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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.

  • 1
    Looking at the [proof on PlanetMath](http://planetmath.org/?op=getobj&from=objects&id=8882), I don't see how replacing the characteristic function of measurable sets with _non-negative_ integrable functions would change the proof significantly.2011-08-11

0 Answers 0