4
$\begingroup$

Let a measure be $\sigma$-finite and suppose $f_n \rightarrow f$ a.e.. Show that there exists $E_k$ and a null-set $F$ partioning $X$ such that $f_n$ converges uniformly to $f$ on each $E_k$.

I was trying to deduce it from Egorov's Theorem http://mathworld.wolfram.com/EgorovsTheorem.html, but I can't work it out, and it looks like it contradicts with the statement that Egorov's Theorem doesn't hold for $\epsilon=0$.

Thanks!

$\endgroup$
  • $\begingroup$ The formulation seems to be lacking a bit. By the partitioning, do you mean $X=F\cup E_1\cup E_2\cup\cdots$? $\endgroup$ – Harald Hanche-Olsen Oct 23 '13 at 17:59
  • $\begingroup$ Yes you are right $\endgroup$ – user53969 Oct 23 '13 at 18:01
4
$\begingroup$

Let $X=F_1\cup F_2\cup F_3\cup\cdots$ where $F_1\subseteq F_2\subseteq F_3\subseteq\cdots$ and $F_k$ has finite measure. Pick $E_k\subseteq F_k$ so that $f_n\to f$ uniformly on $E_k$, and $\mu(F_k\setminus E_k)<2^{-n}$. I think you can build from there.

| cite | improve this answer | |
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.