- #1
QIsReluctant
- 37
- 3
Homework Statement
Let g: ℝ → ℝ be Lebesgue measurable, let μ(A) be a measure of Lebesgue measurable sets defined by
μ(A) = ∫A f dx, where f: ℝ → ℝ is non-negative, with finite integral on compact intervals.
Prove that
∫ℝ g(x)dμ(x)
exists if and only if
∫ℝ g(x)f(x) dx
exists, in which case the two are equal
Homework Equations
N/A[/B]
The Attempt at a Solution
I'm reviewing the definitions/construction of Lebesgue integrals but that is not leading to anything fruitful. I have that the first integral exists iff the integral of |g| exists, which exists iff the sup of all integrals of simple, bounded, measurable, finitely supported functions less than |g| is finite ... and then I am stuck.
[/B]
I suppose that seeing the equation
∫|g|dμ = sup Σai μ(Ai) = sup Σai ∫Ai f dμ
(aiχAi being some simple function approximating |g|) causes me to "get" the reason for the equality -- if we imagine the Ai partitioning ℝ and getting smaller and smaller (small enough to encompass some single point x) we see that the a_i ≈ g(x) and that the integral ≈ f(x)dx. But actually showing the truth of this rigorously is an entirely different matter. Please advise me! Thanks.