If almost uniformly, then a.e. and in measure.
Proof so far.
Now, since almost uniformly, I know that for any , there exists such that and uniformly on , that is, .
First, I want to show that a.e.
Suppose to the contrary that , then there exists such that and that uniformly on , and
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 ?