2
$\begingroup$

Egorov's theorem is used often used to prove the dominated convergence theorem. The proof I am currently reading says that according to Egorov's theorem and absolute continuity of the Lebesgue integral, we can choose a $\delta > 0$ and a set $B$ such that $\mu(B) < \delta$ and $\{f_n\}$ converges uniformly on $C = A \setminus B$ etc.

Should it be obvious that there is uniform convergence not only on $B$, but also on its complement? The theorem itself only mentions that for every $\delta>0$ there exists such a set $B \subset A$ that $\{f_n\}$ converges uniformly on $B$ and $\mu(A\setminus B) < \delta$

Why is it possible to conclude that uniform convergence is preserved on $A\setminus B$?

$\endgroup$
5
  • 1
    $\begingroup$ What do you mean "uniform converges exists not only on 𝐵, but also on its complement"? $\endgroup$ Feb 6, 2019 at 19:07
  • $\begingroup$ You only have uniform converges on $A\backslash B$. If you add $B$ you don't have uniform converges anymore. $\endgroup$
    – Yanko
    Feb 6, 2019 at 19:08
  • $\begingroup$ The example you may want to keep in mind is $f_n=x^n$ with $A=[0,1]$. Then $f_n$ convergence uniformly to $0$ only if you remove an interval around $1$. $\endgroup$
    – Yanko
    Feb 6, 2019 at 19:09
  • $\begingroup$ I will try to clarify my question. I understand Egorov's theorem this way: we have a.e convergence on $A$. For any $\delta > 0$ we can find such a set $B \subset A$ that $\mu(A\setminus B) < \delta$ and $\{f_n\}$ converges uniformly on $B$. The textbook I am currently reading, however, uses this theorem and says the following: "We have a.e convergence on $D$. Then there is $E \subset D$ such that $f_n$ converges uniformly to $f$ on $D \setminus E$. Shouldn't it be uniformly convergent on $E$ instead? $\endgroup$
    – Don Draper
    Feb 6, 2019 at 19:29
  • $\begingroup$ You introduce $C$ and never use it again? Then you mangle the roles of $B$ and $A\setminus B$ in paragraph 2. This indeed has a happy ending, but you need to be more precise in your question. $\endgroup$
    – zhw.
    Feb 6, 2019 at 23:42

1 Answer 1

1
$\begingroup$

Based on your original post and some of the exchanges in the comments, if I am understanding your confusion correctly, the way you normally present Egorov's theorem is as follows:

Accustomed statement of Egorov: Suppose $\mu(A) < \infty$ and $f_n$ is a sequence of measurable functions that converge a.e. to $f$ on $A$. Then for each $\delta>0$, there is a measurable set $B\subset A$ such that $\mu(A\smallsetminus B) < \delta$ and such that $f_n \to f$ uniformly on $B$.

The version you are reading now in this proof is

New version: Suppose $\mu(A) < \infty$ and $f_n$ is a sequence of measurable functions that converge a.e. to $f$ on $A$. Then for each $\delta>0$, there is a measurable set $B\subset A$ such that $\mu(B) < \delta$ and such that $f_n \to f$ uniformly on $A\smallsetminus B$.

Really, these are saying the same thing: in each case, the claim is that for any $\delta>0$ we like, we can find a "small set," i.e., one whose measure is less than $\delta$, such that $f_n\to f$ uniformly on the complement of that small set. In what I am calling the accustomed statement of Egorov's theorem, the small set is $A\smallsetminus B$. In what I am calling the new version, the small set is $B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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