Suppose for each $x\in[0,1]$ I have random variable $f_x$ defined on $(\Omega_x,\mathcal{F}_x,\mathbb{P}_x)$ where $f_x \sim \operatorname{Ber}(x)$; i.e. $f_x=1$ w.p. $x$ and $f_x=0$ w.p. $1-x$.
If I understand this post correctly, can construct a probability space $(\Omega,\mathcal{F},\mathbb{P})$ as an infinite product space of the $(\Omega_x,\mathcal{F}_x,\mathbb{P}_x)$ for all $x\in[0,1]$, and I also believe that $f$ will be a random variable on $(\Omega,\mathcal{F},\mathbb{P})$.
Based on my application, I would like to compute, $$ \mathbb{E}_{\Omega} \left[ \int_0^1 f(x) dx \right]. $$
However, since $f_{\omega}$, $\omega\in\Omega$ is the indicator function of some set on $[0,1]$, it seems possible that $f$ may not be Lebesgue integrable.
Is there another way I can interpret $\int_0^1 f(x) dx$ to avoid this? (i.e. not Lebeguse).
Otherwise, is it possible to compute $\mathbb{P}[ \{f : f \text{ is not Lebeguse measurable} \} ]$?
Perhaps another possible workaround for me would be to define my product space on $[0,1]\cap (\beta\mathbb{Z})$ where $\beta$ is very small so that I can use a sum to approximate the integral?
