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 $m(E \setminus F) < \epsilon$.

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

1 Answer 1


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.)


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