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.