Prove or disprove: if $T$ is rv and $X$ is coordinatewise measurable, then $X\big(T(\omega),\omega\big)$ is rv

Question

Let $\Omega$ ne a non-empty set and $\mathcal{F}$ be a $\sigma$-algebra of subsets of $\Omega$. Let also $T:\Omega \to \mathbb{R}$ is a random variable (rv) and $X:\mathbb{R}\times \Omega \to \mathbb{R}$ is such that:

for all $t\in \mathbb{R}$ function $X(t,\cdot):\Omega \to \mathbb{R}$ is random variable and

for all $\omega\in \Omega$ function $X(\cdot,\omega):\mathbb{R} \to \mathbb{R}$ is continuous.

Prove or disprove: $\omega \overset{g}{\mapsto} X\big(T(\omega),\omega\big)$ is a random variable.

Attempt. I believe the answer is yes. Let $B$ be a Borel set of $\mathbb{R}$, so:

\begin{eqnarray}g^{-1}(B)&=&\{\omega \in \Omega:g(\omega)\in B\}\nonumber\\ &=&\{\omega \in \Omega: T(\omega)=t\in \mathbb{R},~X(t,\omega)\in B\}\nonumber\\ &=&\bigcup_{t\in \mathbb{R}}T^{-1}(\{t\})\cap X(t,\cdot)^{-1}(B).\nonumber \end{eqnarray} Since $T$ and $X(t,\cdot)$ are random variables, sets $T^{-1}(\{t\})$ and $X(t,\cdot)^{-1}(B)$ belong in $\mathcal{F}$. However, $\mathcal{F}$ being a $\sigma$-algebra, is closed under countable operations on sets. How can we turn $\bigcup_{t\in \mathbb{R}}$ to a countable union in order to guarantee that $g^{-1}(B)$ belongs in $\mathcal{F}$?

Thanks in advance.

Answer 1

$X(T(\omega),\omega)=\lim X([nT(\omega)]/n,\omega)$ where $[x]$ is the greatest integer less than or equal to $x$. It is easy to see that $X([nT(\omega)]/n,\omega)$ is measurable by considering the events $\{\omega: k \leq T(\omega) <k+\frac 1 n\}$.

Answer 2

Approximate $T$ with a sequence of simple functions $T_n\to T$. Then, by continuity of $t\mapsto X(t,\cdot)$, $X_{T_n}\to X_T$ and each $X_{T_n}$ is measurable (here $X_{T}(\omega)=X(T(\omega),\omega)$).