# Thread: Converges almost uniformly implies a.e. and in measure.

1. ## Converges almost uniformly implies a.e. and in measure.

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.

2. As it is, take $\displaystyle E_{\epsilon }$ the set that makes $\displaystyle f_n$ converge uniformly on $\displaystyle E_{\epsilon }^c$ then $\displaystyle \mu (A) \leq \mu (E_{\epsilon}) < \epsilon$ thus proving $\displaystyle \mu (A) =0$.

Why can I choose a B in A in that fashion
I don't think you can, since by definition if you want $\displaystyle f_n$ to converge uniformly in $\displaystyle B^c$ you need $\displaystyle A \subseteq B$

3. Indeed you can't write it this way. You could say: There exists $\displaystyle B$ such that $\displaystyle \mu(B)<\mu(A)$ and $\displaystyle (f_n)_n$ converges uniformly to $\displaystyle f$ on $\displaystyle B^c$.

Then you have $\displaystyle A\subset B$, because uniform convergence on $\displaystyle B^c$ implies convergence at all points of $\displaystyle B^c$ (this sentence tells in fact that $\displaystyle B^c \subset A^c$). And this can't be because of what you wrote: the measure of $\displaystyle B$ is less than that of $\displaystyle A$.