Characterization of Almost-Everywhere convergence

Solution 1:

Given I understand you correctly (topologize almost everywhere convergence), we can show that this is not possible. If we have a topological space, then we have convergence of a sequence if and only if every subsequence has a further convergent subsequence and so on.

So pick a sequence that converges in measure but not almost everywhere. There is a theorem that states that every subsequence of this also converges in the same measure. So it has a subsequence that converges almost everywhere (by another theorem). But the original sequence does not converge almost everywhere so we cannot have convergence in a topology.

Should I add more detail?

Solution 2:

I was looking for a proof myself, I found this standard example.

Let $[0,1]$ be given the Lebesgue measure. The vector space $L^\infty([0,1])$ cannot be given a topological vector space structure.

Let $(f_n)$ be a sequence of function which converges in measure to zero but fails to converge a.e.; define the sequence $f_1^1, f_1^2, f_2^2, f_1^3, f_2^3, \dotsc$ where $$ f_m^n(x) = \begin{cases} 1 & \frac{m-1}{n} \leq x \leq \frac{m}{n} \\ 0 & \text{otherwise} \end{cases} $$ and $m$ is enumerated from 1 to $n$. So we have \begin{align*} f_1 &= 1_{[0,1]} \\ f_2 &= 1_{[0,\frac{1}{2}]} , \quad f_3 = 1_{[\frac{1}{2}, 1]} \\ f_4 &= 1_{[0,\frac{1}{3}]} , \quad f_5 = 1_{[\frac{1}{3}, \frac{2}{3}]}, \quad \dotsc \end{align*} Therefore $\mu(f_n \neq 0) \rightarrow 0$ as $n\rightarrow \infty$, hence $f_n$ converges in measure. Note that $\mu$ is a probability measure, and $\sum_n \mu(f_n \neq 0) = \infty$. By Borel-Cantelli, $\mu(f_n \neq 0 \ \text{i.o.}) = 1$. Hence $f_n$ does not converge to zero a.e.

Suppose a topology exists for a.e. convergence. Since $(f_n)$ fails to converge to zero, there must be a neighborhood $U_0$ which $f_n$ is outside i.o. Let $(f_{n_k})$ be a subsequence of terms outside of $U_0$. Any subsequence ${f_{n_k}}$ converges in measure, it has a further subsequence that converges a.e. to zero. But this subsequence is eventually in $U_0$, contradicting the choice that $(f_{n_k})$ is outside $U_0$. Therefore the topology cannot exist.