# Measure theory question about Integral.

Let $f \ge 0$ be a measurable function on measure space $(\Omega, \Sigma)$. Let $\mu$ be a measure. How to prove that if $$\int f \mathrm{d}\mu = 0,$$ then we have $f = 0$ almost everywhere?

This is an exercise 1.4.1 of Probability: Theory and Examples.

-
add comment

## 2 Answers

Hint: let $A_n:=\{x\mid f(x)\geq n^{—1}\}$. What is $\mu(A_n)$? And $\bigcup_{n\geq 1}A_n$? Its measure?

-
add comment

By Markov's inequality

$$\mu\left( \{x : f(x) \geq \frac{1}{n}\} \right) \leq n \int f d\mu = 0.$$

-
add comment