# Measurability and $\mathbb P$-a.s. or everywhere

I have two questions that are quite related:

1. The predictable $$\sigma$$-algebra $$\mathcal P$$ is defined as the smallest $$\sigma$$-algebra making all adapted left-continuous processes measurable. Does left-continuity need to be everywhere or does $$\mathbb P$$-a.s. also work?

2. If a stochastic process $$X:\Omega\times [0,\infty)\to\mathbb R$$ (thus, $$X(\cdot ,t)$$ is a random variable for all $$t$$) is $$\mathbb P$$-a.s. continuous, is it jointly measurable?

I have seen many related questions like this on StackExchange, but often the usual conditions are not stated, hence my questions have not really been adressed. Therefore, assuming the usual conditions, what are the answers to 1. and 2.?

Importantly note: I am aware that $$X$$ is indistinguishable from a jointly measurable map and that $$X:\Omega'\times [0,\infty)\to \mathbb R$$, where $$\Omega'=\Omega\backslash N$$ with $$N$$ as below, is jointly measurable.

A possible rephrasing of 1 and 2 (in more generality): if $$X$$ is $$\mathcal A$$-measurable, and $$Y$$ is indistinguishable from $$X$$, is $$Y$$ also $$\mathcal A$$-measurable? If this does not hold in generality, then let's get back to 2.

My try on 2.: I know $$X$$ is indistinguishable from a process $$Y$$ with continuous paths everywhere. Hence let $$N$$ be the $$\mathbb P$$-null on which $$X$$ and $$Y$$ do not coincide. We can write $$X^{-1}(A)=(X^{-1}(A)\cap (N\times [0,\infty)))\cup (X^{-1}(A)\cap ((\Omega\backslash N)\times [0,\infty))),$$ where the latter is $$\mathcal F\times \mathcal B$$-measurable, but the first part I do not know.

Edit: Do a.s. right-continuous paths imply product measurability has a useful answer to part 2. Then why do we even start with $$\mathbb P$$-a.s. caglad and caglad processes, for instance. If I want to look at the Lebesgue-Stieltjes integral $$\int _0^tHdA,$$ then $$H$$ needs to be jointly measurable, so then we take the version of $$H$$ that is jointly measurable? Seems like an enormous detour. Also, one often shows the inclusions $$\mathcal P\subset \mathcal O\subset \mathcal M\subset \mathcal B\times \mathcal F,$$ the predictably, optional, progressively, and product sigma algebra. But these are the generating $$\sigma$$-algebra where the propery then holds everywhere?

Clarification regarding question 1.:
What is the correct definition of $$\mathcal P$$?

a. $$\mathcal P=\sigma(X:X$$ is an adapted processes with left-continuous paths everywhere);

b. $$\mathcal P=\sigma(X:X$$ is an adapted processes with left-continuous paths a.e.);

Or does a. and b. give the same result? (Do not think so by the way. Can you work with b. in the first place?)

• (1) Remove the singular set $S$, that is the $\mathbb{P}$-nefligible set there the paths of your process $\omega\mapsto X(\omega,\cdot)\in\mathbb{R}^{[0,\infty)}$ is not left continuous. Working woth $\Omega'=\Omega\setminus S$ does not bring any difficulties. (2) yes it is (again, just remove the singular set). May 25, 2021 at 15:17
• @OliverDiaz, thank you for your reply. About (2): you mean it is jointly measurable seen as map $X:\Omega'\times [0,\infty)\to\mathbb R$? But is my original map jointly measurable?
– Mark
May 25, 2021 at 15:20
• Yes the probability space is complete, that is one of the two assumptions of ''assuming the usual conditions''.
– Mark
May 25, 2021 at 16:43
• I agree and I am completely aware this works. What question of mine is your answer supposed to tackle @OliverDiaz?
– Mark
May 25, 2021 at 19:51
• from Kavi Rama Murty's example you can see that if you consider the history of your process including the bad sets, you don't get any thing useful for stochastic Calculus, the process is basically $0$. Maybe someone has concrete example where the bad set needs to be kept ($\Omega$ and filtrations would be very concrete to (an interval perhaps, and the bas set a Cantor set where something interesting happens)? Anyway, hopefully someone will have such thing. May 25, 2021 at 20:01

1. This question is ambiguously worded — what is "or does $$\Bbb P$$-a.s. also work?" supposed to entail? One interpretation: "Is it true that if $$X$$ is indistinguishable from an element of $$\mathcal P$$, then $$X\in\mathcal P$$ as well?" The answer to this is clearly NO, as the example of Kavi Rama Murthy that you cite shows: If $$A\in\mathcal F$$ and $$P(A)=0$$ then any process of the form $$(\omega,t)\mapsto g(t)1_A(\omega)$$ is indistinguishable from the zero process (an element of $$\mathcal P$$), but any $$X\in\mathcal P$$ has the property that each section $$t\mapsto X_t(\omega)$$ is Borel measurable, which fails for the example if $$\omega\in A$$.

2. NO. The example in my reply to 1. is a.s. continuous but not jointly measurable.

Suppose $$H$$ is bounded. To form the (Lebesgue-Stieltjes) integral $$I_t(\omega):=\int_0^t H_s(\omega)dA_s(\omega)$$ you only need $$s\mapsto H_s(\omega)$$ to be measurable (and $$A$$ to be increasing and right-continuous, say). If you want $$I_t$$ to be a random variable, then you need $$H$$ (restricted to $$\Omega\times[0,t]$$) [to be $$\mathcal F\otimes\mathcal B[0,t]$$-measurable. If, in addition, you want $$I_t$$ to be $$\mathcal F_t$$-measurable, then you need $$H$$ (restricted to $$\Omega\times[0,t]$$) [to be $$\mathcal F_t\otimes\mathcal B[0,t]$$-measurable. If you want $$I$$ to be an adapted process, then you need the measurability hypothesis of the previous sentence to be true for all $$t$$; that is, you need $$H$$ to be progressive.

• Thank you for your answer. I agree 1. may be vague. Therefore, I have added a clarification of 1. in my post. If it is still unclear, please tell me. Further, I am aware of the comments you make further on. But this does not really explain why many authors consider cadlag and caglad process with regularity a.s. first, and then talk about these integrals (since they may be undefined due to the fact we have seen that an indistinguishable process may not be measurable).
– Mark
May 25, 2021 at 17:36
• @Makl: When one constructs integration over a $P$-a.s. cad lag local martingale, the singular set were the the cad lag property is not satisfied is removed. It does not make much sense to keep it (unless there is some other process that has some other property of interest with respect another measure that is singular relative to $P$) I have never seen that in applications of stochastic Calculus. When one develops integration with respect say Brownian motions, it is natural to assume that all paths are continuous (the singular set is not important). Work with the nice version of the process. May 25, 2021 at 17:54
• @Mrk: I think that stems from the fact that one first construct the object or prove its existence (fBrownian motions, Levy Porcesses, diffusions with jumps, etc) first according to certain distributional properties and then, after studying other aspects of the processes (regularity of the paths) once can demonstrate that they have this or that property almost surely. Then one works with the "nice" version. May 25, 2021 at 18:17
• In the spirit of your clarification, $\mathcal P_a$ is strictly small than $\mathcal P_b$, using the example provided in my answer. The allowance for continuity (or right continuity, or left continuity) to occur merely a.s. is an acknowledgement of the fact that something like a stochastic integral is only determined a.s. If one is tidy, then $\int_0^t H_s dB_s$ can be chosen to be a version of the stochastic integral that is everywhere continuous. May 25, 2021 at 18:56
• $\mathcal P_a$ is more useful than $\mathcal P_b$ because each of its elements is jointly measurable. And given $X\in\mathcal P_b$ there is a $Y\in\mathcal P_a$ that is indistinguishable from $X$, so just trade in $Y$ for $X$ and be done with it! May 25, 2021 at 19:31