conditional probability on zero probability events and conditional Radon-Nikodym derivatives Consider a stochastic process $\{x_t\}_{t\in T}$ adapted to some filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}\}_{t\in T},\mathbb{P})$ taking values in the state space $(\mathbb{R},\mathcal{B})$
I wish to consider the probability Pr$(x_t\in\mathcal{A}|x_s)$ where $(s<t)$
This should be a sensible question, as I should be able to assign probability to a question of the form "What is the chance I obtain an outcome $x_t \in [0.5,0.6]$ given that last time I got $x_s=0.4$". e.g. a transition probability.
Naturally we have Pr$(A| B)=$Pr$(A\cap B)/$Pr$(B)$ but then we have Pr$(B)=$Pr$(x_s)=0$ for any particular value.
Now I think that this is where we have the notion of regular conditional probability entering which, as I understand it means we write:
$$\text{Pr}(x_t\in \mathcal{A}|x_s=B)=\lim_{\mathcal{B}\to B}\frac{\text{Pr}(x_t\in \mathcal{A}\cap x_s \in \mathcal{B})}{\text{Pr}(x_s\in \mathcal{B})}$$
Vagaries of the meaning of $\lim_{\mathcal{B}\to B}$ aside (which would have to be implementation specific e.g. here $\lim_{r\to 0} \text{Pr}(x_s\in (B-r,B+r)$), is the above correct?
Should I understand this as a Radon Nikodym derivative? It seems related, but not identical.
How do I relate this to, and formulate it in such a way to be consistent to, how conditional probabilities are usually defined? (as I understand it) viz
$$\text{Pr}(x_t\in\mathcal{A}|x_s\in\mathcal{B})=\mathbb{E}_{\mathbb{P}}[1_{\mathcal{A}}(x_t)|\sigma(\mathcal{B})\subseteq\mathcal{F}_s]$$
Surely $$\mathbb{E}_{\mathbb{P}}[1_{\mathcal{A}}(x_t)|\sigma({B})]$$ just wouldn't work? i.e. $\sigma(B)\nsubseteq\mathcal{F}$?
Is it legitimate to construct Radon-Nikodym derivatives out of measures formed from regular conditional probabilities? i.e. is this (heuristically) ok?
$$\frac{d\mathbb{P}(x_t|x_s=B)}{d\mathbb{Q}(x_t|x_s=B)}=\lim_{\mathcal{A}\to\emptyset}\lim_{\mathcal{B}\to B}\frac{\mathbb{P}(x_t\in\mathcal{A}|x_s\in\mathcal{B})}{\mathbb{Q}(x_t\in\mathcal{A}|x_s\in\mathcal{B})}$$
Thanks.
EDIT:
Based on discussion in the comments, the issues appears to boil down to why one can write
$$\text{Pr}(x_t\in \mathcal{A}|B)=\mathbb{E}_{\mathbb{P}}[1_{\mathcal{A}}(x_t)|\sigma({B})]$$
when $B$ is a zero probability event. What I don't understand is what the sigma algebra generated by a zero probability event looks like and why you can condition on it.
Surely the sigma algebra generated by a zero probability event is itself formed from (complements and unions of) zero probability events in $\mathbb{P}$, thus not in $\mathcal{F}$
e.g. $$\sigma(x_s=B)=\{\omega,\omega^c,\emptyset,\Omega\}\nsubseteq\mathcal{F}$$
with $\mathbb{P}(\omega)=0$ such that $\mathbb{E}_{\mathbb{P}}[f(x_t)|\sigma({B})]=\mathbb{E}_{\mathbb{P}}[f(x_t)]$ or $0$?
so why can we condition on these?
What am I getting wrong here?
 A: Just a summary of my comments that may clarify: You "should never" consider the sigma algebra generated by an event (since it has nothing to do with expressions like $E[X|Y=y]$).  You "should" consider the sigma algebra generated by a random variable, and this has lots to do with expressions like $E[X|Y=y]$. Specifically...
Let $S$ be a sample space. Let $(X,Y)$ be a random vector that maps $S\rightarrow \mathbb{R}^2$.   So $X(\omega)$ and $Y(\omega)$ are real numbers for all $\omega \in S$. 
Useful:
$$\sigma(Y) = \mbox{Sigma algebra generated by random variable $Y$}$$
Then $\sigma(Y)$ has many events, including all events of the form $\{Y\leq y\}$ and $\{Y \in [y, y+\delta]\}$.  Now, $E[X|Y]$ (sometimes written $E[X|\sigma(Y)]$) is a random variable with certain properties. There are different "versions," but they all differ only on a set of measure 0.  Let $Z=E[X|Y]$ be a particular version.  It is "$Y$-measurable" and: 
(i)  $Z(\omega) = f(Y(\omega))$ for some function $f$ and for all $\omega \in S$.  
(ii) 
$ \int_H Z dP = \int_H X dP \quad \forall H \in \sigma(Y) $
Existence of such a thing can be proven using Radon-Nikodym concepts. 
Now, $E[X|Y=y]$ can be defined as the value $Z(\omega)$ for any value $\omega$ in which $Y(\omega) = y$. 
Intuitive construction with $Z=E[X|Y]$.
Fix $\delta>0$ with $\delta \approx 0$.  Define $H = \{\omega : Y(\omega) \in [y, y+\delta]\}$. 
Suppose that $Z(\omega) \approx f(y)=E[X|Y=y]$ for almost all $\omega \in H$. Then: 
$$ \int_H X dP = \int_H Z dP \approx \int_H f(y) dP = f(y)P[Y\in [y, y+\delta]] $$
So: 
$$ f(y) \approx \frac{\int_{Y \in [y, y+\delta]} X dP}{P[Y \in [y, y+\delta]]} $$
and so we have an (unrigorous) property that: 
$$ f(y) = \lim_{\delta\rightarrow 0^+} \frac{\int_{Y \in [y, y+\delta]} X dP}{P[Y \in [y, y+\delta]]} $$
There might be some crazy examples where the limit does not exist or does not give the desired result. 
Not useful:
$$\sigma(Y=y) = \{\{Y=y\}, \{Y\neq y\}, S, \phi\} $$
We can formally define $E[X|\sigma(Y=y)]$, but this has nothing to do with $E[X|Y=y]$.  It can be shown that, with prob 1, $E[X|\sigma(Y=y)]=E[X]$.  That is because, for any event $H \in \{\{Y=y\}, \{Y\neq y\}, S, \phi\}$ we have: 
\begin{align}
\int_H E[X] dP = E[X]\int_H dP = E[X]P[H] &= \left\{ \begin{array}{ll}
E[X] &\mbox{ if $H=\{Y\neq y\}$ or $H=S$} \\
0  & \mbox{ otherwise} 
\end{array}
\right.\\
\int_H X dP &=\left\{ \begin{array}{ll}
E[X] &\mbox{ if $H=\{Y\neq y\}$ or $H=S$} \\
0  & \mbox{ otherwise} 
\end{array}
\right.
\end{align}
