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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.

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2 Answers 2

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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?

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By Markov's inequality

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

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