Lebesgue integration over sets of measure zero

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

[tex]
\int_E f d\mu = 0
[/tex]

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 [itex]\mu(E)[/itex], but how? Using Holder's inequality? But don't I need to know that [itex]f\in L^2[/itex] or [itex]f\in L^\infty[/itex] to do that? Do I know either of those? I don't think so...
 

Answers and Replies

  • #2
22,129
3,297
It's fairly easy, but it's tricky. Basically, we put

[tex]\int_E fd\mu\leq \int \|f\|_\infty d\mu=\|f\|_\infty \int d\mu=\|f\|_\infty \mu(E)[/tex]

Note that [itex]\|f\|_\infty=+\infty[/itex] can happen here, that's the tricky part. The thing isd that [itex]0.(+\infty)[/itex] is defined as 0 in measure theory.
 
  • #3
It's fairly easy, but it's tricky. Basically, we put

[tex]\int_E fd\mu\leq \int \|f\|_\infty d\mu=\|f\|_\infty \int d\mu=\|f\|_\infty \mu(E)[/tex]

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

Ah yes! [itex]0 \cdot \infty = 0[/itex]! I had forgotten about that. Thanks micromass.
 

Related Threads on Lebesgue integration over sets of measure zero

  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
6
Views
5K
Replies
3
Views
4K
Replies
6
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
Top