show that $\int_Xfd\mu = (\mu \times \lambda)(\{(x,y)\} \in X \times \mathbb{R};0\leq y\leq f(x))$

Question

The question is:

let f be a non-negative measurable function on a sigma-finite measure space $(X,\mathbb{F},\mu)$ and let $\lambda$ denote the lesbesque measure on $\mathbb{R}$. Show that:

$\int_Xfd\mu = (\mu \times \lambda)(\{(x,y)\} \in X \times \mathbb{R};0\leq y\leq f(x))$

To prove this i'm going to try proving it for indicator functions and simple functions first so i can eventually maybe use the Beppo Levi theorem.

The problem is that the notation is quite confusing if i take $f(x) = 1_A(x)$ i get:

$\int_Xfd\mu = \int_X1_A(x)d\mu = I_mu(A) = \mu(A)$

I don' t see how to go further, how do i get an cartesian product with the lesbesque measure?

Thanks for the help in advance!

Kees

Answer 1

Hint: Write $$\int_X f(x) \, d\mu(x) = \int_X \left( \int_0^{f(x) }1 \, \lambda(dy) \right) \, d\mu(x) = \int_X \left( \int_{\mathbb{R}} 1_{[0,f(x)]}(y) \, \lambda(dy) \right) \, d\mu(x).$$