Proving that $\int 0 d\mu=0$

Question

While looking for a clean proof of $\int 0 d\mu=0$, I encountered a difficulty.

Consider $0$ as a function between $(X,\Sigma, \mu)$ and $({\mathbb R}, \mathcal B)$.

Since $0=1_\emptyset+0\cdot 1_X$, $\int 0 d\mu=\mu(\emptyset)+0\cdot\mu(X)=0\cdot\mu(X)$

Why should $0\cdot\mu(X)$ be $0$ ? Is that some convention ?

It's certainly true when $\mu(X)<\infty$.

Answer 1

In measure theory, espacially when defining the integral, for clearer readiability, one uses the convention, that $$ 0 \cdot \infty = 0 $$ So the answer is: Yes, that's some convention.