Probability, conditional on a zero probability event

Is there a way to resolve probability of an event, given another event that never happens? Mathematically speaking the problem is:

Given that $P(B) = 0$,

$$P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{0}{0}$$

Is this probability vacuously $0$ of $1$? Can we show that it's one or the other?

The comment by Dilip Sarwate points to conditioning on the level of densities which can be interpreted as conditioning on a family of events of probability zero. It is a probabilistic version of Radon-Nikodym derivative.

One can also condition on an individual event of probability zero, if that event admits a natural approximation by events of positive probability. For example, begin with a probability measure $\nu$ on the space $C[0,1]$ of continuous functions such that $\nu(\{f(0)=0\})=0$. Restrict it to the set of functions $f$ such that $|f(0)| \le \epsilon$. Normalize this restriction to a probability measure. If these normalized restrictions converge (in some sense) to a probability measure $\mu$ on $C[0,1]$, we can regard $\mu$ as conditional distribution on the zero-probability event $f(0)=0$.

The definition of conditional probability $$P(A|B)=\frac{P(A \cap B)}{P(B)}$$ means that we can't condition on an event with zero probability; if $P(B)=0$, then $P(A|B)$ is undefined for any event $A$.

Sometimes it makes sense to define conditional probability differently. Let $X$ and $Y$ be jointly continuous random variables; then we can define conditional PDF of $X$ given $Y$ $$f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}$$ for all $y$ such that $f_Y(y)>0$, and using this definition we can define $$P(X\in A|Y=y)=\int_Af_{X|Y}(x|y)dx$$ which implies conditioning on zero probability event $\{Y=y\}$.

The answer to the general OP question is: if $P(B)=0$, the conditional probabilities given event $B$ are undefined, unless a different definition of conditional probability is used.