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.
$$p_Y(y_0) \neq \int_{y_0}^{y_0} p_Y(y)\text{d}y = P(y_0 \leq Y \leq y_0)= P(Y \leq y_0)-P(Y \leq y_0)= 0 \,\,,$$ 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 $X$.