Bivariate continuous density function: How is it possible to define conditional density? Maybe this is a very trivial question for an experienced mathematician, but it is something mysterious for someone from an engineering background, like me; this is bugging me for a long time:

Let's define a bivariate, continuous distribution $p_{x,y}(x,y)$ which is $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}p_{x,y}(x,y)dxdy = 1$. We can define a conditional density which for example is $p_{x|y}(x|y=Y)$ in this setting. The definition of conditional probability is: $p_{x|y}(x|y=Y) = \frac{p_{x,y}(x,y=Y)}{p_{y}(y=Y)}$. Now, something which confuses me is that: $p_y$ is a continuous density function,too and by definition it must be $p_{y}(y=Y) = 0$ since $\int_{Y}^{Y}p_{y}(y)dy = 0$. So this must make the conditional density function $p_{x|y}$ undefined. But this is clearly not the case. How is this contradiction resolved? 

Thanks is advance.
We want to define $p_{X|Y=b}$ such that 
$$ X|Y=b \sim p_{X|Y=b} $$
in the same way that is understood
$$ Z \sim p_Z $$
when $Z$ is random variable.

Let $ x\in \mathbb{R}$, you can think that the probability of $X|Y=b$ belongs to a small interval centered on $x$ is directly proportional to $p(x,b)$. This suggests defining
$$ p_{X|Y=b}(x) = \frac{p_{X,Y}(x,b)}{p_Y(b)} $$
when $p_Y(b) \neq 0$.

In fact, this is enough to give a good definition of $ p_{X|Y=b}$ (since that the set of $b$'s where $p_Y(b)=0$ has Lebesgue measure null). In addition, it is easy to verify that 
$$ \mathbb{P}(a_1&lt;X\le a_2|Y=b) = \int_{a_1}^{a_2} p_{X|Y=b}dx $$
and 
$$ \mathbb{E}[X|Y=b] = \int_{-\infty}^\infty xp_{X|Y=b}(x)dx $$
how we wanted.
$$p_Y(y_0) \neq \int_{y_0}^{y_0} p_Y(y)\text{d}y = P(y_0 \leq Y \leq y_0)= P(Y \leq y_0)-P(Y \leq y_0)= 0 \,\,,$$
for $y_0$ in the domain of $Y$ (necessarily).
In other words, it might be the case that $p_Y(y_0)= P(Y=y_0)=0$, but this is not necessarily true. As a matter of fact, it must be the case that $$p_{X,Y}(x,y_0) = P(X=x,Y=y_0) \leq p_Y(y_0) =P(Y=y_0) \,\,\,\text{for any $x \in D_X$}\,,$$
where $D_X$ is the domain of $X$.