Calculating a product measure with Fubini-Tonelli's theorem

Question

I am kind of lost on the following problem. Let $${f : (X, \mathcal{A}, \mu) \rightarrow (\mathbb{R_{\geq}}, \mathcal{B}(\mathbb{R_{\geq}}), \lambda)}$$ with ${\lambda}$ being the Lebesgue-measure.

1) Show that ${E \in \mathcal{A} \otimes \mathcal{B}(\mathbb{R_{\geq}})}$ with $${E:= \{(x,y) \in X \times \mathbb{R_{\geq}} | x \in X \wedge 0 \leq y < f(x) \}}$$

For this, my approach was $${\mathcal{A} \otimes \mathcal{B}(\mathbb{R_{\geq}}) = \sigma(\mathcal{A} \times \mathcal{B}(\mathbb{R_{\geq}}) \cup \mathcal{B}(\mathbb{R_{\geq}}) \times \mathcal{A})}$$ $${X \times \mathbb{R_{\geq}} \subset \mathcal{A} \times \mathcal{B}(\mathbb{R_{\geq}}) \subset \mathcal{A} \times \mathcal{B}(\mathbb{R_{\geq}}) \cup \mathcal{B}(\mathbb{R_{\geq}}) \times \mathcal{A} \subset \mathcal{A} \otimes \mathcal{B}(\mathbb{R_{\geq}})}$$

at which point, if it is correct, this would be proven. Now the second part is where I am really stuck:

2) Calculate ${\mu \otimes \lambda (E)}$ using Tonelli's theorem, to prove the following equation: $${\int f d\mu = \int_0^\infty \mu(\{f > y\} \lambda(dy))}$$

I cannot comprehend how to solve ${\mu \otimes \lambda (E)}$, since in my understanding it would just be something like ${\mu \otimes \lambda (E) = \mu(E_1) * \lambda(E_2)}$, with ${E_1 \in X}$ and ${E_2 \in \mathbb{R_\geq}}$.

So I started by rewriting the equation to $${\int_0^\infty \mu(\{f> y\}) \lambda(dy) = \int_0^\infty(\int_0^y \chi_{\{ f > y\}} \mu(dx)) \lambda(dy)}$$ so that I can switch the integrals and end up with $${\int_0^\infty \lambda(\{f>y\}) \mu(dx)}$$ Is this a correct approach? Because I cannot see it going anywhere near the needed solution and I do not know how else to apply the theorem of Tonelli/Fubini...

Answer 1

First part is wrong. write $E$ as $\cup_{r\in \mathbb Q} (f>r) \times ([0,r)$ to see that $E$ belongs to the product sigma algebra. For the second part you have started correctly. Just note that $\lambda (\{f(x)>y\})$ is nothing but the Lebesgue measure of the interval $[0,f(x))$ which is $f(x)$.

Answer 2

The other answer covers it, but here is a somewhat different approach.

To do the first part, we use a trick: set $\alpha(x,y)=(f(x),y)$ and $\beta (x,y)=x-y$. Then $\beta\circ \alpha(x,y)=f(x)-y.$ Now, $\beta$ is obviously continuous and it's not hard to show that $\alpha$ is measurable. Then, $E=\{\beta\circ \alpha>0\}$ is measurable.

And for the second part, we can use a result that does not use the full strength of Tonelli. Namely,

$\mu\times \lambda(E)=\int_X\lambda(E_x)d\mu=\int_Xf(x)d\mu=\int^{\infty}_0\mu(E^y)d\lambda=\int^{\infty}_0\mu(\{f>y\})d\lambda$