Half-space comparison of perimeter

Claim: suppose that $E$ is a set of finite perimeter, and $H$ is a half space. Then $P(F\cap H)\le P(F)$. In words: restricting a Caccioppoli set to a half-space will not increase the perimeter.

My question: how to prove this?

I have a feeling it should be very simple, as in Almgren, Taylor, Wang's paper "Curvature-driven flows: a variational appraoch", it is claimed that it follows from "Stokes' theorem and Jensen's inequality". However, I'm having trouble to apply Jensen's inequality to this problem, except in trivial cases that boil down to linearity of the integral.

EDIT: Although no satisfying answer has been given as to what Jensen has to do with this theorem, a complete proof has been given. I've accepted this as an answer.

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I don't think Jensen's inequality is needed. Use Stokes' theorem to say that the flux of a constant vector field perpendicular to $H$ through $F \cap \delta H$ is the same as the flux through $\delta F \backslash H$. So, the area of $F \cap \delta H$ is smaller than the area of $\delta F \backslash H$. – Douglas Zare Feb 21 '11 at 15:26
I agree with Douglas, you also can use the fact that projection to the hyperspace is 1-Lipshitz. Thus if you project the part outside of H to ∂H, the perimeter is decreasing. Clearly the perimeter of the image of F is at lest the perimeter of F∩H. – Anton Petrunin Feb 21 '11 at 21:26

For simplicity I shall assume that $E$ is bounded, i.e. there exists some ball $B_R(0)\subset \mathbb R^n$ strictly containing $E$. W.l.o.g. we may also assume that $H =$ {$x\in \mathbb R^n:x_n < r$} for some $r$. It holds the following (see the book of Giusti for details) $$P(E \cap H)=P(E,H)+\int_{\partial H}\phi_E^+ d\mathcal H^{n-1},$$ where $\phi_E^+$ denotes the inner trace of the characteristic function $\chi_E$ of $E$ on $\partial H$. Now by definition of the trace operator we have (inserting the vector field $X=-\eta e_n$, where $\eta$ is some smooth cutoff function that equals $1$ on $B_R(0)$) \begin{align} 0&=\int_{E\setminus \overline H}div X d\mathcal L^n=-\int_{\mathbb R^n \setminus \overline H} \langle e_n,\nu_E \rangle d\mu_E - \int_{\partial H}\phi_E^{-}\langle e_n,\nu_H\rangle d\mathcal H^{n-1}, \end{align} where $\nu_H$ denotes the outer unit normal to $H$, $\nu_E$ denotes the generalized outer unit normal of $E$ and $\phi_E^-$ is the outer trace of $E$ on $\partial H$. But since $\nu_H=e_n$ we get, using Cauchy-Schwarz' inequality: $$\int_{\partial H}\phi_E^- d\mathcal H^{n-1} \leq P(E,\mathbb R^n \setminus \overline H),$$ and so for a.e. value of $r$, $\phi_E^+=\phi_E^-$ $\mathcal H^{n-1}$-a.e. on $\partial H$, which yields $$P(E \cap H)\leq P(E,H)+P(E,\mathbb R^n \setminus \overline H)=P(E).$$ For an arbitrary value of $r$ just choose a sequence of values $r_k\to r$ for which this is satisfies. Then $E\cap H_{r_k} \to E \cap H$ in $L_{loc}^1(\mathbb R^n)$. Using the lower semi-continuity of the perimeter you get the claim for all values of $r$.