Lebesgue integration over sets of measure zero

[AxiomOfChoice]

Is it true in general that if $f$ is Lebesgue integrable in a measure space $(X,\mathcal M,\mu)$ with $\mu$ a positive measure, $\mu(X) = 1$, and $E \in \mathcal M$ satisfies $\mu(E) = 0$, then

$$\int_E f d\mu = 0$$

This is one of those things I "knew" to be true yesterday, and the day before, and the day before...but now I can't show it! I need to be able to bound that integral, somehow, by $\mu(E)$, but how? Using Holder's inequality? But don't I need to know that $f\in L^2$ or $f\in L^\infty$ to do that? Do I know either of those? I don't think so...

 

[micromass]

It's fairly easy, but it's tricky. Basically, we put

$$\int_E fd\mu\leq \int \|f\|_\infty d\mu=\|f\|_\infty \int d\mu=\|f\|_\infty \mu(E)$$

Note that $\|f\|_\infty=+\infty$ can happen here, that's the tricky part. The thing isd that $0.(+\infty)$ is defined as 0 in measure theory.

 

[AxiomOfChoice]

It's fairly easy, but it's tricky. Basically, we put

$$\int_E fd\mu\leq \int \|f\|_\infty d\mu=\|f\|_\infty \int d\mu=\|f\|_\infty \mu(E)$$

Note that $\|f\|_\infty=+\infty$ can happen here, that's the tricky part. The thing isd that $0.(+\infty)$ is defined as 0 in measure theory.

Ah yes! $0 \cdot \infty = 0$! I had forgotten about that. Thanks micromass.