On the structure of a probability space for the extension of a measure via a martingale

This is a question about theorem 3.2.2 of James Norris's notes on advanced probability (http://www.statslab.cam.ac.uk/~james/Lectures/ap.pdf).

Let $$(\Omega,\mathcal{F}, (\mathcal{F}_n)_{n∈ \mathbb{N}}, \mathbb{P})$$ be a filtered probability space with $$\mathcal{F} = \mathcal{F}_{∞}$$ and let $$(X_n)_{n∈ \mathbb{N}}$$ be a martingale on this probability space. Then the above theorem affirms that by Caratheodory $$\mathbb{E}[X_T] = 1$$ for all "finite" stopping times $$T$$ if and only if there exists a probability measure $$\tilde{\mathbb{P}}$$ on the filtered space $$(\Omega,\mathcal{F}, (\mathcal{F}_n)_{n∈ \mathbb{N}})$$ such that

$$\tilde{\mathbb{P}}(A) = \mathbb{E}[X_n\mathbb{1}_{A}]\qquad \mbox{for all A∈ \mathcal{F}_n}$$

My first question is about the meaning of "finite" here. Is it correct that this should be strictly understood in the sense that $$T:\Omega→ \mathbb{N}$$ (i.e. this is not in the almost-sure sense)?

Secondly, this suggests that the extension solely depends on the statistical properties of $$(X_n)_{n∈ \mathbb{N}}$$: is it correct that in this case the extension does not depend on structure of $$\Omega$$? e.g. say two martingale $$X = (X_n)_{n∈ \mathbb{N}}, X' = (X'_n)_{n∈ \mathbb{N}}$$ are "defined identically" on two different probability spaces $$(\Omega,\mathcal{F}, (\mathcal{F}_n)_{n∈ \mathbb{N}}, \mathbb{P})$$ and $$(\Omega',\mathcal{F}', (\mathcal{F}'_n)_{n∈ \mathbb{N}}, \mathbb{P}')$$, i.e. for all Borel measurable subset $$A$$ of $$\mathbb{R}^{\mathbb{N}}$$,

$$\mathbb{P}(X∈A)=\mathbb{P}'(X'∈ A)$$

and set

$$\tilde{\mathbb{P}}_n(B) = \mathbb{E}[X_n\mathbb{1}_B]\qquad \text{for all B ∈ \mathcal{F}_n}$$

$$\tilde{\mathbb{P}}_n'(B) = \mathbb{E}[X_n'\mathbb{1}_{B'}]\qquad \text{for all B' ∈ \mathcal{F}'_n}$$

Now, say an extension $$\tilde{\mathbb{P}}$$ of $$(\tilde{\mathbb{P}}_n)_{n∈ \mathbb{N}}$$ is achieved by other means (say Kolmogorov extension) on $$\Omega$$. Does that automatically imply that the extension $$\tilde{\mathbb{P}}'$$ of $$(\tilde{\mathbb{P}}_n')_{n∈ \mathbb{N}}$$ can be achieved on $$\Omega'$$?

If not, are there conditions so that this can be done (without any condition of uniform integrability on $$(X_n)_{n∈ \mathbb{N}})$$?

Norris's proof only uses everywhere finite $$T$$ for the "only if" part of the assertion; the "if" assertion is true even for a.s. finite $$T$$.
As to your second question, consider the following example in the simplest case $$X_n=1$$ for all $$n$$. Let $$\Omega:=\Bbb R^{\Bbb N}$$, with coordinate maps $$Y_k(\omega) = \omega_k$$ for $$\omega=(\omega_1,\omega_2,\ldots)\in\Omega$$. Let $$\mathcal F_n:=\sigma(Y_1,\ldots,Y_n)$$. Let $$\Bbb P$$ be the probability measure on $$(\Omega,\mathcal F)$$ under which $$Y_1,Y_2,\ldots$$ are i.i.d. standard normal random variables. For $$\Omega'$$ I take $$\{\omega\in\Omega:\lim_k\omega_k=0\}$$, and then $$\mathcal F'_n$$ to be the trace of $$\mathcal F_n$$ onto $$\Omega'$$. For each $$n\in\Bbb N$$ there is a unique probability measure $$\Bbb P'_n$$ defined on $$(\Omega',\mathcal F'_n)$$ under which $$(Y'_1,\ldots,Y'_n)$$ are i.i.d. standard normal. (Here $$Y'_k$$ is the restriction of $$Y_k$$ to $$\Omega'$$.) The family ($$\Bbb P'_n)$$ is consistent in the sense of the Kolmogorov extension theorem. But there is no probability measure on $$\mathcal F'_\infty:=\sigma(\cup_{n\in \Bbb N}\mathcal F'_n)$$ whose restriction to $$\mathcal F'_n$$ is equal to $$\Bbb P'_n$$. If there were such a probability, call it $$\Bbb P'$$, then we would have $$\Bbb P'(\Omega')=1$$, while clearly $$\Bbb P(\Omega')=0$$, which leads to a contradiction.
• Hi John, thank you very much for your answer :) However, I feel like your couter-example doesn't really answer the second question in the first place as there is no original probability measure on $\Omega'$ and hence $\mathbb{P}'_n$ is not really defined via a martingale with respect to this probability measure. I will edit the post to clarify that. Commented Jul 26, 2023 at 10:08
• The lack of a $\Bbb P'$ defined on $\mathcal F_\infty'$ is sort of the point. Because the $\Bbb P'_n$s are consistently defined, there is an additive set function, call it $\tilde{\Bbb P}$ defined on $\cup_n\mathcal F'_n$ such that $\tilde{\Bbb P}(A) = \Bbb P_n(A)$ for each $A\in\mathcal F'_n$ and for each $n$. But $\tilde{\Bbb P}$ doesn't extend to $\mathcal F'_\infty = \sigma\left(\cup_n\mathcal F'_n\right)$ as a probability measure. The extension issue thus depends on the measurable space one's working on. Commented Jul 26, 2023 at 14:24
• I see your point. However, I guess the question then if the following. If $\Omega'$ has enough structure such that $(\Omega', \mathcal{F}'_{∞}, \mathcal{F}_n', \mathbb{P}')$ is a filtered probability space, does that imply that the extension for $\tilde{\mathbb{P}}'$ can be done? Commented Jul 27, 2023 at 11:19