Proving that the conditional probability of a continuous distribution on a discrete distribution taking a value in a given interval is an integral I've been trying to prove that Bays' Theorem still works when you condition a continuous distribution on a discrete one and I've hit a snag that can be accurate described as "I need to prove that the conditional probability of a continuous distribution taking a value in a given interval, given that the discrete distribution that it is conditioned on has taken a particular value, is an integral", or in notational terms: For Y continuous, X discrete, and k an arbitrary real number I need to prove that P(y$<$Y$<$y+k | X=x) is an integral (from y to y+k).
Is there any way to go about this? The only information that I have which I can think of that I haven't already stated explicitly is that X and Y form a joint distribution of unknown properties.
Finally, I will pre-emptively state that I am not very clued up on measure theory, so if possible I'd like to avoid any answers which are in terms of metric spaces, although with as little information as I have, I fear that they may be necessary.
 A: You are asking whether or not $P(y<Y<y+k|X=x)$ can be written as the integral of some function. Here is a simple proof that it can. 
Since $Y$ has a density, for all* $y$, we have that the limit
$$
\lim_{h\to0}\frac{P(y<Y\le y+h)}h
$$
exists. We want to prove that the same is true for the conditional distribution, that is, we want to show
$$
\lim_{h\to0} \frac{P(y<Y\le y+h|X=x)}h 
=\lim_{h\to0} \frac{P(y<Y\le y+h\,\cap \,X=x)}{P(X=x)\cdot h}
$$
exists. To prove the last limit exists, note that
$$
0\le \frac{P(y<Y\le y+h\,\cap \,X=x)}{P(X=x)\cdot h}\le \frac{P(y<Y\le y+h)}{P(X=x)\cdot h}
$$
since the event in the middle numerator is a subset of the event in the right numerator. Both the left and right limits exist as $h\to0$, so by the squeeze theorem, the limit in the middle does as well. 
This shows that the conditional cumulative distribution function $P(Y\le y|X=x)$ has a derivative, so by the fundamental theorem of calculus, $P(y<Y\le y+k|X=x)$ is the integral of that derivative. 

 *Technically, this need not exist for all $y$. For example, the function $F(y)=P(Y\le y)$ could be piecewise differentiable, so the derivative exists at all but a few sharp corners. To get into more detail, you do need measure theory. 
