Does this uniquely define a conditional expectation? I know how we define conditional expectations in  probability theory without measure theory (i.e. with density functions on R). My question is about defining it on general probability spaces using measure theory.
Assume a probability space $(\Omega, \sigma, P)$
If we want to define a conditional expectation $E(X|G)$ where G is a subset of Omega (i.e. Y is in sigma), then there is no problem whatsoever.
But What if instead we want to find $E(X|Y=y)$ where y is a single value of the random variable Y, so that it will have measure zero? then we cant calculate the conditional expectation the standard way.
So I came up with the following formula:
$$\lim_{\epsilon\to 0}\int x dP(x|Y_\epsilon)$$
where $Y_\epsilon=\{\omega : |Y(\omega)-y|<\epsilon\}$.
Does this correctly and uniquely define the conditional expectation of X given a measure zero realization of Y?
 A: Another problem arises if $\lim_{\epsilon\rightarrow 0} P[X\leq 1| Y_{\epsilon}]$ does not exist.  For example, 
let $Y$ be a nonnegative random variable, let $y=0$,  and assume $P[Y=0]=0$. Suppose $(X,Y)$ satisfies the following for $k \in \{1, 2, 3, ...\}$: 
If $k$ is even: 
\begin{align}
P\left[X \leq 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (1/2)\frac{1}{2^{k^2}} \\
P\left[X > 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (1/2)\frac{1}{2^{k^2}} 
\end{align}
If $k$ is odd: 
\begin{align}
P\left[X \leq 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (3/4)\frac{1}{2^{k^2}} \\
P\left[X > 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (1/4)\frac{1}{2^{k^2}} 
\end{align}
Then $Y_{1/k} = \{\omega : 0 \leq Y < 1/k\}$  and since $1/2^{k^2}$ decreases so rapidly, we have for large $k$ that $\sum_{i=k}^{\infty} 1/2^{i^2} \approx 1/2^{k^2}$ and so 
$$P[Y_{1/k}] \approx P\left[Y \in \left(\frac{1}{k+1}, \frac{1}{k}\right]\right] = \frac{1}{2^{k^2}} $$
and
\begin{align}
P[X \leq 1, Y_{1/k}] &\approx (1/2)\frac{1}{2^{k^2}} \quad, \mbox{if $k$ even} \\
P[X \leq 1, Y_{1/k}] &\approx (3/4)\frac{1}{2^{k^2}} \quad, \mbox{if $k$ odd} 
\end{align}
So
$$\lim_{m\rightarrow\infty} P[X\leq 1 | Y_{1/(2m)}] = 1/2 $$
but
$$ \lim_{m\rightarrow\infty} P[X \leq 1 | Y_{1/(2m+1)}] = 3/4 $$
