Topology of almost everywhere convergence - What's wrong with my construction?

Question

I understand that the a.e. convergence is not topologizable, as there is contradiction, for instance with the fact that $L^p$ convergence does not imply a.e. convergence.

So I am asking what is wrong with this construction, it is supposed to yield a topology whose convergence is a.e. convergence.

Let $\Omega$ be a measure space, and let $X=L^p(\Omega)$ for some $p\in [1,\infty)$.

Definition. A subset $S\subset X$ is called closed if for all $\left\{f_n\right\}\subset X$ with $f_n\to f$ a.e., we have $f\in S$.

Now let $\tau$ be the collection of all closed sets. We need to prove that it satisfies the topology axioms (for closed sets).

1. It is clear that $ \emptyset, X\in \tau$.

2. Let $ \left\{S_j\right\}_{j\in J}\subset \tau$. Then if $\left\{f_n\right\}\subset S:=\bigcap_{j\in J}S_j$, and $ f_n\to f$, since $ \left\{f_n\right\}\subset S_j$ for all $ j\in J$, we have $ f\in S_j$ for all $j$, and thus $ f\in S$. Therefore, $ S$ is closed. 

3. Let $ S_1,S_2\in \tau$. Then if $ \left\{f_n\right\}\subset S_1\cup S_2$ with $ f_n\to f$, there is a subsequence of $ \left\{f_n\right\}$ that is entirely contained on either $ S_1$ or $ S_2$. Such subsequence converges again to $ f$, so $ f\in S_1\cup S_2$, and $ S_1\cup S_2$ is closed. 

Thus the complementary of $\tau$ is a topology on $X$, and I expect that in this topology we should have $f_n\to f$ iff $f_n\to f$ a.e:

If $f_n\to f$ a.e., then let $U$ be an open (i.e., $U^c$ is closed) neighborhood of $f$. By contradiction if $f_n\not\to f$, there would be a subsequence $\left\{f_{n_k}\right\}$ which stays aways from $U$, yet $f_{n_k}\to f\in U$ a.e., which contradicts the fact that $U^C$ is closed.

If $f_n\to f$, then I am struggling to prove that $f_n\to f$ a.e. Is this is it, then? It is not true that the convergence in the topology just defined implies a.e. convergence?

(Attempt: Suppose $f_n\to f$. If $f_n\not \to f$ a.e., then there is a subsequence $F:=\left\{f_{n_k}\right\}$ such that $f\notin \bar{F}$. I would like to say that there is a neighborhood $U$ of $f$ such that $U\cap \bar{F}= 0$, a contradiction with $f_n\to f$; but this is not possible unless I prove some separability properties of $(X,\tau)$. )

Answer 1

You have just defined the topology of convergence in measure, at least for finite measures. For $\sigma$-finite measures, it would amount to the topology of local convergence in measure.

One can characterize convergence in measure by saying that $\langle f_n\rangle$ converges to $f$ if every subsequence of $\langle f_n\rangle$ has a further subsequence that converges almost everywhere to $f$.

We show your topology has the same closed sets as the topology of convergence in measure. First note that convergence almost everywhere implies convergence in measure, so every set closed in the topology of convergence in measure is closed in your topology. Conversely, suppose a set $S$ is not closed in the topology of convergence in measure. Then there is a sequence $\langle f_n\rangle$ in $f$ converging in measure to $f$ but $f\notin S$. A subsequence will converge almost everywhere to $f$. Since all the terms of the subsequence are in $S$, $S$ is not closed in your topology.