# partioning of $X$ such that $f_n$ converges uniformly to $f$

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!

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

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.