TITLE: Proof of the Ham-Sandwich theorem
QUESTION [6 upvotes]: I have doubts about the proof of the Ham-Sandwich theorem descibed on planetmath (http://planetmath.org/proofofhamsandwichtheorem) and wikipedia (http://en.wikipedia.org/wiki/Ham_sandwich_theorem): 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?

REPLY [2 votes]: 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.