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…
Best Answer
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 $f(x)$.