Can (Should?) we somehow extend meaning of $P(A|B)$ to when $P(B)=0$? The expression $P(A|B)$, where $(\Omega, \mathcal{A}, P)$ is a probability space and $A, B \in \mathcal{A}$ is only defined for when $P(B)>0$.
Suppose we have the following case: $([0,1], \mathcal{B} \cap [0,1], P)$, where $P(B)$ is the usual length of the measurable set $B$. Intuitively, we are picking a random real from $[0,1]$, where the likehood of it coming from any partition into equal parts is the same. So, it would make sense intuitively if $P(\{1\} | \{0,1\}) = \frac{1}{2}$. Is there any natural way to extend the concept conditional probability that agrees with this?
 A: When $B$ is an event, there isn't a way to do this [as far as I know].
However, we may [sort of] do this when conditioning on a sigma algebra (as in the measure-theoretic development of conditional expectation).  There we might write $P(A|X)$ where $A$ is an event and $X$ is a random variable to denote $P(A|\sigma(X))$.  If we assume $X$ has a density, then knowing $X$ is in a sense conditioning on a probability zero event.  We may even formally write $P(A|X=x)$ (or even worse, e.g., $P(A|X=0)$) which makes it seem like we're back to conditioning on an event (and intuitively, we often think of it as such), but we're really not.
A: Measure theory teaches you that probabilities are in fact (generally, Lebesgue) measures (which means length in 1D, area in 2D, volume in 3D, n-hypervolume in nD). Your encompassing space $\Omega$ has measure (probability) 1. The probability of A knowing that B boils down to "what would the measure of A be, if B was a set of measure 1, instead of some fraction of $\Omega$ ?". It's a way of re-setting the problem so that B is the encompassing space.
There is no way to "scale" (or "ratio", I don't know which is the better term) a space of measure $0$ to make it into a space of measure $1$. That would require solving the equation $0x = 1$.
Of course, all of this depends on your measure. If your singleton set has a nonzero measure, nothing's preventing you from having fun with it. Then again, this means that $P(B) \neq 0$
