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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?

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1 Answer 1

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?

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Thanks a lot! If I remember correctly $g$ was a smooth non-negative function. –  flavio Dec 20 '12 at 12:19
    
Hmm, still not sure if this would ensure that $\lambda(\{g=a\})=0$ for all $a$. –  Stefan Hansen Dec 20 '12 at 12:21

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