Bayes Theorem - Both Events Need Nonzero Probability? Bayes' theorem:
$$
P(A|B)  = \frac{P(B|A)P(A)}{P(B)}.
$$
Clearly, $P(B)>0$ is required.  However, 
$$
P(B|A) := \frac{P(B \cap A)}{P(A)},
$$
so if $P(A)=0$ we would have
$$
P(A|B)  = \frac{\frac{P(B \cap A)}{0}0}{P(B)},
$$
which is undefined.
So, even though I usually see Bayes' theorem written with the condition $P(B)>0$, it seems we also need $P(A)>0$.  Am I missing something?
 A: You're not missing anything!
Read the formula literally: $\operatorname{Pr}\left(A\ |\ B\right)$ is the fraction of $\operatorname{Pr}\left(B\right)$ that is contributed by $\operatorname{Pr}\left(A \cap B\right)$. This is what the Venn diagram in Hamed's answer depicts.
If $\operatorname{Pr}\left(B\right)$ is zero, there's nothing to contribute to. Think of it this way: Jill has \$0. What fraction of that \$0 did Jack contribute? The fraction is undefined.
That's what Xi'an was saying in his comment.
A: Conditional probability actually confines the space to the smaller space. Let give you an example. Assume you are looking for the probability $P(B|A)$. That is the probability of red section in plot to probability of  $A$ (sum of yellow and red section), $P(B|A)=P(B\cap A)/P(A)$. So assume $A=\phi$ then  $B\cap \phi=\phi$ and $P(B\cap \phi)=P(\phi)=0$ and $P(A)=0$. Then probability is not defined. 
For the second part of your question, we need to consider that conditional probability is defined if denominator is non-zero(positive). That is the formula $P(A|B)=\frac{P(A\cap B)}{P(B)}$ is defined if $p(B)>0$ then assuming that $P(A)=0$ in your case is not true. I hope I could explain that clearly.

