# Show that Egoroff's Theorem continues to hold if the convergence is pointwise a.e. and $f$ is finite a.e. on $E$

First off, Egoroff's theorem states that if $E$ is a set of finite measure and $\{f_n\}$ is a sequence of measurable functions that converges pointwise on $E$ to the real valued function $f$, then there exists a closed set $F$ that is roughly the same size as $E$ on which $\{f_n\}$ converges uniformly.

In symbols, for each $\epsilon > 0$, there exists such $F$ such that $m(E \setminus F) < \epsilon$.

• What are $Z_n$ and $S_n$?
– user140541
Commented Nov 4, 2016 at 0:16
• Sets of zero measure on which the sequence does not converge pointwise and $f$ is not finite, respectively. Commented Nov 4, 2016 at 0:33
• As @tomasz pointed out, you need only one set on which $f_n$ does not converge to $f$...
– user140541
Commented Nov 4, 2016 at 1:24

Hint: If $N$ is a null set (with respect to an outer regular measure $\mu$, e.g. a Lebesgue measure), then for any $\varepsilon>0$, there is an open set $U\supseteq N$ such that $\mu(U)<\varepsilon$. Apply the theorem for $f_n\cdot \chi_{U^c}$ for appropriate $N$.
(About your attempt, it's a little bit confusing: why do you need a sequence of $Z_n$ and $S_n$? Just one set is enough. Anyway, you need to be a little more careful if you want the set you get in the end to be open.)