2
$\begingroup$

I want to prove following statement in Folland: If $f_n \to f$ almost uniformly, then $f_n \to f$ a.e. and in measure.

This is what I did: For $k \in \mathbb{N}$, we choose $F_k$ s.t. $\mu(F_k) < \frac{1}{k}$ and $f_n \to f$ uniformly on $F_k^c$ (so I can do this, how can I verify that there exists such $F_k$?). Then, take $E = \bigcup_1^{\infty} F_k$. Then $f_n \to f$ on $E^c$ (is it uniform or not?), and I try to verify $\mu(E) = 0$ (but have problems also), so if this is valid, things are OK for a.e. convergence.

What can I further do for convergence in measure?

This is a homework question, so if you give reasonable hints, I will be very happy. Thanks!

$\endgroup$
3
$\begingroup$

Your first step is good. We can remove a set of measure less than $1/k$ and get uniform convergence on the compliment -- this is true by the definition of almost uniform convergence.

Now, we will prove convergence in measure. Pick $\epsilon > 0$. We want to show that

$$m(\{x:|f_n - f| \geq \epsilon \}) \to 0 $$

Well, on the complement of $F_k$, we have some integer $N_{k,\epsilon}$ such that for $n \geq N_{k,\epsilon}$

$$ |f_{n} - f| < \epsilon$$

Why is this the case? Can you show that the set

$$\{x :|f_{n} - f| \geq \epsilon\}$$

is contained in $F_{k}$ for $n \geq N_{k,\epsilon}$?

Use that to show

$$m(\{x:|f_n - f| \geq \epsilon \}) \to 0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy