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?
Best Answer
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 $$\mu (\lbrace \omega \in \Omega : f(\omega) = \infty \rbrace) = 0$$