# Measurability of an uncountable union

Let $\{X_i\}_{I \in I}$ be a family of real-valued random variables on $(\Omega,\mathcal{A},P)$ for an uncountable index set $I$, which is the sample space of $(I,\mathcal{F},Q)$.

On the product space $(I\times\Omega,\mathcal{F}\otimes\mathcal{A},P\times Q)$ define $$Y((i,\omega)):=X_i(\omega), \ (i,\omega)\in I\times\Omega.$$

Is $Y$ a random variable, i.e. is $Y$ $(\mathcal{F}\otimes\mathcal{A})$-measurable?

• To clarify my problem, assume $B \in \mathcal{B}(\mathbb{R})$. Then I expect $Y^{-1}(B)=\bigcup_{i \in I}\{i\}\times X_i^{-1}(B)$. Assuming that each singleton $\{i\} \in \mathcal{F}$ this is still an uncountable union of measurable sets. – Daniel Lingohr Aug 23 '18 at 14:41

Let $I$ be the interval $[0,1]$ with Lebesgue measure.
Let $\Omega$ be a single point.
Let $A$ be a non-measurable subset of $[0,1]$, and let $X_i={\bf 1}(i\in A)$. Then $Y^{-1}(\{1\})=A\times \Omega$, so $Y$ is not measurable.
• @DanielLingohr Here's what I know; if $I$ is an interval, and the stochastic process $(X_t)_{t\in I}$ is separable, then $X_t^*:=\limsup_{s\to t^+} X_s$ is measurable (proof). The separability condition is not so restrictive; as Wikipedia mentions, any process can be modified to be separable. But for this Doob's result to be of use you need $X_t$ to be right continuous, say. – Mike Earnest Aug 23 '18 at 16:12