If P(A|B)=1, then is it correct that P(A|B,C)=1 under the assumption that P(B,C)>0? If $P(A\mid B)=1$, then what can we say about $P(A\mid B\cap C)$? Is it 1 or not?
The condition says that if B occurs, then A occurs a.s.
Then, if B and C occur, this implies that B has occurred and due to the condition A must occur a.s..
Of course C should be a possible event. This is my intuition. But when I try to prove it by using Bayes' Theorem, I cannot complete the proof:
$P(A\mid B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}=\frac{P(C\mid B\cap A)P(A\mid B)P(B)}{P(C\mid B)P(B)}=\frac{P(C\mid B\cap A)}{P(C\mid B)}...$
Then, it seems to me that I am in a cycle.
 A: First of all, $P(A|B)=1$ implies that $P(A\cap B)=P(B)$. Since $P(B)=P(A\cap B)+P(A^c\cap B)$, we conclude $$P(A^c\cap B)=0.$$
Assuming that $P(B\cap C)>0$, then we can say that
$$
P(A|B\cap C)={P(A\cap B\cap C)\over P(B\cap C)}={P(B\cap C)-{P(A^c \cap B \cap C)}\over P(B \cap C)}
$$
Finally, since $A^c\cap B\cap C$ is a subset of an event with a probability of zero, namely $A^c \cap B$, we conclude that $P(A^c \cap B \cap C)=0$ as well, so the above fraction simplifies to $1$.
$\tag*{$\square$}$
A: Suppose we have a variable $X$ distributed uniformly on $[0,1]$. Let $A$ be the event $X\ne\frac12$, $B$ be the event $X\le\frac12$, and $C$ be the event $X\ge\frac12$.
Then $P(A\mid B)=1$ but $P(A\mid B\cap C)=0$.
$B$ and $C$ are a possible events, with non-zero probability; and $B\cap C$ is also a possible event (albeit with zero probability). But $P(A\mid B\cap C)$ is not equal to $1$. So it's not surprising that you can't prove that it is. You also need that the probability of $B\cap C$ is non-zero.
