Conditional Probability on Joint Uniform Distribution This is a very basic question, but I somehow manage to confuse myself all the time. So any help is greatly appreciated. Suppose we have two random variables $X$ and $Y$ with joint distribution function $f$ where $f$ is uniform. That is 
$$ 
f(x,y)= \begin{cases} 1 & \text{if $0\leq x\leq 1$ and $0\leq y \leq 1$}, \\ 0 & \text{elsewhere}. \end{cases}
$$
If I want to determine the conditional probability $\Pr(Y>q\mid X=p)$ I get confused since 
$$
\Pr(Y>q\mid X=p)=\frac{\Pr((Y>q)\cap (X=p))}{\Pr(X=p)}
$$
but then isn't the denominator a zero probability event? In general what is the answer of this conditional probability?
 A: the question on how to condition on a set of the form $\{X=p\}$ which has zero probability used to bug as well. It is defined via the conditional expectation given the random variable $X$.
Recall that for a random variable $Y$ and a measurable function $h$ such that $h(Y)$ is integrable the conditional expectation of $h(Y)$ given $X$,  $\mathbb{E} [h(Y)| X]$, is a $\sigma(X)$-measurable random variable (with certain other properties). From the $\sigma(X)$-measurability we can deduce that there is a function $\psi$ (which depends on $h$ and $Y$) such that
$$
\mathbb{E} [h(Y)\mid X]= \psi(X).
$$
Then we can define
$$
\mathbb{E} [h(Y)\mid X=p]:= \psi(p).
$$
In your case, if we choose $h(Y)=1_{\{Y>q\}}$ we get that
$$
\mathbb{E} [h(Y)\mid X=p] = \mathbb P (Y>q\mid X=p) =\psi(p).
$$
A: You should use the probability density functions.
$$\begin{align}\mathsf P(Y>q\mid X=p) &= \int_{y>q} f_{\small Y\mid X}(y\mid p)~\mathsf d y\\[2ex]&= \dfrac{\int_{y>q}f_{\small X,Y}(p,y)\,\mathsf d y}{\int_{\Bbb R}f_{\small X,Y}(p,y)\,\mathrm d y}\\[2ex]&=\left(\dfrac{\int_q^1\mathrm d y}{\int_0^1\mathrm dy}\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1} \\[2ex]&=\left((1-q)\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1}\end{align}$$
