# Show that $\int f=0$ implies $f=0$ a.e, given that $f$ is a nonnegative measurable function.

I am trying to show that $\int f=0$ implies $f=0$ a.e, given that $f$ is a nonnegative measurable function.

But I search my head from what I have learn I have no any clue of solving the problem. That is why I brought it to this room with the hope that somebody will give me a hint.

Thanks

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– Jonas Meyer May 21 '12 at 20:45

Hint: $\{x: f(x)>0\}=\bigcup_n \{x: f(x)\geq \frac 1n\}$. So it suffices to show that $\{x: f(x)\geq\frac 1n\}$ has measure $0$.
Is the measure of the set $\{x:f(x) \ge \frac {1}{n}\}$ zero? – Hassan Muhammad May 21 '12 at 20:55
The integral of $f$ over that set is $\ge 1/n$ times the measure of that set. So that integral is positive and hence $\int f$ is positive, if the measure of that set is positive. – Michael Hardy May 21 '12 at 21:52
Hint: let $A_n:=\{f(x)\geq 2^{-n}\}$. Write $0\geq \int_{A_n}f(x)d\mu\geq 2^{-n}\mu(A_n)$. What about $\mu\left(\bigcup_{n\geq 1}A_n\right)$?