If two probability measures $P$ and $Q$, defined on the same space $(\Omega, \mathcal{F})$, are equivalent, then their conditional probabilities (provided these are regular) merge in total variation almost surely, i.e. $$d(P(\cdot \mid \mathcal{F_n}) , Q(\cdot \mid \mathcal{F}_n)) : =\sup_{A \in \mathcal{F}}|P(A \mid \mathcal{F_n}) - Q(A \mid \mathcal{F}_n)| \to 0 \ \ \text{a.s.},$$ where $\{\mathcal{F}_n : n \in \mathbb{N} \}$ is a filtration with $\mathcal{F}_n \uparrow \mathcal{F}$. See this question.
I want to show that the converse holds in the following sense.$^{[*]}$
Suppose $Q \mid_{\mathcal{F}_n} \ll P \mid_{\mathcal{F}_n}$, i.e. that $Q$ is absolutely continuous with respect to $P$ on $\mathcal{F}_n$. Also suppose that the conditional probabilities $Q_n := Q(\cdot \mid \mathcal{F}_n)$ and $P_n := P(\cdot \mid \mathcal{F}_n)$ are regular, so that $Q_n(\omega)$ and $P_n(\omega)$ are probability measures on $\mathcal{F}$ for all $\omega \in \Omega$. Finally, suppose that with $Q$-probability $1$ $$d(P_n, Q_n) \to 0.$$ Then,
Conjecture. $Q \ll P$.
Idea. My idea was to use the following result about Lebesgue decomposition. Let $X_n = \frac{dQ|_{\mathcal{F}_n}}{dP|_{\mathcal{F}_n}}$, and let $X = \limsup_n X_n$. Then, for all $A \in \mathcal{F}$, $$Q(A) = \int_A X dP + Q(A \cap \{X = \infty \}).$$ This result can be found in Durrett, Chapter 4, 3.3. The leading idea of the proof is that $\{X_n\}$ is a non-negative martingale with respect to $P$ and therefore the martingale convergence theorem implies that $X_n \to X$ a.s. ($P$) and $P(X = \infty)=0$.
Now, I could conclude that $Q \ll P$ if I could show that $Q(X = \infty)=0$, but I haven't been able to do this. I've been trying to argue as follows. Suppose $Q(X = \infty) = \epsilon > 0$, and let us try to contradict the assumption that $d(P_n,Q_n) \to 0$ a.s ($Q$). We know that for all $n$ $$\epsilon = Q(X = \infty) = \int Q_n(X = \infty)dQ,$$ so for all $n$, there exists a set $B_n$ with positive $Q$-probability on which $Q_n(X = \infty) \geq \epsilon$. I thought maybe I could show that with positive $Q$-probability $$|Q_n(X=\infty) - P_n(X = \infty)| \geq \epsilon$$ infinitely often and get the desired contradiction. But I'm stuck because I don't know how $P_n(X = \infty)$ behaves with respect to $Q$; I only know that $P_n(X = \infty) = 0$ a.s. ($P$). And I can't guarantee that $\cap_n B_n$ has positive $Q$-probability.
Any suggestions or hints are appreciated.
$^{[*]}$ In this paper (Theorem 2), the converse is shown for the special case in which each $\mathcal{F}_n$ is generated by a countable partition. The proof is fairly elementary and proceeds by approximating an event for which absolute continuity fails by a sequence of finite dimensional events. My hope here is to relax the assumption that $\mathcal{F}$ is countably generated and find a proof using martingales.
Note that $P_n(X=\infty)$ is an $\mathcal F_n$-measurable random variable, and so
$$Q(P_n(X=\infty)>0)=Q\big|_{\mathcal F_n}(P_n(X=\infty)>0)=\int_{\{P_n(X=\infty)>0\}}\frac{dQ|_{\mathcal F_n}}{dP|_{\mathcal F_n}}\,dP=0$$
since $P(P_n(X=\infty)>0)=0$. Hence there exists $\Omega_n\in\mathcal F_n$ with $Q(\Omega_n)=1$ such that $P_n(X=\infty)=0$ on $\Omega_n$. This implies
\begin{align*} \int|Q_n(X=\infty)-P_n(X=\infty)|\,dQ &\ge\int_{\Omega_n}|Q_n(X=\infty)-P_n(X=\infty)|\,dQ\\ &=\int_{\Omega_n}Q_n(X=\infty)\,dQ\\ &=\epsilon \end{align*} for every $n$. But this integral must converge to zero by the bounded convergence theorem, so we have a contradiction.