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Consider a smooth function $g(x) \colon \mathbb{R}^n \to \mathbb{R}$ such that $\nabla g > 0$ entrywise. Let $M_t = \{ x \mid g(x) = t \}$. Assume that $\{M_t\}$ don't intersect pairwise and their union gives $\mathbb{R}^n$. Let $dg \wedge \omega = dx_1 \wedge ... \wedge dx_n \equiv dx.$ How to show than that $$ \int\limits_{\mathbb{R}^n} f(g(x),x) dx = \int\limits_{\mathbb{R}} dt \int\limits_{M_t}f(t,x)\omega $$

share|cite|improve this question I think I have to deal with this theorem, but how to specify a form corresponding to the Hausdorff measure? Is it always possible? – Nimza Feb 6 '12 at 20:36
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  1. Using the coarea formula we obtain $$ \int\limits_{\mathbb{R}^n} f(g(x),x) dx = \int\limits_{\mathbb{R}} dt \int\limits_{g(x)=t} \frac{f(t,x)}{| \nabla g(x) |} dH^{n-1}(x) $$
  2. If $dS$ is a volume form on $\{ g(x) = t \}$ then $$ \int\limits_{g(x) = t} \frac{f(t,x)}{| \nabla g(x) |} dH^{n-1}(x) = \int\limits_{g(x) = t} \frac{f(t,x)}{| \nabla g(x) |} dS $$
  3. A form $\alpha = \frac{dS}{| \nabla g(x) |}$ satisfies $dg \wedge \alpha = dx$. Then for any form $\omega$ such that $dg \wedge \omega = dx$ we have $$ \int\limits_{g(x) = t} \frac{f(t,x)}{| \nabla g(x) |}dS = \int\limits_{ g(x) = t} f(t,x) \omega $$ because it's restriction to the tangent bundle of $\{ g(x) = t \}$ is the same as for $\alpha$.
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