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strong text In GT's book(1998 Edition) Chapter9 P223, Let $g$ be a nonnegative, locally integrable function in $\mathbb{R}^n$ and $u\in C^2(\Omega)\bigcap C^0(\bar\Omega)$.

How to prove

$\int_{Du(\Omega)}{g(p)}dp\le \int_{\Gamma^{+}}{g(Du)|\det D^2u|}dx$?

where $\Gamma^{+}=\{y\in\Omega|u(x)\le u(y)+ Du(y)\cdot(x-y), for \forall x\in \Omega\}$

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More precisely it is the area formula, see 3.7 of Frank Morgan, Geometric Measure Theory for a sketch of the proof, or 3.2.3 of Federer, Geometric Measure Theory for a complete proof.

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  • $\begingroup$ I know that $\int_{(Du-\varepsilon I)(\Omega)}{g(p)}dp= \int_{\Gamma^{+}}{g(Du-\varepsilon I)|\det (D^2u-\varepsilon I)|}dx$. But I don't know how to use approximation. $\endgroup$
    – 张德凯
    Commented Apr 7, 2015 at 4:02
  • $\begingroup$ I'm not sure what approximation you're talking about. Presumably there's no such approximation in Gilbarg-Trudinger. $\endgroup$
    – Fan Zheng
    Commented Apr 7, 2015 at 4:31

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