Do sequences of full conditional probabilities converge? Setup.
Let $\Omega$ be a set, and let $\mathcal F$ be its powerset. A full conditional probability is a function $P(\cdot \mid \cdot): \mathcal F \times \mathcal F \setminus \emptyset \to [0,1]$ that satisfies:

(1) For all non-empty $F \in \mathcal F$, $P(\cdot \mid F)$ is a finitely additive probability measure on $(\Omega, \mathcal F)$;
(2) For all non-empty $F \in \mathcal F$, $P(F\mid F)=1$;
(3) For all $A,B \in \mathcal F$ and all non-empty $C \in \mathcal F$ for which $B \cap C$ is also non-empty,
$$P(A \cap B \mid C) = P(B \mid C)P(A \mid B \cap C)\tag{1}.$$

If $P$ is a full conditional probability, we write $P(F)$ instead of $P(F \mid \Omega)$.
Note that if $P(B)>0$, then $P(A \mid B) = P(A \cap B)/P(B)$. This is obtained from (1) with $C = \Omega$. In other words, full conditional probabilities extend the usual notion of conditional probability given an event to allow for conditioning on events with probability zero.
Question. Let $F_1 \supset F_2 \supset ...$ be a decreasing sequence of non-empty subsets of $\Omega$.

Is it the case that $P(A \mid F_n)$ converges for all $A \in \mathcal F$?

Observations. Let $F = \bigcap_n F_n$. If $P(F)> 0$, so that $P(A \mid F_n) = P(A \cap F_n)/P(F_n)$, then $P(A \mid F_n)$ converges. If $P(\cdot \mid \Omega)$ is countably (and not merely finitely) additive, then the limit is equal to $P(A \mid F)$. Without countable additivity, however, $P(A \cap F_n)$ might not converge to $P(A \cap F)$, and $P(F_n)$ might not converge to $P(F)$ (they converge to something, though, because they are both bounded and non-increasing).
 A: Not necessarily!
Here is a counter-example:
Let $\Omega:=\mathbb{N}$.  For $\varnothing\neq F\subseteq\Omega$, let $P(\cdot\,|\,F):=\delta_{\min(F)}$ be the Dirac measure with a unit mass at the smallest element of $F$.
Note that $P(\cdot\,|\,\cdot)$ satisfies the axioms.  Namely, for $F\neq\varnothing$, $P(\cdot\,|\,F)$ is a (countably additive) probability measure concentrated at $F$.  For $A,B,C\subseteq\Omega$ with $B\cap C\neq\varnothing$,
\begin{align*}
   P(B\,|\,C)P(A\,|\,B\cap C)=1 \quad
      &\Longleftrightarrow\quad
      \text{$B\ni\min(C)$ and $A\ni\min(B\cap C)$} \\
      &\Longleftrightarrow\quad
      \text{$B\ni\min(C)$ and $A\ni\min(C)$} \tag{$*$} \\
      &\Longleftrightarrow\quad
      \text{$A\cap B\ni\min(C)$} \\
      &\Longleftrightarrow\quad
      P(A\cap B\,|\,C)=1 \;,
\end{align*}
where ($*$) is true because when $B\ni\min(C)$, we have $\min(B\cap C)=\min(C)$.  Hence, the third axiom is also satisfied.
Now, let $A$ be the set of even numbers and $F_n:=\{n, n+1, n+2, \ldots\}$.  Clearly, $F_1\supset F_2\supset\cdots$ but $P(A|F_n)$ does not converge.
