Hi math stackexchangers,

I have a question about the difference between two math statements (for reference they can be found in Royden Fitzpatrick pages 64-65).

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

Lemma 10: Under the assumptions of Egoroff's Theorem, for each $\eta > 0$ and $\delta > 0$, there is a measurable subset $A$ of $E$ and an index $N$ for which $$|f_n - f| < \eta\text{ on }A\text{ for all }n \geq N \text{ and }m(E \setminus A) < \delta.$$

My question is this. What is the difference between Egoroff's Theorem and Lemma 10? Am I misunderstanding uniform convergence? To me it looks like Lemma 10 provides uniform convergence. Is the difference that Egoroff's Theorem ensures that $F$ is closed but Lemma 10 doesn't ensure that $A$ is closed?


  • $\begingroup$ hmm, yeah if $\| f_n - f \| := \sup_{x \in A} \| f_n(x) - f(x) \|$, then it sure looks like the only difference is closedness $\endgroup$ Commented Jul 15, 2012 at 0:31
  • $\begingroup$ The difference is that in Egoroff's Theorem, $F$ is not merely measurable, but closed. The Lemma only guarantees the existence of a measurable set (which could be open, closed, or neither). $\endgroup$ Commented Jul 15, 2012 at 0:36
  • $\begingroup$ Thanks guys. I wasn't sure because it seemed to me unnecessary to split off the Lemma into a separate part of the proof and then come back just to get closedness! $\endgroup$
    – Alex
    Commented Jul 15, 2012 at 0:38
  • $\begingroup$ @AlexOlssen: Depends on the what has been done before. I expect that Royden has some results on approximating measurable sets with open or closed sets that he can apply once he knows the result for "measurable". $\endgroup$ Commented Jul 15, 2012 at 0:43
  • $\begingroup$ @uncookedfalcon: That is not the only difference. Lemma 10 does not assert uniform convergence, because $A$ depends on $\eta$. $\endgroup$ Commented Jul 26, 2012 at 6:30

1 Answer 1


A significant difference was not mentioned in the comments: In the statement of Lemma 10, $A$ depends on $\eta$. Let us forget about closedness for a minute (I'll come back to that later) and compare the two statements:

Egoroff: $$(\forall \varepsilon>0) (\exists A\subseteq E), m(E\setminus A)<\varepsilon \boldsymbol{\large( \forall \eta>0)}$$ $$(\exists N\in\mathbb N)(\forall x\in A)(\forall n\geq N) |f_n(x)-f(x)|<\eta.$$

Lemma 10: $$(\forall \varepsilon>0) \boldsymbol{\large( \forall \eta>0)} (\exists A\subseteq E), m(E\setminus A)<\varepsilon $$ $$(\exists N\in\mathbb N)(\forall x\in A)(\forall n\geq N) |f_n(x)-f(x)|<\eta.$$

I know this isn't a pretty way to write it, but I hope it makes clear the important difference between the two. Notice that in Egoroff's theorem, $A$ had to be chosen once and for all to work for all (subsequently chosen) $\eta>0$, whereas in Lemma 10 $A$ only needs to work for a previously fixed $\eta$.

I remember when I was first learning measure theory from Royden doing an exercise showing how Egoroff's theorem can be proved using Lemma 10 (or whatever it was numbered in my edition). It was pretty straightforward because the exercise told me what sequences of $\delta$s and $\eta$s to use.

As for closedness, note that Egoroff's theorem without the assumption of closedness applies to all finite measure spaces (where there need not even be a topology in general) and the closedness can be added using inner regularity. For example, a subset $A$ of $\mathbb R$ is Lebesgue measurable if and only if for all $\varepsilon>0$ there exists a closed set $F\subseteq A$ such that $m^*(A\setminus F)<\varepsilon$ (where $m^*$ denotes Lebesgue outer measure). Uniform convergence on $A$ implies uniform convergence on F, and $E\setminus F = (E\setminus A)\cup (A\setminus F)$.

  • 2
    $\begingroup$ Good point. Concerning the formatting with \Huge: maybe something like $$(\forall \varepsilon>0) \, (\exists A\subseteq E),\, m(E\setminus A)<\varepsilon, \; {\large \color{#C00}{( \forall \eta>0)}}$$ versus $$(\forall \varepsilon>0) \, {\large \color{#C00}{( \forall \eta>0)}}\, (\exists A\subseteq E), \; m(E\setminus A)<\varepsilon $$ might look a bit more pleasant to the eye and have the same effect? $\endgroup$
    – t.b.
    Commented Jul 26, 2012 at 6:45
  • $\begingroup$ @t.b.: Maybe; it didn't occur to me, but I didn't think about it much, and was being almost whimsical with the formatting choice. I am reluctant to use a format that is useless to color-blind readers, but am open to suggestions. Making your alternative suggestion available in your comment already helps, and for that I am grateful. $\endgroup$ Commented Jul 26, 2012 at 6:48
  • $\begingroup$ I see. I think the difference in size using \large or \Large instead of \Huge would be enough. Another option would be to make it bold using \boldsymbol{...}. Anyway, it's not so important. $\endgroup$
    – t.b.
    Commented Jul 26, 2012 at 6:58
  • 1
    $\begingroup$ @JonasMeyer: most colorblind people can tell red from black... And I agree the \Huge formatting is both disturbing and slightly annoying. $\endgroup$
    – PseudoNeo
    Commented Jul 26, 2012 at 9:13
  • $\begingroup$ @t.b.: Thank you, I've edited to make it smaller and bold. $\endgroup$ Commented Jul 27, 2012 at 4:25

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