Suppose we have a compact set $X \subset \mathbb{R}^n$. Then for every unit vector $\hat{v} \in S^{n-1}$ there exists an affine plane $P_{\hat{v}} \subset \mathbb{R}^n$ such that $P_{\hat{v}}$ is orthogonal to $\hat{v}$ and $P_{\hat{v}}$ divides $X$ into two pieces of equal measure.
This was mentioned in my topology class but I have no idea why it is true and why compactness is required.
On
To mathguy's answer, I would add the following simple counterexample which illustrates the necessity of the hypothesis that the set is compact (or at least of finite measure). Take $n=1$ and $X=[0,\infty)$. Then no matter how you cut $X$, one half will have finite measure and the other half will have infinite measure.
Why it's true: A continuity argument. The set is bounded, so if you move the orthogonal hyperplane far enough in either direction, on one side you will have measure zero and the set $X$ will be on the other side. If you take the difference, it will go from $-\mu(X)$ to $\mu(X)$. Also as you move the hyperplane, the difference is continuous, also because $X$ is bounded, so by the intermediate value theorem there is a position where the difference is 0.
You don't need compact but you do need bounded and measurable.