If $\int_\Omega f d\mu < \infty$ and $f$ is non-negative, then $\mu(\{\omega\in\Omega:f(\omega)=\infty\})=0$ So I started reading about measure theory, and encountered the following question:

Suppose $\int_\Omega fd\mu<\infty, f$ measurable and non-negative. Prove: $\mu\left(\{\omega\in\Omega:f(\omega)=\infty\}\right)=0$.

I assumed for a contradiction that $\mu(\{\omega\in\Omega:f(\omega)=\infty\})=c>0$. I defined a sequence of functions $f_n$ s.t:
$f_n(\omega)=
\begin{cases} 
     f(\omega) & f(\omega)\neq\infty\\
      n & f(\omega)=\infty 
   \end{cases}
\
$
Obviously, $f_1\leq f_2\leq...$ and $f_n\rightarrow f$ pointwise. Now I define $h_n=n\chi_{\{\omega|f(\omega)=\infty\}}$. So $h_n$ is a simple function s.t $h_n\leq f_n$ and therefore $\int_\Omega f_nd\mu\geq\int_\Omega h_nd\mu=nc$.
So by monotone convergence we have:
$\int_\Omega fd\mu=\lim\int_\Omega f_nd\mu\geq\lim nc=\infty$ - a contradiction.
Only thing I wasn't able to show is that each $f_n$ is measurable.
Is my proof correct or am I missing something?
 A: Your proof is right, a faster one is to prove the contrapositive statement, that is: if $\mu(\{\omega \in \Omega : f(\omega )=\infty \})=c>0$ then $\int fd\mu=\infty $, what is trivial, as
$$
\begin{align*}
\int_{\Omega }fd\mu&=\int_{\{\omega : f(\omega )=\infty \}}fd\mu+\int_{\{\omega: f(\omega )\neq \infty  \}}fd\mu\\
&\geqslant \int_{\{\omega : f(\omega )=\infty \}}fd\mu\\&=\infty \cdot \int_{\{\omega : f(\omega )=\infty \}}d\mu\\& =\infty\cdot c \\&=\infty
\end{align*} 
$$
∎
P.S.: someone, strangely, downvoted this answer, but the above is completely correct.
A: An alternative proof using Markov inequality :
One has
\begin{align*}
\mu (\lbrace \omega \in \Omega : f(\omega) = \infty \rbrace)& = \mu \left( \bigcap_{n \in \mathbb{N}}\lbrace \omega \in \Omega : f(\omega) > n \rbrace\right) \\
&= \lim_{n \rightarrow +\infty} \mu (\lbrace \omega \in \Omega : f(\omega) > n \rbrace)
\end{align*}
because the intersection is decreasing. But Markov's inequality gives
$$\mu (\lbrace \omega \in \Omega : f(\omega) > n \rbrace) \leq \frac{1}{n} \int_{\Omega} f d\mu$$
so you get directly $$\mu (\lbrace \omega \in \Omega : f(\omega) = \infty \rbrace) = 0$$
