TITLE: If f is integrable, is it finite almost everywhere?
QUESTION [11 upvotes]: If $\int_\Omega f d\mu<\infty$, and $f$ is non-negative, can we conclude that $f$ is finite a.e. on $\Omega$?
Is being finite a.e. the same as having a finite essential supremum, i.e. there exists an $M$ such that $f\leq M$ a.e.?
I know that $f$ integrable does not mean that $f$ has a finite essential supremum, from the counter example $f=1/\sqrt x$ on $(0,1]$.
Thanks for help.

REPLY [15 votes]: Let $E_n=f^{-1}((n,\infty))$. Then $n\mu(E_n)\leqslant \int_{E_n} f\ \mathsf d\mu$ so $$\mu(E_n)\leqslant \frac1n\int_{E_n}f\ \mathsf d\mu \leqslant \frac1n\int_\Omega f\ \mathsf d\mu.  $$ Since $\int_\Omega f\ \mathsf d\mu<\infty$, this implies that $\mu(E_n)\stackrel{n\to\infty}\longrightarrow 0$. Since
$$f^{-1}(\{\infty\}) = \bigcap_{n=1}^\infty E_n, $$
$E_n\supset E_{n+1}$, and $\mu(E_N)<\infty$ for some $N$, by continuity from above we have
$$\mu(f^{-1}(\{\infty\}) = \lim_{n\to\infty} \mu(E_n) = 0. $$
It follows that $f$ is finite a.e.