# Convergence in Egorov's theorem

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 $$A\setminus B$$?

• What do you mean "uniform converges exists not only on 𝐵, but also on its complement"? Feb 6, 2019 at 19:07
• You only have uniform converges on $A\backslash B$. If you add $B$ you don't have uniform converges anymore. Feb 6, 2019 at 19:08
• 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 $1$. Feb 6, 2019 at 19:09
• 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? Feb 6, 2019 at 19:29
• 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.
– zhw.
Feb 6, 2019 at 23:42

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 $$B$$.
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 $$B$$.