As it is, take the set that makes converge uniformly on then thus proving .
I don't think you can, since by definition if you want to converge uniformly in you needWhy can I choose a B in A in that fashion
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.
Let
Claim:
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 ?
Thank you.
As it is, take the set that makes converge uniformly on then thus proving .
I don't think you can, since by definition if you want to converge uniformly in you needWhy can I choose a B in A in that fashion
Indeed you can't write it this way. You could say: There exists such that and converges uniformly to on .
Then you have , because uniform convergence on implies convergence at all points of (this sentence tells in fact that ). And this can't be because of what you wrote: the measure of is less than that of .