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?

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

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