Let f : X → R be a continuous function on a compact metric space...

landdenaw

landdenaw

Answered

2022-06-28

Let f : X R be a continuous function on a compact metric space X. Assume that a Borel probability measure μ is absolutely continuous with respect to Lebesgue measure Leb.
Is it true that if f ( x ) < 0 for Leb a.e. x, then f d μ < 0 ??
I think it should be true as μ << Leb.
Attempt: Assume that f d μ 0. Then f 0 μ a.e. x. That means μ ( { x : f ( x ) > 0 } ) > 0. That implies Leb ( ( { x : f ( x ) > 0 } ) > 0 which is not true.

Answer & Explanation

klemmepk

klemmepk

Expert

2022-06-29Added 16 answers

The compactness, metrizability and other conditions are unnecessary. All that is needed is that f is measurable and μ λ, where I use λ for Lebesgue measure.
λ { x X : f ( x ) 0 } = 0 μ λ μ { x X : f ( x ) 0 } = 0
And immediately we have that, as X is not μ-null:
X f d μ < 0
I'm afraid we do not quite have:
X f d μ 0 f 0 μ  - a.e.
Since a function that is both positive and negative can simply cancel, e.g.
R sin x x d x
Famously evaluates to π > 0, but of course sin ( x ) x is negative on a set of infinite measure.
Otherwise, what you said is fine.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?