Conditioning on a negligible event is often done for an absolutely continuous random variables, say $X$. It is not uncommon to encounter expressions such as $\mathbb{P}(a<Y<b|X=x)$ for some other random variable $Y$ (assume it absolutely continuous too for simplicity). This expression can be made sense of as a limit: $\lim_{\epsilon\longrightarrow 0}\mathbb{P}(a<Y<b|x-\epsilon\leq X\leq x+\epsilon)$.
The computation of this limit goes through expressing the numerator and denominator defining the conditional probability as integral of the joint density of $X,Y$ over the set $[x-\epsilon,x+\epsilon]\times[a,b]$ and of the density of $X$ over $[x-\epsilon,x+\epsilon]$ respectively.
My questions are: why would this limit always exist (in $[0,1]$ that is)? Given that it does, would it be the same as that obtained by replacing $[x-\epsilon,x+\epsilon]$ with $[x,x+\epsilon]$ for example? Cheers.
The existence of the limit through Radon–Nikodym theorem and Lebesgue's differentiation theorem.
Fix a borel set $A$ and define the following measure on $\mathbb{R}$. $$\mu_{A}(C)=P(Y\in A \; , X \in C)$$
By absolutely continuity of $X$ we know that $\mu_{A}<<\mathcal{L}$ where $\mathcal{L}$ is the Lebesgue measure. Now we apply Radon–Nikodym to find a measurable function $f$ so that for every Borel set $C$ we have $$ \mu_{A}(C)=\int\limits_{C}{f dt} $$
We now apply Lebesgue differentiation theorem to get that for almost every $x$ we have $$\lim\limits_{|B|\to 0}\frac{\int\limits_{B}{fdt}}{|B|}=\lim\limits_{|B|\to 0}\frac{P(Y\in A\; , X \in B)}{|B|}=f(x)$$ Where the limit is taken over balls $B$ centerd at $x$ and $|B|$ is their length.
We have $$\frac{P(Y \in A \; , X \in B)}{P(X \in B)}=\frac{|B|}{P(X \in B)}\frac{P(Y \in A \; , X \in B)}{|B|}$$ Then $$\lim\limits_{|B|\to 0}\frac{P(Y \in A \; , X \in B)}{P(X \in B)}=\frac{f(x)}{f_{X}(x)}$$
This is the conditional expectation of the event $Y \in A$ given $X\;$(Defined as this ). To see this note that for any event $X^{-1}(C) \in \sigma(X)$ we have $$\int\limits_{X^{-1}(C)}{\frac{f(X)}{f_{X}(X)}dP}=E(\chi_{C}(X)\frac{f(X)}{f_{X}(X)})=\int\limits_{C}{\frac{f(x)}{f_{X}(x)}f_{X}(x)dx}=P(Y \in A \; X \in C)=\int\limits_{X^{-1}(C)}{\chi_{A}(Y)dP}$$
This might not mean much if you've never read or done a course on measure theory but this is the 'formal' approach.
Radon-Nikodym Lebesgue's differentiation theorem