Let $E \subset \Re$ be measurable with $\mu(E) > 0$. Show that there are disjoint sets $B$ and $C$ such that $E=B \cup C$ and $\mu(B) = \mu(C) = \frac12\mu(E)$.
Solution:
Assume $\mu(E) = \infty$. $f(t) = \mu(E \bigcap (-\infty, t))$.
$\lim\limits_{t \rightarrow -\infty} f(t) = 0$
$\lim\limits_{t \rightarrow \infty} f(t) = \mu(E)$
Somewhere in between is $\frac12\mu(E)$.
If $\mu(E)<+\infty$, then consider function $f(t)=\mu(E\cap(-\infty,t))$. Then $$ \lim\limits_{t\to-\infty}f(t)=0\qquad\qquad\lim\limits_{t\to+\infty}f(t)=\mu(E)\tag{1} $$ Note that for $t_1<t_2$ we have $|f(t_2)-f(t_1)|=\mu(E\cap([t_1,t_2))\leq\mu([t_1,t_2))=t_2-t_1$. Hence $f$ is Lipschitz function, and as a consequence continuous. Then using $(1)$ we conclude that there exist $t_0\in\mathbb{R}$ such that $f(t_0)=0.5\mu(E)$. It remains to set $B=E\cap(-\infty,t_0)$, $C=E\setminus B$.
If $\mu(E)=+\infty$. Consider the sets $E_n=E\cap[n,n+1)$ for each $n\in\mathbb{N}$. These sets are disjoint sets of finite measure whose union is $E$. From the previous case we have for each $n\in\mathbb{N}$ a pair of disjoint sets $B_n$, $C_n$ such that $E=B_n\cup C_n$ and $\mu(B_n)=\mu(C_n)=0.5\mu(E_n)$. Now consider $$ B=\bigcup\limits_{n=1}^\infty B_n\qquad\qquad C=\bigcup\limits_{n=1}^\infty C_n $$ It is easy to check that they are disjoint and $E=B\cup C$, $\mu(B)=\mu(C)=\infty=0.5\mu(E)$.
So, I should also provide a solution for when $0 < \mu(E) < \infty$?
– Jake Casey May 08 '13 at 17:48