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

Question

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.

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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.

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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?

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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?)

Answer 1

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.