I have a question from Rick Durrett's Probability Theory and Examples. In that book,
Lemma 2.2.8 If $Y \ge 0$ and $p > 0$, then $E(Y^p) = \int^{\infty}_{0} py^{p-1}P(Y>y)\,dy $
Proof. Using the definition of expected value, Fubini's theorem (for nonnegative random variables), and calculating the resulting integrals gives
\begin{array} $\int py^{p-1}P(Y>y)dy &= \int^{\infty}_0 \int_{\Omega} py^{p-1} 1_{(Y>y)} dP dy\\& = \int_{\Omega}\int^{\infty}_0 py^{p-1} 1_{(Y>y)} dy dP \\&=\int_{\Omega}\int^{Y}_0 py^{p-1} dy dP = \int_{\Omega}Y^p dP = EY^P \end{array}
So, my question is about the application of Fubini's theorem here. As far as I know, if we are going to use Fubini's theorem, the integrand function $f(y,\omega) = py^{p-1} 1_{(Y(\omega)>y)} $ has to be measurable when it is nonnegative. (Here, $Y$ is a Random Variable on the probability space $(\Omega, F, P)$.)
So, we should check that for $\forall a \in \mathbb{R}$, $\{(y,\omega); py^{p-1} 1_{(Y(\omega)>y)}<a\} $ has to be measurable on the product measure space. Since $py^{p-1}$ is trivially a measurable function, we can only check $1_{(Y(\omega)>y)}$ is a measurable function, in other words, $\forall a \in \mathbb{R}$, $\{(y,\omega); 1_{(Y(\omega)>y)}<a\} $ is measurable.
When $a > 1$, it is trivial since $\{(y,\omega); 1_{(Y(\omega)>y)}<a\} = \mathbb{R} \times \Omega $. However, when $a \le 1$, I cannot show that $\{(y,\omega); 1_{(Y(\omega)>y)}<a\}$ is measurable. So could anyone tell me how to show $f(y,\omega) = py^{p-1} 1_{(Y(\omega)>y)} $ is measurable on the product measure space?
When $a \leq 1$ what you need is measurability of $\{Y \leq y\}$. Equivalently we have to show measurability of $\{Y > y\}$. Write this as $\cup_{q \in \mathbb Q} \{Y>q>y\}$ and note that $\{Y >q \} \cap \{y<q\}$ is measurable on the product.