Functions finite a.e. are bounded outside of a set of arbitrarily small measure Let $E$ be a measurable set with $m(E)<\infty$ and $f$ be a measurable function on $E$ that is finite almost everywhere on E. For each $\epsilon>0$, show that there is a measurable set $F$ contained in E such that $f$ is bounded on $F$ and $m(E$ \ $F)<\epsilon$.
My attempt:
Since $E$ is measurable there is a closed set $F$ such that $F\subseteq E$ and $m(E$ \ $F)<\epsilon$. Is $f$ bounded on this set? How do I show it? Or should I try a different approach?
 A: Your attempt won't work in general as $f$ may not be bounded on $F$. For example, for $f = \infty\chi_{\mathbb{Q}}|_{[0,1]}$. Any set on which $f$ is bounded is a subset of $[0, 1]\setminus\mathbb{Q}$ so any closed set on which $f$ is bounded contains $\overline{[0,1]\setminus\mathbb{Q}} = [0, 1]$ and therefore must be $[0, 1]$, but $f$ is unbounded on this set.
Hint: There is a measure zero set $N \subseteq E$ such that $f$ is finite on $E\setminus N$. Now note that $E\setminus N = \bigcup\limits_{n=0}^{\infty}E_n$ where $E_n = f^{-1}([n, n+1))$. Use the fact that $\mu(E\setminus N) = \sum\limits_{n=0}^{\infty}\mu(E_n) < \infty$.
A: The Xiao's idea is pretty simple and straight.
You can write 
$$\{ f = \infty\} = \bigcap_{n=1}^{\infty} \{f>n\}$$
Note that $\{f>n+1\}\subset\{f>n\}$. Then, using that $m(E) < \infty$ and that $f$ is finite almost sure,
$$m(\{ f = \infty\}) = \lim_{n \rightarrow \infty} m(\{f>n\})=0.$$ 
Now, given $\epsilon > 0$ there is a $n_0$ such that $m(\{f>n\}) < \epsilon$. If you let $F= \{f\le n_0\}$ you have $m(E - F) < \epsilon$ and obviously $f$ is bounded in $F$.
