I try to reduce my problem to a more general statement from which I want to know whether this is true in general.

I have a sequence of continuous-time stochastic processes $X_t^{(n)}, t \geq 0$ with values in some Polish space $E$ for which I know that they all are stochastically continuous and jointly measurable. In particular, the paths are Borel measurable. As $n \to \infty$ this sequence convergences in distribution to a stochastic process $Y_t$ which is not necessarily stochastically continuous any more.

  1. Is the limit process $Y_t$ jointly measurable?
  2. If 1. is not true, is it then at least true that $Y_t$ (or some version) has Borel measurable sample paths (or Lebesgue measurable)?

In general, there are of course processes $Y_t$ such that $Y_t$ has non-measurable sample paths, e.g. taking $Y_t \in \{ 0, 1 \}$ uniformly distributed and independent for each $t$. Moreover, this process is not jointly measurable. However, I have a process $Y_t$ that arises as a limit of processes with nice properties.

I hope to have found some suggestions for an answer to question 2 in "Probability With a View Towards Statistics" by Hoffman-Jorgensen, Exc. 9.3-9.6

(i) A set $A \subseteq E^{[0, \infty)}$ is called thick if $E^{[0, \infty)}$ is the only measurable set in the product $\sigma$-algebra $\mathscr{B}(E)^{\otimes [0, \infty)}$ that contains $A$.

(ii) The set $M([0, \infty), E) := \{ \omega : [0, \infty) \to E \ | \ \omega \text{ measurable} \}$ is thick (and also the set of non-measurable paths is thick).

(iii) If $A$ is a thick set then every stochastic process has a version with paths in $A$. In particular, every stochastic process has a version with measurable sample paths (and also a version with non-measurable sample paths).

So, it only remains then to check whether 1. is true in general.

  • $\begingroup$ From my reading of Hoffman-Jorgensen, it seems that one can choose a version of the process in math.stackexchange.com/a/2713056 such that its sample paths are measurable. This seems to be in contradiction to the claim in this answer: math.stackexchange.com/a/1572893. $\endgroup$ – S.Surace Mar 29 '18 at 10:37
  • $\begingroup$ @S.Surace Can you explain, why Hoffman-Jorgensen's statement is a contradiction to math.stackexchange.com/a/1572893 ? If $X(t, \omega)$ is measurable in $t$ for each $\omega$ (or just for almost all $\omega$) then $Y(\omega) := \int h(t) X(t, \omega) \, dt$ is defined (for a $t$-measurable function $h(t)$) for all $\omega$, but $Y(\omega)$ need not be measurable in $\omega$. In other words, $Y$ need not be a random variable and therefore not of interest for a further probabilistic treatment. $\endgroup$ – yadaddy Apr 3 '18 at 8:29
  • $\begingroup$ I'm not sure about this, but isn't the linked answer, second paragraph, saying that the integral is not defined because of the lack of joint measurability? If my understanding of Hoffman-Jorgensen is right, one could fix up all the processes that are not jointly measurable by finding a suitable version so that comment would be moot. But how would those versions look like? This would probably make for a nice separate question. $\endgroup$ – S.Surace Apr 3 '18 at 21:11
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    $\begingroup$ @S.Surace In my eyes, the integral is defined (pointwise in $\omega$, i.e. for almost all $\omega$), but the so resulting function $Y(\omega)$ needs not be measurable in $\omega$ and therefore not of further interest. If $X(t, \omega)$ is jointly measurable, then Fubini-Tonelli guarantees that $Y(\omega)$ is indeed a random variable. $\endgroup$ – yadaddy Apr 4 '18 at 6:14
  • $\begingroup$ Thanks, I think this is the proper way of looking at it! $\endgroup$ – S.Surace Apr 4 '18 at 16:54

I believe 1. is not true in general. Take $X^{(n)}_t$ to be the Ornstein-Uhlenbeck process that is the solution to $dX^{(n)}_t=-n X^{(n)}_t dt+\sqrt{2n} dW_t$. Then $X^{(n)}_t$ is stochastically continuous and jointly measurable, but it converges in probability to a familiy $(Y_t)_{t\geq 0}$ which consists of iid standard normal random variables. This process $(Y_t)_{t\geq 0}$ is not jointly measurable, as shown by an argument in Section 19.5 of the book Counterexamples in Probability by Stoyanov (1987), which goes back to Example 1.2.5 of Kallianpur's Stochastic Filtering Theory (1980).

  • $\begingroup$ Thank you. I have also already found that example in your mentioned references. Here is a similar example: Consider a sequence of Markov chains $X_t^{(n)}$ on two states $\{ 0, 1 \}$ with transition rates $q_{01}^{(n)} = n$ and $q_{10}^{(n)} = n$ and any initial distribution. Each $X_t^{(n)}$ is stochastically continuous and jointly measurable and this sequence converges in distribution to an iid process $Y_t$ on $\{ 0, 1 \}$ with distribution $P(Y_t = 0) = \frac{1}{2} = P(Y_t = 1)$ for each $t$. Such a process is never jointly measurable (similar arguments as in Stoyanov or Kallianpur). $\endgroup$ – yadaddy Apr 3 '18 at 8:37

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