Let $V$ be the vector space of all real valued Borel measurable functions on $[0,1]$. Then I want to show that convergence a.e. is not given by a semimetric. We know that if $f_n \to f$ in measure then there is a subsequence of $f_n$ converging to $f$ a.e., and the hint is to use this theorem to show that convergence in measure implies convergence a.e., a contradiction.

I don't know how to proceed. I first let $d$ be such a semimetric. If $f_n \to f$ in measure, then we get a subsequence $f_{n_k}$ converging to $f$ a.e., and hence $d(f_{n_k},f) \to 0$, but I'm not sure how this implies $f_n \to f$ a.e.



Assume convergene a.e. is given by some metric $d$.

Consider any $(f_n)$ such that $f_n\to f$ in measure but $f_n\not{\to}f$ a.e. From asumption $f_n\not{\to} f$ in metric, so there exist subsequence $(f_{n_k})$ such that $d(f_{n_k},f)\geq C$ for some $C>0$. Clearly $f_{n_k}\to f$ in measure, so there exist subsequence $(f_{n_{k_l}})$ such that $f_{n_{k_l}}\to f$ a.e. This means that $d(f_{n_{k_l}},f)\to 0$. On the other hand $d(f_{n_{k_l}},f)\geq C$ as $(f_{n_{k_l}})$ is a subsequence of $(f_{n_k})$. Contradiction.

A small modifications shows that convergence a.e. can not be described in terms of any topology!


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