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Let $f \in L^{1}(\mathbb{R})$ such that $f(x) >0$ for almost all $x \in \mathbb{R}$. Let A be a Lebesgue measurable set such that $\int_{A} f =0$. Prove $m(A) = 0$.

wlog we can assume $f(x)>0$ everywhere define $A_{n}={x \in A: f(x) > \frac{1}{n}}$ so since $f(x) > 0$ everywhere, $\int_{A_n} f dm \leq \int_{A} fdm = 0$. Then it is easy to see that A is a countable union of sets of measure 0, so A has measure 0 in this case.

My question is: why do we need A to be Lebesgue measurable? doesn't the above proof works for any measurable set? I still don't see the difference.


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What do you think the difference is between Lebesgue measurable and measurable in this context? – Mariano Suárez-Alvarez Nov 10 '10 at 18:41
@Mariano: that is it endowed with the Lebesgue measure, so I see that the argument works in general with any measure. – student Nov 10 '10 at 18:45

When they write $L^1(\mathbb{R})$, the measure on $\mathbb{R}$ hasn't been specified, so it's assumed to be Lebesgue measure. If you had $L^1(\mu)$ for an arbitrary measure $\mu$ your argument would work just as well.

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Yes. this will work for any measure $\mu$ provided $f \in \mathcal{L}^{1}(\Omega,\mathcal{F},\mu)$ – user17762 Nov 10 '10 at 19:51

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