Is Bayes' theorem, is the denominator always assumed to be non-zero?

Question

Is Bayes' theorem, is the denominator always assumed to be non-zero or, in certain circumstances, you can prove that it is actually never zero?

Answer 1

The statement of Bayes theorem explicitly includes "and $P(B) > 0$." So yes, the theorem only applies when $P(B) > 0$.

Answer 2

Let $A$ and $B$ be two random variables taking values in $\Omega_X$ and $\Omega_Y$. Then:

\begin{equation} P(A=a | B=b) = \frac{P(A=a, B=b)}{\sum\limits_{a' \in \Omega_X} P(A=a', B=b)} \end{equation}

In this form I think it is easier to see why the denominator shouldn't be zero: it can be interpreted as a normalization factor. If it is zero it means that every term in the sum is zero since probabilities are non negative, that is in no circumstances the events $A=a$ and $B=b$ can occur together. If this is the case, the probability of $A=a$ given $B=b$ is ill defined