Let $\displaystyle f:(X,\Sigma, \mu)\to (\bar{\mathbb R_+}, \mathcal B)$ be a measurable non-negative function that may assume the value $+\infty$.
Suppose that $\int f d\mu$ is finite. Prove that $\{x\in X, f(x)=+\infty\}$ has measure $0$.
Here's my proof.
Let $A=\{x\in X, f(x)=+\infty\}$.
Suppose for contradiction that $\mu(A)>0$ and let $M>0$.
Then $f\geq 1_{A}M$. Hence $\int f d\mu\geq M \mu(A)$
Letting $M$ go to $\infty$ yields a contradiction.
Since I'm new to measure-theory, I'd like to know if there are shorter, different or more conventional proofs.
A different proof: By Markov's inequality,
$$\mu(\{f \geq R\}) \leq \frac{\int f \, d\mu}{R}.$$
Hence, by the continuity of the measure $\mu$,
$$\mu(\{f=\infty\}) = \lim_{R \to \infty} \mu(\{f \geq R\}) = 0.$$
Another one: From Tonelli's Theorem it is not difficult to see that
$$\int f \, d \mu = \int_{(0,\infty)} \mu(\{f \geq R\}) \, dR$$
for any non-negative (measurable) function $f$. Since the left-hand side is finite and $R \mapsto \mu(\{f \geq R\})$ decreasing, this implies
$$\lim_{R \to \infty} \mu(\{f \geq R\})=0;$$
hence, $\mu(\{f=\infty\})=0$.