For multiplication with conditional probabilities, why can't $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$? My book says this for the formula of multiplication with conditional probabilities:
For any events $A_1, A_2, ..., A_n$ with $A_1 \cap A_2 \cap ... \cap A_n \neq \emptyset$
$P(A_1 \cap A_2 \cap ... \cap A_n) = P(A_1)P(A_2 |A_1)P(A_3|A_1\cap A_2)...P(A_n|A_1\cap A_2  \cap ...\cap A_{n-1})$
This makes intitutive sense to me as $P(A_1 \cap A_2 \cap ... \cap A_n)$ is just equivalent to $P(A_1)$ occurs and then given that $A_1$ occurs, we need $A_2$ to occur too, and then,
given that $A_1$ and $A_2$ occur, we need $A_3$ to occur too, and then,
given that $A_1$ and $A_2$ and $A_3$ occur, we need $A_4$ to occur too and so on till we reach $P(A_n|A_1\cap A_2  \cap ...\cap A_{n-1})$ and stop because then all the events in the intersection in the LHS occurred!
My question is, why isn't $A_1 \cap A_2 \cap ... \cap A_n \neq \emptyset$ allowed? Wouldn't that just make $P(A_1 \cap A_2 \cap ... \cap A_n) = 0$ which is a completely legimately value for probability? Is any mathematical law broken with $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$ that makes it not allowed?
Thank You!
 A: The right condition you want is $P(A_1 \cap \ldots \cap A_{n-1}) \neq 0$, however, this is not implied by $A_1 \cap \ldots \cap A_{n-1} \neq \emptyset$ (unless maybe your book considers discrete probability spaces and requires no $0$-sets).
If $A_1 \cap \ldots \cap A_{n-1} = \emptyset$ (or even $P(A_1 \cap \ldots \cap A_{n-1}) = 0$), then the last probability would not be defined since
$$P(A\vert B) = \frac{P(A\cap B)}{P(B)}$$
However, if $P(A_1 \cap \ldots \cap A_{n}) = \emptyset$ and $P(A_1 \cap \ldots \cap A_{n-1}) \neq 0$ then the formula is well defined and true since the LHS is $0$ and the last enumerator in the last factor is also $0$.
A: Every probability measure verifies that $P(\emptyset)=0$
So $A_1 \cap A_2 \cap ... \cap A_n = \emptyset$ $\implies$ $P(A_1 \cap A_2 \cap ... \cap A_n) = P(\emptyset)=0$ is trivial
The most important property that probability measures have (that other measures doesn't necesarily verify) is that if two events A, B are independent then $P(A \cap B) = P(A) P(B)$
