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

<br /> \int_E f d\mu = 0<br />

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.