The following question is taken from Royden's Real Analysis $4$th edition, Chapter $3,$ question $28,$ page $67:$

Question: Show that Egoroff's Theorem continues to hold if the convergence is pointwise a.e. and $f$ is finite a.e., that is.

Assume $E$ has finite measure. Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise almost everywhere on $E$ to the real-valued function $f$ which is finite almost everywhere. Then for each $\varepsilon>0,$ there is a closed set $F$ contained in $E$ for which $$\{f_n\}\to f \text{ uniformly on }F \text{ and }m(E\setminus F)<\varepsilon.$$

My attempt:

Let $A = \{ x\in E: f_n\not\to f \text{ pointwise} \}$ and $B=\{ x\in E: |f|=\infty \}.$ By assumption, $$m(A)=m(B)=0.$$ So $A$ and $B$ are measurable. Note that $$m[E\setminus (A\cup B)] = m(E) - m(A\cup B) = m(E).$$ Fix $\varepsilon>0.$ Since $\{f_n\}$ converges to $f$ pointwise on $E\setminus (A\cup B)$ to the real-valued function $f,$ by Egoroff's Theorem, there exists a closed set $F$ contained in $(E\setminus (A\cup B))\subseteq E$ for which $$f_n\to f \text{ uniformly on } F \text{ and }m[(E\setminus(A\cup B)) \setminus F] < \varepsilon.$$ Since $E$ has finite measure and $F\subseteq E,$ by monotonicity, $F$ has finite measure. As $F$ is closed, it is also measurable. Therefore, by Excision property, we have
$$m(E\setminus F) = m(E) - m(F) = m[E\setminus(A\cup B)] - m(F) = m[(E\setminus(A\cup B)) \setminus F]<\varepsilon.$$

Is my proof correct? I ask for verification because in this post Kenny's comment about countable union of null sets. I do not use this anywhere in my proof. So I wonder whether my proof miss out something.

  • $\begingroup$ Looks good to me. $\endgroup$ Commented Nov 29, 2017 at 5:30
  • $\begingroup$ I’m not sure I understand which part of the linked document you are referring to. Maybe I missed something, but I didn’t see anything about Egoroff’s theorem in there. $\endgroup$ Commented Nov 30, 2017 at 3:43
  • $\begingroup$ What's up with the title? The title seems to suggest that you want to prove a version of Egoroff on a domain with infinite measure, but in the body of your question, only finite-measure domains are considered. $\endgroup$ Commented Nov 30, 2017 at 3:46
  • $\begingroup$ But in your proof you assume that $m(E) < \infty$! And you'll have to clarify what it is that you want to prove, because the obvious statement for Egoroff on an infinite-measure domain is plainly false and there are easy counterexamples. $\endgroup$ Commented Nov 30, 2017 at 3:49
  • $\begingroup$ @NateEldredge: I apologize for mixing up between Egoroff's Theorem and Lusin's Theorem. For this post, I want to prove that Egoroff's Theorem holds for almost everywhere pointwise convergence and $f$ is finite almost everywhere. $\endgroup$
    – Idonknow
    Commented Nov 30, 2017 at 3:51

1 Answer 1


The proof looks fine. Kenny's cited comment ("Can't you remove from $E$ the subsets on which $f_n$ doesn't converge pointwise and the subset on which the limit is not finite? A countable union of null sets is null") does not really need to mention countable unions, since all you need here is a finite union of (two) null sets.


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