If $\displaystyle f_n \rightarrow f $ almost uniformly, then $\displaystyle f_n \rightarrow f $ a.e. and $\displaystyle f_n \rightarrow f $ in measure.

Proof so far.

Now, since $\displaystyle f_n \rightarrow f $ almost uniformly, I know that for any $\displaystyle \epsilon > 0 $, there exists $\displaystyle E \subset X $ such that $\displaystyle \mu (E)< \epsilon $ and $\displaystyle f_n \rightarrow f $ uniformly on $\displaystyle E^c $, that is, $\displaystyle | f_n(x)-f(x)| \rightarrow 0 \ \ \ \ \forall x \in E^c $.

First, I want to show that $\displaystyle f_n \rightarrow f $ a.e.

Let $\displaystyle A = \{ x : f_n(x) \ not \ \rightarrow f \} $

Claim: $\displaystyle \mu (A)=0 $

Suppose to the contrary that $\displaystyle \mu (A) > 0 $, then there exists $\displaystyle B \subset A $ such that $\displaystyle \mu (B) < \frac { \mu (A) }{2} $ and that $\displaystyle f_n \rightarrow f $ uniformly on $\displaystyle A \backslash B $, and $\displaystyle \mu (A \backslash B ) > \frac { \mu (A) }{2} > 0 $

But since fn do not converge to f on A , that is a contradication.

My question is: Why can I choose a B in A in that fashion and to get fn converges uniformly to A for all x in $\displaystyle A \backslash B $?

Thank you.