Divide a finite measure set into two part of equal measure using one hyperplane in R^n.

Question

The term "measure" here are all refer to Lebesgue measure.

A is a measurable set in $\mathbb R^n$ with finite measure, the goal is to prove that there exists a real number $r$ with $m(A \cap\{x:x_n<r\})=m(A\cap\{x:x_n>r\})$, where $x=(x_1,...,x_n)$ in $\mathbb R^n$.

I've tried to prove that measure of finite measure sets is a continuous function and use intermediate value theorem. I've done this for closed cubes, but encountered some problem when trying to prove this for general measurable sets. Also, I was wondering if there's a simpler way to do this. Any help or hint are very appreciated.

Answer 1

What you want to achieve is equivalent to$${m(A \cap\{x:x_n<r\})\over m(A)}=\frac 12$$

To be able to use the IVT, it is enough to prove that $$\lim_\limits{r\to -\infty} m(A \cap\{x:x_n<r\})=0$$ and $$\lim_\limits{r\to +\infty} m(A \cap\{x:x_n<r\})=m(A)$$ but by symmetry any of these equalities will imply the other.

Now the second equality comes from the fact that the series $$\sum_\limits{k=-\infty}^\infty m(A\cap \{x:k<x_n\leq k+1\})$$

converges to $m(A)$ by the continuity from below, and the increasing function of $r$ $$m(A \cap\{x:x_n<r\})$$ can be sandwiched between partial sums of the above series.

Answer 2

Assuming your set is bounded, any hyperplane passing through the center of gravity of your set divides it into two parts of equal measure.

If it is unbounded, it depends if the center of gravity has finite coordinates.