There is a plane bisecting a compact set in $\Bbb R^3$ Let $A$ be a compact set in $\Bbb R^3$, and let $u\in S^2$ be a unit vector. I want to show that there is a plane that is normal to $u$, and that bisects $A$ into two pieces of equal measure.
At least intuitively, this is obvious: image continuously sliding an affine plane along the line determined by $u$ and apply the intermediate value theorem. But I can't give or find a rigorous proof of this. Any hints?
(This result is needed to prove the ham sandwich theorem)
 A: Let $f:\Bbb R\to\Bbb R$ be defined as follows: for any real number $t$, we set $f(t)$ to the the measure of the intersection of $A$ and $\{x\in\Bbb R^3\mid x\cdot u\leq t\}$.
We have that $f(t)=0$ for any small enough $t$, and equal to the measure of $A$ for any large enough $t$. And $f$ is continuous. So by the intermediate value theorem, there must be a $t_0$ for which $f(t_0)$ has value half the measure of $A$. Then $x\cdot u=t_0$ is the plane you're looking for.
A: Apply Fubini Tonelli Theorem. More precisely
W.l.o.g. we can assume that $u$ is vertical. Then define $f(t)$ as the area of $A\cap\{z=t\}$. Then define $F(z)=\int_{-\infty}^z f(t)df$.
Since $A$ is compact, so is $A\cap \{z=t\}$, which is therefore measurable, hence $f(t)$ is well defined and measurable on $t$. It follows that $F(z)=\int_{\infty}^z f(t)df$ is continuous in $t$. By Fubini Tonelli $F(s)$ is just the volume of $A\cap\{z\leq s\}$. The function $G(z)=Volume(A)-F(t)=Volume(A\cap\{z\geq t\})$ is continuous. The function $G(z)-F(z)$ is continuous, is positive for $z$ small enough and negative for $z$ big enough because $A$ is compact, hence bounded. The intermediate valued theorem now applies.
