Suppose we have a measurable space $(\Omega,\mathcal{F})$ and an $\mathbb{R}$-valued continuous-time (but not necessarily continuous) stochastic process $X$. $X$ is jointly measurable if it is measurable with respect to the product $\sigma$-algebra $\mathcal{B}(\mathbb{R}^+)\otimes \mathcal{F}$.

An example of a process which is not jointly measurable is $X(t,\omega)=\mathbf{1}(t\in V)$, where $V$ is some non-Borel-measurable set. All left or right continuous processes are jointly measurable.

As a matter of curiosity, I tried constructing a martingale which is not jointly measurable, but didn't get very far. How might I do it?



The reason I don't think there is any answer to this question that will be very enlightening is this: the use of $\mathcal{B}(\mathbb{R}^+)$ is arbitrary. This $\sigma$-algebra is not used in the definition of the stochastic process, nor in the definition of a martingale. It's just a 'nice' and common $\sigma$-algebra on the positive reals.

If you look at the proof that Brownian motion is jointly measurable you will see that you don't do much work with $\mathbb{R}^+$, most of the argument is done with sets and measurability in the probability space $\Omega$. The relevance of $\mathcal{B}(\mathbb(R)^+)$ is pretty much another way of expressing the continuity of Brownian motion.

So my 'cheating' answer is this: replace $\mathcal{B}(\mathbb{R}^+)$ by some artificially smaller $\sigma$-algebra, such as that generated by intervals of the form $(x,y)$, where $x \in [0, 1]$ and $y \in [2, \infty)$. Then you can show that say, Brownian motion is not jointly measurable with respect to the product of this $\sigma$-algebra and $\mathcal{F}$.

Obviously this is quite unsatisfactory, especially if you have been searching for clever tricks to answer your question 'properly'. But I would argue that any proper answer is just going to be doing the same thing I have done here: using one $\sigma$-algebra in the definition of jointly measurable, thereby claiming that it is of natural importance in the definition of stochastic processes, and then using a set not in this $\sigma$-algebra in the definition of your process, contradicting the claim that it was natural. This is what your non-martingale non-jointly measurable process does.

  • $\begingroup$ Thanks for taking a look. Hopefully your comments will inspire someone to provide a concrete answer. $\endgroup$ – Ben Derrett Dec 22 '13 at 18:52

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