Let $B_1, \ldots B_n$ be Borel sets in $\mathbb{H}^n$, and $\mu$ (absolutely continuous w.r.t) the hyperbolic volume. Is there a horosphere $H$ that cuts each of the $B_i$'s into two parts with equal $\mu$-measure ?
(For Euclidean space with hyperplanes in place of horosphere this is the folklore Ham Sandwich theorem)
At first I thought the same proof would do, namely apply Borsuk-Ulam theorem to a well-chosen sphere, but I am stuck on this (perhaps wrong) track: the space $\mathcal{H}^+$ of horoballs in $\mathbb{H}^n$ is topologically $S^n$ with two points removed, as well as the space $\mathcal{H}^-$ of complementary sets of horoballs ; glue $\mathcal{H}^+$ and $\mathcal{H}^-$ together by adding the points $\mathbb{H}^n$ and $\emptyset$ and equip the result (say $X$) with the $\mathbb{Z}_2$ action defined by taking the complementary. A solution will not occur near the gluing points, so one can perturb slightly $X$ near them to turn it into $S^n$. Sadly this $S^n$ does not have the adequate $\mathbb{Z}_2$ action needed for Borsuk-Ulam theorem...
Nice question.This fails however already in the case $n=2$ though, taking $B_1, B_2$ two concentric balls of suitable radii, $r_1, r_2$ with $r_1$ is sufficiently small. To prove this consider first the case when $r_1=0$ and the measure is positive concentrated at one point, the center. Once you understood this case, argue by contradiction for $r_1>0$, taking a suitable limit as $r_1\to 0+$.