This comes from Chapter 2, Real Analysis, by Folland
Proposition 2.20 – If $f\in L^{+}$ and $\int f < \infty$ then $\{x:f(x) = \infty\}$ is a null set and $\{x:f(x) > 0\}$ is $\sigma$-finite
proof (1st part): Let $E = \{x:f(x) = \infty\}$, then $E$ is measurable. Define a simple function $\phi_n = n1_{E} \ \ \forall n\geq 1$ with $0 \leq \phi_n \leq f$, so $$\int f \geq \int \phi_n = n\mu(E)$$ Thus, $$\frac{1}{n}\int f \geq \mu(E) \ \ \forall n\geq 1$$ Since, $0\leq \int f < \infty$ it follows that $\mu(E) = 0$
Proof (2nd part): Now, set $$\{x:f(x) > 0\} = \bigcup_{n\in\mathbb{N}}\{f(x) > 1/n\}$$ For each $n$, set $\phi_n = \frac{1}{n}1_{\{f(x) > 1/n\}}$ with $0\leq \phi_n \leq f$, so $$\int f \geq \int \phi_n = \frac{1}{n}\mu(\{f(x) > 1/n\})$$ Thus, $$n\int f \geq \mu(\{f(x) > 1/n\})$$ Since, $0\leq \mu(\{f(x) > 1/n\}) \leq n\int f < \infty \ \ \forall n$ so, $\{x:f(x) > 0 \}$ is $\sigma$-finite.
Best Answer
Note that if $\{f(x) = \infty\}$ has positive measure, then $\int f d\mu $ is infinite. To show that $\{f(x) > 0\}$ is $\sigma$-finite, try to use
$$\{ f(x) >0\}= \bigcup_{n\in \mathbb N} \{ f(x) > 1/n\}.$$