Is $1_{[0,X(\omega)]}(y)$ measurable?

Question

About https://staff.fnwi.uva.nl/s.g.cox/mtp_2017.pdf problem 5.11

" Let $F$ be a distribution function of a non-negative random variable $X$ and $\alpha>0$. Show that \begin{align} EX^{\alpha} = \alpha\int_0^{\infty} x^{\alpha-1}(1-F(x))dx \end{align} "

The hint they give is to write $EX^{\alpha} = Ef(X)$ with $f(x) = \int_0^x \alpha y^{\alpha-1}dy$.

So $EX^{\alpha} = \int \int_0^X \alpha y^{\alpha-1} dy dP$.

Now I want to use Fubini, but to do so, we need that $[0,\infty)\times \Omega: (y,\omega)\mapsto 1_{[0,X(\omega)]}(y)$ is measurable. Because then I can switch $dP$ and $d\lambda$ at $\int\int \alpha y^{\alpha-1} 1_{[0,X)}(y) d\lambda(y) dP$, where $\lambda$ is here the Lebesgue measure restricted on $[0,\infty)$.

If $y$ is given, then $1_{[0,X)}(y)$ is just a composition of an indicator function and $X$, so that is measurable. And if $\omega$ is given, then we also get an indicator function, but now dependable of $y$. But this says nothing about measurability in two dimensions.

I don't see a specified $\sigma$-algebra, so maybe we can just make one such that this function satisfies it (?)

Has anobody an idea how to deal with this? I thank in advance because I can't do that in the comments and of course I want to be grateful.

Answer 1

1. Show that the mapping $$\mathbb{R}^2 \ni (x,y) \mapsto g(x,y) := 1_{[0,x]}(y)$$ is (Borel)measurable. (Hint: It suffices to show that $g^{-1}(\{1\})$ is Borel.)
2. Show that the mapping $$\Omega \times \mathbb{R} \ni (\omega,y) \mapsto h(\omega,y) :=(X(\omega),y) \in \mathbb{R}^2$$ is measurable. (Here, $\Omega \times \mathbb{R}$ is endowed with the product $\sigma$-algebra.)
3. Conclude that $$(\omega,y) \mapsto 1_{[0,X(\omega)]}(y)$$ is measurable.