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The classical ham sandwich theorem says that given $n$ measurable sets in $\mathbb{R}^n$, it is possible to divide all of them in half (with respect to their measure, i.e. volume) with a single $(n − 1)$-dimensional hyperplane.

Does the theorem remain true if we replace measurable sets with measurable integrable real-valued functions on $\mathbb{R}^n$ and we wish for the hyperplane to divide the integrals of the functions in half? The classical theorem is the case where the functions are characteristic functions of sets in $\mathbb{R}^n$.

More precisely. let $f_1,\dots,f_n$ be real-valued functions in $L^1 (\mathbb{R}^n,m)$ (where $m$ is Lebesgue measure). Can we find an $(n − 1)$-dimensional hyperplane dividing $\mathbb{R}^n$ into two half-spaces $L_1,L_2$, such that $\int_{L_1} f_i dm= \int_{L_2} f_i dm$ for all $i=1,\dots,n$?

My feeling is that this is false, but I'm having trouble finding an elegant counterexample.

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If attempting to adapt the proof of the ham sandwich theorem in a straightforward manner does not work, it may be instructive to study why it does not work, and whether this hurdle can be overcome or leads to intuition for finding a counterexample. (I do not see why it would not work.) –  Jonas Meyer Aug 11 '11 at 19:34
    
What is the question exactly if you replace measurable sets by measurable integrable real-valued functions? I don't think the problem is precisely well defined (in the sense that there could be many ways to generalize this question), or perhaps I am just tired, but that would be good to clear up. Other question : are you only considering Lebesgue measure or arbitrary measure? –  Patrick Da Silva Aug 11 '11 at 19:57
    
@Jonas: I am not familiar with the proof, nor with the tools of algebraic topology topology required to understand it. In this case I'm more interested in the answer than in a proof of it. Patrick: I've edited the question. The measure under consideration is always Lebesgue measure, and all functions are assumed to be Lebesgue measurable and Lebesgue integrable. –  Mark Aug 11 '11 at 20:22
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Just use the proof of the ham sandwich theorem, but adapt it by utilizing integrals of $f$ over space on one or the other side of hyperplanes instead of measures of the halves of bisected sets. It should work just as easily so long as there's no technical snag I'm not catching. The HST becomes a special case with the $f$'s being indicator (characteristic) functions of $n$ sets. –  anon Aug 11 '11 at 20:33
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Looking at the proof on PlanetMath, I don't see how replacing the characteristic function of measurable sets with non-negative integrable functions would change the proof significantly. –  robjohn Aug 11 '11 at 20:49
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