$(\Omega, \mathcal{F}, P)$ is a probability triple. Suppose $B_{i} \in \mathcal{F}$ with $P\left(B_{i}\right)>0$. Let $P_1(A) = P(A|B_1)$; define $P_{n}(A)=P_{n-1}\left(A \mid B_{n}\right)$. What is the probability measure $Q(A)=\lim _{n \rightarrow \infty} P_{n}(A)$ for $A \in \mathcal{F}$? (What is the condition if the measure $Q$ exists?)
Based on my understanding, $\displaystyle P_n(A) = \frac{P(A \cap(\cap_{i=1}^n B_i))}{P(\cap_{i=1}^n B_i)}$. If $\{B_i\}_{i=1}^n$ is a decreasing sequence, $B_n \downarrow B$, $B > 0$ according to the assumption, $P_n(A)$ is either 0 or 1. What about the more general situation? Does the limit exist when both numerator and denominator approach zero? Do we have something similar to L'Hospital's rule to calculate the limit?
Edit
Not sure whether the following analysis is reasonable or not.
Let $C_k = \cap_{i=1}^k B_k$, $C_k$ is a decreasing sequence; suppose $C_n \downarrow C$, $C$ is nontrivial.
if $A \cap C \neq \emptyset$, $Q(A) = \frac{P(A \cap C)}{P(C)}$;
if $A \cap C = \emptyset $, $Q(A) = 0$;
The limiting probability is $P\left(A \mid C_{n}\right) \rightarrow P(A \mid C)$ if $P(C) \neq 0$.