I Finding the pdf of a transformed univariate random variable

[Hamiltonian]

TL;DR Summary

Confused as to how to obtain the cdf of a transformed random variable.

The above theorem is trying to find the pdf of a transformed random variable, it attempts to do so by "first principles", starting by using the definition of cdf, I don't understand why they have a $f_X(x)$ in the integral wouldn't $\int_{\{x:r(x)<y\}}r(X) dx$ be the correct integral for the cdf of Y.

 

[FactChecker]

No. The way $r(x)$ gets into the equation is only in the limit of the integral, $r(x) \lt y$. Suppose $r(x) \lt y$. What is the probability (density) of that? It is the probability (density) of that associated $x$ value, which is $f_X(x)$. So those are the probabilities that we want to total for the integral. (Notice that the integral is with respect to $dx$, not $dy$)

PS. I may have ignored the situation where multiple $x$ values give the same $r(x)$. That still works out because the integral limit allows the associated $x$ densities to be summed

 

[Trollfaz]

Suppose $Y=f(X)$ and we know the pdf of X. To get the pdf of Y, we can find the CDF of Y $P(Y<y)$ then differentiate it wrt to y to get the pdf.
$$P(Y<y)=P(f(X)<y)$$
And depending on X you have to do an appropriate manipulation to get the cdf of Y. Here's and example, deriving the pdf of a $\chi^2$ function with a deg of freedom of one. This variable here is just $Z^2$, Z is a standard normal distribution. The pdf of Z is a bit complex but you can find it here.
https://www.thoughtco.com/normal-distribution-bell-curve-formula-3126278 lets call this function f(x).
And cdf of f is $\int_{-\infty}^{x}f(x) dx$ and called F(x). Let X be a standard normal variable, and $\chi=X^2$. So
$$P(\chi<y)=P(-\sqrt{y}<X<\sqrt{y})$$
which is $F(\sqrt{y})-F(-\sqrt{y}$ differentiating wrt y
$$P(\chi=y)\frac{1}{2\sqrt{y}} f(\sqrt{y})+\frac{1}{2\sqrt{y}} f(\sqrt{y})=\frac{1}{\sqrt{y}} f(\sqrt{y})$$

 

[Trollfaz]

Apologies the website gave gives the formula for the general normal distribution for the standard normal, take $\sigma=1,\mu=0$ in the equation