Conditional distributions allowed pdf to take on single value? My question is about Conditional probability distributions. From what I have learned, PDF's aren't allowed to take on singular values, yet I find that this definition seems to go out the window when we are talking conditional probability. Looking for some type of justification for this. 
Why is it that when we are talking conditional, pdf's they are allowed to take on a single value. 
I.E: For x and Y continuous:
$P[Y > y | X = x] =    \frac{f_x(x) \cap F_y[Y >y]}{f_x(x)}$
Why?
 A: Your expression is a horrendous mismash, but what you seem to mean is that is:$$\begin{align}\mathsf P(Y{>}y\mid X{=}x) ~=&~ \dfrac{\int_y^\infty f_{X,Y}(x,t)\operatorname d t}{f_X(x)} \\[1ex] =&~ \int_y^\infty f_{Y\mid X}(t\mid x)\operatorname d t\end{align}$$
This is okay.   We can handle probability density functions of continuous random variables in a mostly analogous manner to probability mass functions of discrete random variables.
For a given value, $x$, of random variable $X$, the (conditional) probability that random variable $Y$ exceeds a value, $y$, can be calculated if we know either both the joint probability density function and the marginal density function ( $f_{X,Y}, f_{X}$ ), or the conditional probability density function ( $f_{Y\mid X}$ ).


From what I have learned, PDF's aren't allowed to take on singular values,

The probability that a continuous random variable will realise any particular value has zero measure (by definition).   This is not the same as saying it is impossible to do so; it is certain that the random variable will realise some value in the support.
