Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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 – uncookedfalcon Jul 15 '12 at 0:31
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). – Arturo Magidin Jul 15 '12 at 0:36
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! – Alex Jul 15 '12 at 0:38
@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". – Arturo Magidin Jul 15 '12 at 0:43
@uncookedfalcon: That is not the only difference. Lemma 10 does not assert uniform convergence, because $A$ depends on $\eta$. – Jonas Meyer Jul 26 '12 at 6:30
up vote 4 down vote accepted

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 $E\setminus F = (E\setminus A)\cup (A\setminus F)$.

share|cite|improve this answer
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? – t.b. Jul 26 '12 at 6:45
@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. – Jonas Meyer Jul 26 '12 at 6:48
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. – t.b. Jul 26 '12 at 6:58
@JonasMeyer: most colorblind people can tell red from black... And I agree the \Huge formatting is both disturbing and slightly annoying. – PseudoNeo Jul 26 '12 at 9:13
@t.b.: Thank you, I've edited to make it smaller and bold. – Jonas Meyer Jul 27 '12 at 4:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.