# split a measurable set in $\mathbb{R}^2$ into two equally measured subsets

Suppose that $$E\subseteq\mathbb{R}^2$$ is a measurable set with $$m(E)<\infty$$ ($$m$$ is the lebesgue measure). Show that there exists a line from the origin that splits $$E$$ into two equally measured subsets.

Well, if $$m(E)=0$$ the statement is trivial, so lets assume that $$m(E)>0$$. Just for comfort, I define the set: $$E_\alpha =\{(x,y)\in E : y>\alpha x\}$$ for every $$\alpha \in \mathbb{R}$$.

Now let’s define the function:

$$f(\alpha) = \frac{m(E_\alpha)}{m(E)}$$

My idea is to show that $$f$$ is continuous and use the Intermediate value theorem. However, there are two things that I am struggling with:

1. Proving that $$f$$ is continuous.
2. Proving that $$[a,b]\subseteq f(\mathbb{R})$$ for some $$a\leq\frac{1}{2}\leq b$$

Hints will be appreciated

• How about parametrizing by an angle $\theta$? When you rotate by $\pi$, the two sides are interchanged. Commented Sep 13, 2022 at 18:23
• @GEdgar I’m not sure that I understand what you mean… can you explain it a bit more? Commented Sep 13, 2022 at 18:48
• See "Ham Sandwich Theorem" en.wikipedia.org/wiki/Ham_sandwich_theorem Commented Sep 13, 2022 at 19:02

1. It follows from the continuity of the Lebesgue measure (with respect to monotone unions and intersections). Indeed, let $$H:=\left\{x >0\right\}$$. If $$\alpha_n \rightarrow \alpha^+$$ then $$m(E_\alpha \cap H)=m(\bigcup_n E_{\alpha_n} \cap H)=\lim_n m(E_{\alpha_n} \cap H)$$ and similarly, since $$\left\{y=\alpha x\right\}$$ and $$\left\{x=0\right\}$$ have null measure $$m(E_\alpha \cap H^C)=m(\bigcap_n E_{\alpha_n} \cap H^C)=\lim_n m(E_{\alpha_n} \cap H^C)$$ Note that in the second case we need to use $$m(E)<\infty$$ to pass to the limit. In conclusion $$m(E_\alpha)=\lim_n m(E_{\alpha_n})$$ If $$\alpha_n \rightarrow \alpha^-$$ the same argument works, "switching" the two cases. So $$f$$ is continuous.
2. Note that as $$\alpha \rightarrow -\infty$$ (reasoning as in point 1) $$m(E_\alpha) \rightarrow m(E \cap H)$$ while as $$\alpha \rightarrow +\infty$$ $$m(E_\alpha) \rightarrow m(E \cap H^C)$$ Now, if $$\displaystyle m(E \cap H)=\frac{m(E)}{2}$$ then we are done, since $$E$$ is split in two by the vertical axis. Otherwise, we have $$\lim_{\alpha \rightarrow -\infty} f(\alpha)+ \lim_{\alpha \rightarrow +\infty} f(\alpha)=1$$ with exactly one limit being $$< 1/2$$. So by the intermediate value theorem we can find a suitable $$\alpha$$ s.t. $$f(\alpha)=1/2$$.