Probability of event conditional to null event In general, it is not possible to define $P(A|B)$ when $P(B)=0$.  But I think there are situations, where the notation makes intuitively sense, for example in the product space $\Omega\times\Omega$, if $B\subseteq\Omega$ is a non-empty null event and $A\subseteq\Omega$ not, I'd like to say $P(A\times\Omega|\Omega\times B)=P(A)$.
So does there exist a framework which makes it possible to assign values to the probability of events condidional to some null event?
I thought, if there is a topology $\tau$ on $\Omega$ and $B$ some null event such that


*

*any open set $B'\supseteq B$ has non-zero probability

*and the net $(P(A|B'))_{B'\in\tau,B'\supseteq B}$ converges to some value $p$,


one could define $P(A|B)$ to equal $p$ but I don't know if this is well thought out.
 A: The Radon-Nikodym theorem should put your ideas into a firm frame-work. If $(\Omega,{\cal B},\mu)$ is a probability space with a $\sigma$-algebra and ${\cal F}\subset {\cal B}$ is a sub-sigma algebra then given any integrable random variable  $X$ we may form the conditional expectation $U=E(X|{\cal F})$ with respect to ${\cal F}$. It is  an ${\cal F}$ measurable  r.v. such that if $F\in {\cal F}$ then
 $$ \int_F X \;d\mu = \int_F U \; d\mu $$
You may view $U$ as an average over ${\cal F}$ measurable sets. When ${\cal F}$ has no point masses you are in the null-set situation you describe. But note that you may redefine $E(X|{\cal F})$ on sets of null-measures without affecting the above identity.
Example: Consider a uniform distribution on $[0,1]^2$. Let $Y(\omega_1,\omega_2)=\omega_2$. Then 
$${\cal F}=Y^{-1} ({\cal B}_{[0,1]}) = \{\; [0,1]\times A: A \subset [0,1] \}$$  (with $A$ measurable in $[0,1]$) defines a sub-$\sigma$ algebra. Let $X(\omega_1,\omega_2)=\omega_1+\omega_2$. Then 
$$ U(\omega)=E(X|{\cal F}) (\omega_1,\omega_2)=\frac{1}{2} + \omega_2 $$
It is  ${\cal F}$ measurable (you should check) and verifies for measurable $A$:
$$ \int_{[0,1]\times A} (x_1+x_2) \;dx_1\;dx_2= \int_{[0,1]\times A} (\frac{1}{2}+x_2) \;dx_1\;dx_2 $$
It is unique in the sense that if $U'$ is another such r.v. then $E(|U-U'|)=0$
