# In Egorov's theorem, remove the condition $\mu(E) < \infty$ and let the sequence be convergent in measure. The conclusion holds for subsequence

Let $(X,\mathscr{F},\mu)$ be a measure space, $E \in \mathscr{F}$, $\{f_n\}$ is a sequence of measurable functions on $E$, and the sequence converges to function $f$ in measure. Show that $\exists \{f_{n_k}\}$, such that $\forall \delta > 0$, there exists $E_\delta \subset E$, such that $\mu(E-E_\delta) < \delta$ and $\{f_{n_k}\}$ converges uniformly to $f$ on $E_\delta$.

Since the assumption that $\mu(E) < \infty$ is removed I found the proof of Egorov's theorem, as given on wiki, cannot be applied to this problem. I tried to consider the subsequence $\{f_{n_k}\}$ that converges almost everywhere to $f$ on $E$(And the subsequence also converges in measure), while no progress is made. But I think it's the sequence desired since if $\mu(E) < \infty$, the statement follows directly from Egorov's theorem. Any help will be appreciated.

Extract $(f_{n_k})_k$ a subsequence such that for each $k\geqslant 1$, $$\mu\{x, |f_{n_k}(x)-f(x)|>k^{-1}\}\leqslant 2^{-k},$$ which is possible thanks to the assumption of convergence in measure. Define $A_k:=\{x, |f_{n_k}(x)-f(x)|>k^{-1}\}$. For a fixed $\delta$, take $K$ such that $\sum_{k\geqslant K}2^{-k}\lt\delta$ and $E_{\delta}:=\bigcap_{k\geqslant K}X\setminus A_k$.