If f is integrable, is it finite almost everywhere?

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.

$E_n=\{x\in \Omega, f(x)\geq n\}$ then $f\geq n\chi_{E_n}$ so $\int_\Omega f d\mu \geq\int_{E_n} n d\mu=n\mu({E_n})$

For any $m>0$ $$ \mu(\{f(x)>m\}=\int_{f>m}d\mu\le\frac1m\int_{f>m}f\,d\mu\le\frac1m\int_\Omega f\,d\mu. $$ Now let $m\to\infty$.