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

**tttcomrader** · November 11th 2009, 04:46 PM

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

Proof so far.

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

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

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

Claim: $\mu (A)=0$

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

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 $A \backslash B$ ?

Thank you.

**Jose27** · November 11th 2009, 07:17 PM

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

Why can I choose a B in A in that fashion

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

**Laurent** · November 12th 2009, 07:02 AM

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

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

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