Let, $(\Omega, A, \mu)$ be a sigma-finite measure space and $f : \Omega \to \Bbb R_{\ge 0}$ . Consider, $(\Omega \times [0,+\infty),A \otimes \beta([0,+\infty)), \mu \otimes \lambda)$ and the set $C:=\{(\omega,t) \in \Omega \times [0,+\infty): f(\omega) \ge t\}$.
Then prove that,(i) $C \in A \otimes \beta([0,+\infty))$ and (ii) $\mu \otimes \lambda(C)=\int f d\mu$
My attempt:
Could do nothing with (i).
Tried (ii) : $$\mu \otimes \lambda(C)=\int_{\Omega} \lambda(C_{\omega}) d\mu=\int_{\Omega} \lambda\{t \in [o,+\infty):f(\omega)\ge t\} d\mu=\int_{\Omega}\int_{0}^{f(\omega)} d\lambda \text{ }d\mu$$ $$=\int_{\Omega} \int_{0}^\infty1_{f(\omega)\ge t} d\lambda \text{ } d\mu= \int_{0}^\infty (\int_{\Omega}1_{f(\omega) \ge t} \text{ } d\mu) d\lambda $$ $$=\int_{0}^\infty\mu(f(\omega) \ge t)d\lambda$$
I have reached here!
Thanks in Advance for help!
You already got $(\mu \otimes \lambda)(C)=\int_{\Omega} \int_0^{f(\omega)} d\lambda d\mu(\omega)$. Isn't this $\int_{\Omega} f(\omega) d\mu(\omega)$ or $\int f d\mu$?. To prove measurability of $C$ write $C^{c}$ as $\cup_{r\in \mathbb Q} \{(\omega ,t): f(\omega) <r)\}\cap \{(\omega ,t): r<t\}$.