Why $P(B)$ needs to be greater in Bayes' Theorem? I know that in the Bayes' formula, $P(A\mid B)$, $P(B)$ needs to be greater than $0$, because of the denominator. I also know that $P(B)$ being null implies that the $B$ event is impossible, and does not make sense to calculate the probability of $A$ given $B$. However, if $B$ is an empty event and $A$ is an empty event? Could $P(B)$ be $0$? Is there another event that allows $P(B)$ be zero?
Thanks.
 A: 
I know that in the Bayes formula, P(A|B), P(B) needs to be greater than 0, because of the denominator.

$P(A\mid B)$ is not Bayes' formula, nor is it the Bayes' Theorem referred to in the title of the question.  $$P(A\mid B) = \frac{P(A\cap B)}{P(B)} ~~ \text{provided that }
 P(B) > 0, \tag{1}$$
is the definition of the conditional probability of $A$ given that
$B$ occurred.

I also know that P(B) being null implies that the B event is impossible, and does not make sense to calculate the probability of A given B.

No, $P(B)=0$ implies exactly what it says, that $B$ is an event of probability $0$.  It does not imply that $B$ is the empty set which is the impossible event.  An event of probability $0$ can occur on a trial of the experiment while the empty set can never occur on a trial of the experiment.  As fblundun's comment on the OP's questions point out, events of probability $0$ arise when one is dealing with continuous random variables.
While one cannot use $(1)$ to define $P(A\mid B)$ when $B$ is an event of probability $0$, note that when $B$ is an event of probability $0$, the event $A\cap B$, which is a subset of $B$, and so necessarily has probability that is no larger than $P(B)$, also is an event of probability $0$ and so $(1)$ is necessarily of the form $\frac 00$ when $B$ is an event of probability $0$. It is, nonetheless, possible to get something useful from this indeterminate result.  The analogy is less than perfect, but think of L'Hopital's rule in calculus for evaluating limits of ratios of functions that converge to the form $\frac 00$: we take the ratio of the derivatives of the two functions as the limit. Thus, for continuous random variables, we get things like
$$f_{X\mid Y}(x\mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$ for the conditional density of $X$ conditioned on the (zero-probability but by no means impossible) event that $\{Y=y\}$.
A: You can have non-zero $P(A|B)$ even if $P(B)$ is zero. This is called a vacuous truth.
It will just make Bayes formula undetermined
$$P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{P(B|A) P(A)}{ P(B|A) P(A) +P(B|\neg A) P(\neg A) } = \frac{0}{0}$$
If $P(B)$ is $0$ then $P(B|A) P(A) =P(B,A)$ is necessarily $0$ as well (since $P(B) = P(B,A) + P(B, \neg A)$, and you get an undetermined division of 0 by 0.
If you like you could still write the formula in the form without division $$P(A|B)P(B) = P(B|A)P(A)$$
Here you see why it is undetermined. If $P(B) = 0$ then the value of $P(A|B)$ can be any value while the left side remains zero. So, in that case, you can not determine $P(A|B)$ by knowing the other three variables in this equation of four variables. The multiplication with $0$ can not be inversed because multiplication with zero is non-injective and can not be inverted (and you could see the Bayesian rule as the formula describing this inversion).

Why P(B) needs to be greater in Bayes Theorem?

Because Bayes Theorem will be equivalent to division by 0 (or inversing multiplication by 0) which is indeterminate.
