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

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.).

• Can you clarify what is meant by the word "version" here? Different authors use it in different ways. – Nate Eldredge Apr 10 '18 at 16:31
• Off the top of my head, it seems like the Kolomogorov 0-1 law would imply that if $Y$ is measurable then it is zero a.s. That probably leads to a contradiction. – Nate Eldredge Apr 10 '18 at 16:42
• I added the definition that is used in the book of Hoffman-Jorgensen. – S.Surace Apr 10 '18 at 17:02
• So, "version" here means that for any measurable subset $A$ of $\mathbb{R}^{[0,1]}$, we have $P(X \in A) = P(\tilde{X} \in A)$? Thanks for the clarification, because that wasn't either of the two definitions that I had in mind. – Nate Eldredge Apr 10 '18 at 18:16
• That‘s correct. – S.Surace Apr 10 '18 at 18:47

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

• A couple of questions: 1) Since we already know that $X$ is not jointly measurable, we cannot use Fubini to exchange integrals. Are you using some other theorem that applies even when joint measurability fails? 2) Assuming that we can show $Y=0$, could you please elaborate on the step leading to $X=0$? 3) Do the equalities hold for all $\omega$ and $t$ respectively or for almost all of them? Are you claiming, essentially, that if $Y$ is measurable and $X$ is bounded, then $Y=0$ a.s. and $X_t(\omega)=0$ for all $\omega$ and almost every $t\in[0,1]$, which is a contradiction to Var($X_t$)=1? – S.Surace Apr 11 '18 at 10:13
• Also, I assume that all the arguments are intended for $\tilde X$, the version with measurable sample paths (the time integral is not defined for $X$). – S.Surace Apr 11 '18 at 10:15
• What happens if $Y$ is not square integrable? – S.Surace Apr 12 '18 at 21:16
• I think everything about my suggestion might be wrong and I really don't know. For example, the conclusion that the $Y$'s all have the same distribution. For any process in my experience that change change of variable argument works, but I am just not sure about this thing. But, almost wlog you might assume $X_t$ is bounded, since you can bound and recenter if necessary without changing the essential feature, and so the Y's would be as well. My apologies if I've just wasted your time. – user83457 Apr 13 '18 at 7:35
• No I think the second part of the answer is very good. If we could show that the Y‘s are zero without invoking variance we would be done. As for the wlog part: do you want to use a sigmoidal to transform the X to an RV taking values in [0,1], or use a cutoff? – S.Surace Apr 17 '18 at 13:22

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."

• If I'm not mistaken, this is essentially the same argument as given in Stoyanov, which proves that the family of RV's can't be jointly measurable, iid, and have non-zero variance simultaneously. So, if one insists on the independence and $X_t$ having unit variance, the process cannot be jointly measurable. Maybe my question is not phrased clearly enough. But I would like to prove that, if we are willing to give up joint measurability, we cannot define a path-wise integral of the original process as a random variable. But maybe this Proposition provides just that and I fail to see it. – S.Surace Apr 12 '18 at 16:50
• @S.Surace In this version, the argument shows that we cannot find a jointly measurable version of the process with measurable sample paths. There are also more direct arguments why a continuum of iid variables will have sample realizations that are not measurable, There are also ways to overcome the issue by working with larger probability spaces and a proper extension of product measures, see here. – Michael Greinecker Apr 12 '18 at 19:48
• Yes, I understand that it shows that we cannot find a jointly measurable version of the process with measurable sample paths. But we also know, as per my question, that there is a version that fails joint measurability, but still has measurable sample paths. Does it follow from this Proposition that the pathwise integral is not measurable? Why is giving up joint measurability not a fix (which you seem to be implying when you say that there is no fix)? – S.Surace Apr 12 '18 at 21:06
• @S.Surace Can you tell me where in Hoffman-Jorgensens book(s) one can find the modification? my scepticism is based on the idead that iid and Lebesgue measurability are largely incompatible. By Lusin's theorem, meaasurable functions are continuous on a very large set. But continuity means that the value of a function depends on the value of close-by points, which violates independence. – Michael Greinecker Apr 12 '18 at 22:54
• It's in volume 2, page 108, Exc. 9.3-0.6. See also math.stackexchange.com/questions/991413. – S.Surace Apr 12 '18 at 23:10