I have doubts about the proof of the Ham-Sandwich theorem descibed on planetmath () and wikipedia (): There you fix one of the $n$ sets in $\mathbb R^n$ to be bisected and for each "direction" $p\in S^{n-1}$ and $t\in\mathbb R$ you consider all hyperplanes with normal vector $p$ containing $tp$. By the mean value theorem and the continuity properties of the Lebesgue measure you get $t\in \mathbb R$ such that the corresponding hyperplane cuts the set in two parts of equal mass. However, there might be a whole interval of such $t$ and both references cited above claim that you get a continuous function $t(p)$ if you choose the midpoint of that interval. (The $n-1$-dimensional Borsuk-Ulam theorem then finishes the proof.)
This is the point I do not see. The proof on planetmath strongly suggests that this "midpoint-continuity" holds for each continuous function $f: S^{n-1} \times \mathbb R \to \mathbb R$ which is increasing in the second variable and such that the level sets $\lbrace t\in \mathbb R: f(p,t)=0 \rbrace$ are not empty and compact intervals.
A counterexample to this claim is $f(p,t)= \|p-e\| t + \varphi(t)$ where $e$ is any fixed element of $S^{n-1}$ and $\varphi(t)$ is an increasing function whose $0$-set is $[0,1]$ (this is to make the level sets compact). The level sets are then singletons $\lbrace 0\rbrace$ for $p\neq e$ and $[0,1]$ for $p=e$.
I believe that something similar to this may really happen in the Ham-Sandwich situation.
I know that there are other proofs using the $n$-dimensional Borsuk-Ulam theorem. But the one discussed here has the advantage that you get the Pancake theorem from the $1$-dimensional Borsuk-Ulam which is so much simpler than the higher dimensional cases.
Do I misunderstand something?
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$\begingroup$Let $A$ be the set to be bisected, and let $H(p,t)=\{x\colon x\cdot p=t\|p^2\|\}$. Further, let $K$ be the essential support of $A$, i.e., $x\in K$ if every neighbourhood of $x$ meets $A$ in a set of positive measure. Note that $K$ is compact.
Further, for each $p\in S^{n-1}$ let $t_+(p)$ and $t_-(p)$ be the largest and smallest values of $t$ so that $H(p,t)$ bisects $A$ into pieces of equal measure.
Claim: $t_\pm(p)$ are continuous functions of $p$.
At a point $p$ for which $t_-(p)<t_+(p)$, it should be clear that $H(p,t)$ (with $t_-<t<t_+$) divides $K$ into disjoint compact sets $K_\pm$ (this is where boundedness comes in). Further, $t_\pm(p)$ is characterized by $H(p,t)$ just touching either of these sets. The continuity of $t_\pm$ is then easily established.
At a point $p$ where $t_-(p)=t_+(p)$, there are thin slices of $A$ to either side of $H(p,t)$ of positive measure. If you perturb $p$ a little, a small change in $t$ is sufficient to include the entire slice to one or the other side of $H(p,t)$. This will establish the continuity of $t_\pm$ at such a point.
$\endgroup$ 5 $\begingroup$If one simply wishes to prove the ham sandwich theorem (for bounded sets) I think one can avoid this discussion by a limiting argument. Superpose on $A$ a thin gas of uniform density $\rho$ filling a ball of radius (say) twice the diameter of $A$. Now the corresponding function $H(p,t)$ is continuous in $t$ with a positive derivative bounded away from zero, and it easily follows that $t(p)$ is unique and depends continuously on $p$.
Use this and the standard Borsuk-Ulam argument to solve the HST for the modified $A$ and the original $B$ and $C$. Now let the density $\rho$ of the gas go to zero. The set of potential solution planes is compact (a ball times a sphere) so we can extract a convergent subsequence of these planes as $\rho\to 0$. A limit of this subsequence will solve the original problem.
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