Law of total probability and continuous random variables

I am not able to find an intuitive explanation of the density function of the product of two random variables.

A general argument for the mistake I was doing is as follows.

If the random variables (let's say $$X$$ and $$Y$$) are discrete, then we can use the law of total probability (which seems intuitive to me) to find a function of these two, which becomes -

If $$Z = f(X, Y)$$ and $$X = g(Z, Y)$$, then

$$P(Z = z) = \sum_{y} {P(X = g(z, y), Y = y)}$$

But when it comes to the case of continuous random variables, we seem to adjust by adding a Jacobian. I am not able to understand this very point from the perspective of only probability theory. What cases are we not considering if we say that

$$f_Z (z) = \int_{y} {f_{X,Y} (g(z, y), \ y) \ dy}$$

\begin{align}\dfrac{\mathsf d ~\mathsf P(Z\leqslant z)}{\mathsf dz} ~&=~ \int_{\Bbb R} \begin{vmatrix}\dfrac{\mathsf d^2~\mathsf P(Z\leqslant z, Y\leqslant y)}{\mathsf dz~\mathsf d y}\end{vmatrix}~\mathsf dy\\ f_Z(z)~ &=~ \int_\Bbb R \begin{Vmatrix}\dfrac{\partial~[g(z,y),y]}{\partial~[z,y]}\end{Vmatrix}f_{X,Y}(g(z,y),y)~\mathsf dy\end{align}
• I understand that but I am looking for an explanation from probability theory and not calculus. What is wrong in saying that I am just summing (or integrating) up all the cases (since they are mutually exclusive and exhaustive)? In order to make $Z = f(X, Y)$, we can sum all the mutually exclusive cases by considering different values of $y$. – Kartik Sharma Sep 30 '18 at 9:41
• Because there are uncountably many such cases, therefore you cannot just "sum"; it is an integration. You have to use calculus. So if you want to change the derivative function from a measure over $Z,Y$ to a measure over $X,Y$, therefore you need to use a change of variables theorem (the Jaccobian). – Graham Kemp Sep 30 '18 at 11:55