If $\mu(E)=0$ then $\int_E f d\mu=0$

Question

Prove that if $f,g\in\mathscr{M}^+$ then if $\mu(E)=0$ then $\int_E f d\mu=0$ even when $f(x)=\infty$ for all $x\in E$.

This is a little bit contradictory. If $\mu$ is the Lebesgue measure in the space $(\mathbb{R},\mathscr{B}_{\mathbb{R}})$ then I can define a measurable function for example $x=5$ whose integral is infinite and the Lebesgue measure of its domain is $0$.

Question:

Could someone provide me a hint for this proof?

Is the problem I raised a contradiction? If not. Why not?

Thanks in advance!

Answer 1

By definition, $\int_E f \, d\mu = \int \chi_E \,f \, d\mu = \sup \{ \int \psi \, d\mu \mid \psi \text{ simple function with } 0 \leq \psi \leq \chi_E \, f \}.$

What is the value of $\int \psi \, d\mu$ if $\psi$ is a simple function with $0 \leq \psi \leq \chi_E \, f$?

What makes you think that $\int_E 5 \, d\mu = \infty$ if $\mu(E) = 0$?