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:

*

*Proving that $f$ is continuous.

*Proving that $[a,b]\subseteq f(\mathbb{R})$ for some $a\leq\frac{1}{2}\leq b$
Hints will be appreciated
 A: 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. $f(\alpha)=1/2$.
(Note: This is basically the argument suggested by @GEdgar in the comments)
