Setwise convergence with martingales

Question

Can someone please tell me whether the following reasoning is correct and/or help me conclude?

Let $(\Omega, \mathcal{F}, P)$ be a probability space equipped with a filtration $\{\mathcal{F}_n \}_n \uparrow \mathcal{F}$, and let $\{P_n\}_n$ be a sequence of probability measures that is uniformly absolutely continuous with respect to $P$.

Suppose the following condition holds: for all $n$ and $F \in \mathcal{F}_n$, \begin{equation}\tag{1} P_n(F) = P_{n+1}(F). \end{equation}

I want to show that $\{P_n\}_n$ converges setwise to a probability measure that is absolutely continuous with respect to $P$.

Let $X_n = \frac{dP_n|_{\mathcal{F}_n}}{dP|_{\mathcal{F}_n}}$ be the Radon-Nikodym derivative of $P_n$ with respect to $P$ restricted to $(\Omega, \mathcal{F}_n)$. Then $\{X_n \}_n$ is a nonnegative martingale because each $X_n$ is integrable, $\mathcal{F}_n$-measurable and if $F \in \mathcal{F}_n \subset \mathcal{F}_{n+1}$, then (1) implies $$\int_F X_n dP = P_n(F) = P_{n+1}(F) = \int_F X_{n+1}dP.$$ Therefore, $\{X_n\}_n$ converges almost surely to a nonnegative integrable limit $X_\infty$. So $P_\infty$ defined by $P_\infty(A)=\int_A X_\infty dP$ is a measure.

Now let $F \in \cup_n \mathcal{F}_n$. For all sufficiently large $n$, $$P_n(F) = \int_F X_n dP,$$ so $$\lim_{n \to \infty} P_n(F) = \lim_{n \to \infty} \int_F X_n dP = \int_F X_\infty dP = P_\infty(F),$$ where the interchange of limit and integral is justified by the uniform integrability of $\{X_n \}_n$, which is equivalent to the uniform absolute continuity of $\{P_n \}_n$. It's clear from the above that $P_\infty(\Omega)=1$. We have shown that $\{ P_n\}_n$ converges setwise to $P_\infty$ on $\cup_n \mathcal{F}_n$, and I'd like to conclude by asserting that since $\cup_n \mathcal{F}_n$ is a pi-system that generates $\mathcal{F}$, $\{P_n \}_n$ converges setwise to $P_\infty$ on all of $\mathcal{F}$. I think that this follows from the pi-lambda theorem, but it would be great if someone could confirm that.

Answer 1

Define

$$\mathcal{D} := \{F \in \mathcal{F}; \lim_{n \to \infty} P_n(F) = P_{\infty}(F)\}.$$

If we can show that $\mathcal{D}$ is a Dynkin system, then it follows that $\mathcal{D} = \mathcal{F}$ since the $\cap$-stable generator $\bigcup_n \mathcal{F}_n$ is contained in $\mathcal{D}$.

Since $P_{\infty}(\Omega) = P_n(\Omega)=1$, we clearly have $\Omega \in \mathcal{D}$. Moreover, if $F \in \mathcal{D}$, then

$$\lim_{n \to \infty} P_n(F^c) = 1- \lim_{n \to \infty} P_n(F) = 1- P_{\infty}(F) = P_{\infty}(F^c),$$

and so $F^c \in \mathcal{D}$. Now let $(F_k)_{k \in \mathbb{N}} \subseteq \mathcal{D}$ be a sequence of pairwise disjoint sets and define

$$F := \bigcup_{k \geq 1} F_k.$$

We have to show $F \in \mathcal{D}$. By the uniform absolute continuity of $(P_n)_{n \in \mathbb{N}}$ we can choose for fixed $\epsilon>0$ some $\delta>0$ such that

$$P_n(A) \leq \epsilon \qquad \text{for all $n \in \mathbb{N}$, $A \in \mathcal{F}$ such that $P(A)\leq \delta$}. \tag{1}$$

Choose $K \in \mathbb{N}$ sufficiently large such that

$$P \left( F \backslash \bigcup_{k=1}^K F_k \right) \leq \delta \quad \text{and} \quad P_{\infty} \left( F \backslash \bigcup_{k=1}^K F_k \right) \leq \epsilon. \tag{2}$$

Clearly,

$$|P_n(F)-P_{\infty}(F)| \leq \left|P_n \left( F \backslash \bigcup_{k=1}^K F_k \right) - P_{\infty} \left( F \backslash \bigcup_{k=1}^K F_k \right) \right| + \left| P_n \left( \bigcup_{k=1}^K F_k \right)-P_{\infty} \left( \bigcup_{k=1}^K F_k \right) \right|.$$

It follows from $(1)$ and $(2)$ that the first term on the right-hand side is less or equal than $2 \epsilon$. On the other hand, the second term on the right-hand side can be estimated above by

$$\sum_{k=1}^K |P_n(F_k)-P_{\infty}(F_k)|$$

which converges to $0$ as $n \to \infty$. Hence,

$$\limsup_{n \to \infty} |P_n(F)-P_{\infty}(F)| \leq 2 \epsilon.$$ Since $\epsilon>0$ is arbitrary, this shows $F \in \mathcal{D}$.

Answer 2

Remember, we have $P_n\ll P$, not just $P_n|_{\mathcal F_n}\ll P|_{\mathcal F_n}$. So if we instead look at $\overline X_n=\frac{dP_n}{dP}$ we can let the martingale convergence theorem do all of the work for us. Indeed, $\{\overline X_n\}$ is still a nonnegative martingale, and by assumption is uniformly integrable. Hence by the martingale convergence theorem, $\overline X_n\to \overline X_\infty$ almost surely and in $L^1$ for some nonnegative integrable random variable $\overline X_\infty$. As before, let $P_\infty(F)=\int_F\overline X_\infty dP$. Then for any $F\in\mathcal F$ we have

$$|P_n(F)-P_\infty(F)|\le\int_F|\overline X_n-\overline X_\infty|dP\to0.$$

This implies $P_\infty$ is a probability measure and $P_n\to P_\infty$ setwise.