Non-measurability of time integral of non-jointly measurable process

Question

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.).

Answer 1

First, as per Nate Eldredge, the rv would have to be zero. Assume, mostly for convenience, that the $X_t$ are bounded. Here are a couple of arguments: With $Y = \int_0^1X_t dt$,compute the variance.
$$E Y^2 = \int_0^1 \int _0^1EX_tX_s dt ds = \int \int 1_{(t=s)}(t,s) dtds = 0 $$ As the rvs are all bounded there is no issue with interchange of order of integration.$$$$ Secondly, use scaling. It is easy to see that for any interval $I$ $$\frac 1 {|I|}\int_IX_tdt $$ has the same distribution as $Y$, and that for disjoint $I_1, I_2$ the corresponding integrals are independent. With $Y_1 = 2\int_0^{\frac 1 2}X_t, Y_2=2\int_{\frac 1 2}^1 X_t$ $$Y = \frac {Y_1 + Y_2} 2$$ where all $Y$'s are i.i.d. They are also bounded. Take the variance of both sides to conclude $Y$ is 0 $$$$ Second, It follows that $\frac 1 {|I|}\int_IX_tdt = 0$, and therefore by the lebesgue differentiation theorem $X_t = 0$. That can be done using a fixed partition such as the dyadic partition that uses only countably many intervals so I don't think there is an issue with throwing out sets of measure 0.

addressing points raised by op: 1. not jointly measurable. good point, I thought that the second argument was superfluous but maybe not.
Lebesgue differentiation argument. I had in mind that if f is integrable on [0,1] then $E(f| F_n) \rightarrow f$ a.e. where $F_n$ is nth dyadic partition $\lbrace\frac j {2^n} \rbrace$, which certainly converges and I think converges to $f$. The only issue is, do the dyadic interals generate the borel field, and I believe they do, but I didn't prove it. Then, given that $\int_I X_t = 0$ a.e. for all I in the dyadic partitions, you throw out countably many sets of measure 0 where $\int_I X_t \ne 0$ for some dyadic interval, and for $\omega$ the remaining set $X_t(\omega) = 0$ a.e. t

Answer 2

The following is from Proposition 1.1 of this paper by Yeneng Sun, which is quite close to the argument given by Stoyanov. It shows that there exists no fix. By a standard Borel isomorphism argument, $f$ could take values in any Polish space (and need not be bounded).

Proposition: Let $(I,\mathcal{I},\mu)$ and $(X,\mathcal{X},\nu)$ be probability spaces with (complete) product probability space $(I\times X,\mathcal{I}\otimes\mathcal{X},\mu\otimes\nu)$ and $f$ be a jointly measurable bounded function from $I\times X$ to $\mathbb{R}$ such that for $\mu\otimes\mu$-almost all $(i,j)$ the functions $f(i,\cdot)$ and $f(j,\cdot)$ are independent. Then for $\mu$-almost all $i$, the function $f(i,\cdot)$ is constant.

Proof: It suffices to show that $\int_A f(i,x)-\int f(i,\cdot)~\mathrm d\nu ~\mathrm d\mu=0$ for every $A\in\mathcal{I}$ and $\nu$-almost all $x$. For then $(i,x)\mapsto f(i,x)-\int f(i,\cdot)~\mathrm d\nu$ is a Radon-Nikodym derivative of the zero-measure with respect to $\mu\otimes\nu$ and therefore almost surely equal to zero so that the proposition follows.

Now, we get from Fubini's theorem that $$\int_X\bigg(\int_A f(i,x)-\int f(i,\cdot)~\mathrm d\mu\bigg)^2~\mathrm d\nu$$ $$=\int_X \bigg(\int_A f(i,x)-\int f(i,\cdot)~\mathrm d\mu\bigg)\bigg(\int_A f(j,x)-\int f(j,\cdot)~\mathrm d\mu\bigg)~\mathrm d\nu$$ $$=\int_X \bigg(\int_A f(i,x)-\int f(i,\cdot)~\mathrm d\mu\bigg)\bigg(\int_A f(j,x)-\int f(j,\cdot)~\mathrm d\mu\bigg)~\mathrm d\nu$$ $$=\int_{A\times A} \int_X \bigg(f(i,x)-\int f(i,\cdot)~\mathrm d\mu\bigg) \bigg(f(j,x)-\int f(j,\cdot)~\mathrm d\nu\bigg)~\mathrm d\nu~\mathrm d\mu\otimes\mu$$ and the independence condition implies that the last expression is zero. $\Box$

Clearly, the independence condition is preserved between "diffeent versions."