0
$\begingroup$

Let $f_n,f\in L^1(\mu,\mathbb{R})$ s.t. $f_n\to f$ in measure. Then there is a subsequence $(f_{n_k})$ of $(f_n)$ s.t. $f_{n_k}\to f$ pointwise a.e. and almost uniformly, i.e., $\forall\,\varepsilon>0$, $\exists\,E$ s.t. $\mu(E)<\varepsilon$ and $f_{n_k}\to f$ uniformly on $E^c$.

I think the above is true, and it follows directly from the standard proof that convergence in measure implies having subsequence converging a.e. pointwise. For example, as in this answer, we can just take $$A_k := \{x \in X; |f_{n_k}(x)-f(x)| > 2^{-k}\},$$ $$B_m:=\bigcup_{k=m}^\infty A_k.$$ Then $\lim\mu(B_m)=0$ and $f_{n_k}$ converges uniformly in $B_m^c$.

Is this proof correct?

$\endgroup$
  • $\begingroup$ Yes, it looks fine. $\endgroup$ – Kavi Rama Murthy Jan 11 at 5:32

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.