then $x_n \to x$"

But he went on and said we can't conclude that convergence in $L^1$ implies convergence almost everywhere (and there are indeed counterexamples), because "convergence almost everywhere is not induced by any metric". I am not sure about the last quote, but I remember he said there was some problem with the fact that I needed a metric space in order to use the theorem above. Can someone please explain why the theorem above cannot be used in this situtation? I know it must be true, I have counterexamples, but I don't see why.

Answer 1

The answer may depend on the measure space $\left(S,\mathcal F,\mu\right)$ we consider. If $S$ and $\mu\left(S\right)$ are finite then almost everywhere convergence is equivalent to convergence in $L^1$.

However, suppose that $\left(S,\mathcal F,\mu\right)$ is such that there exists a sequence $\left(f_n\right)_{n\geqslant 1}$ which converges in $L^1$ to $0$ but not almost everywhere. Suppose that there exists a metric $d$ on the space of $\mathcal F $-measurable functions such that for each sequence $\left(g_n\right)_{n\geqslant 1}$, $g_n\to g$ almost everywhere is equivalent to $d\left(g_n,g\right)\to 0$. For each subsequence of $\left(f_n\right)_{n\geqslant 1}$, due to convergence in $L^1$, you can extract a subsequence which converges almost everywhere to $0$. Using the quoted fact in the opening post, we derive that $d\left(f_n,0\right)\to 0$ hence convergence of $\left(f_n\right)_{n\geqslant 1}$ almost everywhere to $0$, a contradiction.

Answer 2

The problem is that the set of measure zero depends on the subsequence. For every subsequence $\{n_k\}$ there is a set $E(n_k)$ of measure zero such that $f_{n_{k}}(x) \to f(x)$ provided $x $ is not in this set. The union of these sets of measure 0 need not have measure 0, so you cannot, in general, find one set $A$ of measure zero such that every subsequence of $\{f_n\}$ has a further subsequence converging to $f$ at every point outside $A$.