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Given the vector space $$V = \{\text{measurable function on $[0,1]$}\},$$ from the following lemma

Lemma Let $X$ be a topological space and $\mathbf{x}=(x_n)_{n\in \mathbb{N}}$ be a sequence of elements of $X$. If every subsequence of $\mathbf{x}$ contains a subsequence convergent to $x$ then $x_n \to x$.

we know there does not exist a topology on $V$ that would capture a.e. pointwise convergence.

Take any sequence $f_n \rightarrow f$ in measure but $f_n$ does not converge to $f$ pointwise a.e., for each subsequence $f_{n_k}$ we have $f_{n_k} \rightarrow f$ in measure thus there exists a further subsequence $f_{n_{k_l}} \rightarrow f$ pointwise a.e.

On the other hand, we know there is a topology for pointwise convergence, where we look at $V$ as a subspace of $\mathbb{R}^{[0,1]}$ with the product topology. More details here.

So is there any intuition behind this subtle difference? And is there any other examples of this phenomenon?

Thank you.

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    $\begingroup$ I think 'a.e.' makes all the difference. If, instead of Lebesgue measure on $[0,1]$ you had a measure space where every singleton had positive measure so that a.e. convergence is same as convergence at every point then the product topology would be consistent with a.e. convergence. $\endgroup$ – Kavi Rama Murthy Mar 9 '18 at 7:25

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