Change of variables in integration Under certain conditions on the functions $g:\mathbb{R}^n \to \mathbb{R}$ and $f:\mathbb{R} \to \mathbb{R}$ involved I have seen formulas such as
$$
\int \limits _{\mathbb{R}^n} f\circ g (x) \,dx = \int f(t) \, d \nu (t) ,
$$ 
where 
$$
\nu (t) = - \int \limits _{g(x) \ge t} \,dx .
$$
Sometimes this formula comes without the minus sign and the inequality reversed, which I suppose is so that it is an increasing function. Is the integration in $t$ for all $t$ or just positive $t$?
It appears clear that this is related to the general change of variables formula 
$$
\int f\circ g \, d\mu = \int f \, d\nu  
$$ 
where $\nu (B) = \mu (g^{-1}(B))$, but I must admit I can't see the exact chain of equalities linking them together. 
Is there any good reference explaining these things, or is it really easy to see? 
 A: Here's my take on it. Let $\lambda$ denote the Lebesgue measure on $\mathbb{R}^n$. The function
$$
t\mapsto \nu(t)=-\int_{g(x)\geq t}\,\mathrm dx=-\lambda(\{g\geq t\})
$$ is an increasing right-continuous function on $\mathbb{R}$, and therefore gives rise to a Lebesgue-Stieltjes measure $\mu$ characterized by
$$
\mu(]a,b])=\nu(b)-\nu(a),\quad a,b\in\mathbb{R},\;a<b.
$$
What I take the integral $\int f(t)\,\mathrm d\nu(t)$ to mean is exactly $\int f\,\mathrm d\mu$. So we have to prove that
$$
\int f\,\mathrm d\mu=\int f\circ g\,\mathrm d\lambda
$$
but since $\int f\circ g\,\mathrm d\,\lambda = \int f\,\mathrm d (\lambda\circ g^{-1})$ it is enough to show that $\mu=\lambda\circ g^{-1}$. So let $a,b\in\mathbb{R}$, $a<b$, then
$$
\mu(]a,b])=\nu(b)-\nu(a)=\lambda(\{g\geq a\})-\lambda(\{g\geq b\})=\lambda(\{g\in [a,b[\})=\lambda\circ g^{-1}([a,b[),
$$
but here I am stuck unfortunately. I really want to conclude that $\lambda\circ g^{-1}([a,b[)=\lambda\circ g^{-1}(]a,b])$, but I don't think that this is true for a general $g$. Maybe the assumptions you are mentioning will ensure this?
