I'm teaching a seminar on probability theory and I want to motivate why joint measurability of a stochastic process is important. The following seems to be the canonical counterexample for a process that is not jointly measurable (see Section 19.5 of the book Counterexamples in Probability by Stoyanov (1987)).

Let $X=(X_t)_{t\in [0,1]}$ be a stochastic process consisting of mutually independent random variables with zero mean and unit variance. We can show that $X$ cannot be jointly measurable, i.e. when regarded as a map
$\Omega \times [0,1]\to \mathbb{R}$. However, by a discussion in Probability With a View Towards Statistics by Hoffman-Jorgensen, there is a version $\tilde X$ of $X$ with Lebesgue-measurable sample paths. In that book, *version* is defined as follows (paraphrased):

Let $X,\tilde X: \Omega\to S^T$ be $\mathcal{F}$/$\Sigma^T$-measurable maps defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, where $(S,\Sigma)$ is a measurable space and $T$ an index set. Then $\tilde X$ is said to be a version of $X$ if they are equal in law, i.e. $\mathcal{L}_X=\mathcal{L}_{\tilde X}$, where $\mathcal{L}_X=X_*\mathbb{P}$, $\mathcal{L}_{\tilde X}=\tilde X_*\mathbb{P}$ are the measures induced on $(S^T,\Sigma^T)$.

Let

$$ \begin{align} Y: \Omega&\to\mathbb{R}, \\ \omega&\mapsto\int_0^1\tilde X_t(\omega)\mu(dt), \end{align}$$ where $\mu$ is the Lebesgue measure on $[0,1]$.

I would like to show that **$Y$ is not measurable** and hence not a random variable.

(if $X$ were jointly measurable, the Fubini-Tonelli theorem would guarantee that $Y$ is also measurable, which serves as a motivation for why joint measurability might be interesting. But unless we can prove the non-measurability of $Y$ above, we may not be convinced that it is important.).