In the context of probability theory, density functions and distribution functions I stumbled upon the following.
Let's assume that $\int\limits_{-\infty}^{\infty}f(x)dx=1$ and $g:\mathbb{R}\to \mathbb{R}$ is a bijective continuously differentiable function. We want to show that $$ \int\limits_{-\infty}^{\infty}f(g^{-1}(y))(g^{-1}(y))'dy=1. $$ First, we start with integration by substitution, where $x:=g^{-1}(y)$. We define the preimages by $b:=g^{-1}(\beta)$ and $a:=g^{-1}(\alpha)$ and get $$ \int\limits_{a}^{b}f(x)dx=\int\limits_{\alpha}^{\beta}f(g^{-1}(y))(g^{-1}(y))'dx. $$ Now, letting $\beta\to\infty$ and $\alpha\to-\infty$ I am not sure how to correctly argue that this implies $a\to-\infty$ and $b\to\infty$. Intuitively, it makes sense because by letting $\beta\to\infty$ and $\alpha\to-\infty$ I cover the whole image set of $g$. This means that I need to gather all the preimages/the whole domain of $g$ which is $\mathbb{R}$ or in other words $(-\infty,\infty)$. But this sounds very sloppy. Do you have any suggestions to make this a bit more formal?