0
$\begingroup$

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.

Thanks

$\endgroup$
2
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.