Show that $f$ is integrable $\implies $ $f$ finite a.e.

Question

Show that $f:\mathbb R^d\longrightarrow \mathbb R$ is integrable $\implies $ $f$ finite a.e.

My attempts I wanted to use Borel-Cantelli lemma, but my construction doesn't look good for it since I did like this:

Let $A=\{x\mid |f(x)|=\infty \}$ and suppose that $m(A)>0$. Define $A_n=\{x\mid |f(x)|>n\}$. We have that $A_{n+1}\subset A_n$ for all $n$, and thus $$\lim_{n\to\infty }m(A_n)=m\left(\bigcap_{n\in\mathbb N}A_n\right)=m(A)>0.$$

How can I continue ?

Answer 1

You are allready there. Observe that: $$\int |f|dm\geq\int |f|1_Adm=(+\infty)\times m(A)=+\infty$$ Showing that $f$ is not integrable. This is proved on base of the assumption that $f$ is not finite a.e..