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 $A\setminus B$? 
 A: 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 $B$.
