Partition a set in $\mathbb R^2$ into equal measure sets

Question

1. Given a finite measure measurable set $E$ in $\mathbb R^2$ can we find a subset $A$ such that $m(A) = m(E \setminus A)$?

2.The question given to me is : given a positive $L^1$ function $f$ on $\mathbb R^2$ can we find a subset $A$ such that integral of $f$ over $A$ = integral of $f$ over $\mathbb R^2\setminus A$?

Defining a new measure as integral of $f$, the sum reduces to proving part 1. Now what?

Note : Everything above is wrt Lebesgue measure.

Answer 1

Theorem. Let $m$ be any measure on $\mathbb R^2$ such that $m(L)=0$, for every vertical line $L⊆\mathbb R^2$ (such as any measure that is absolutely continuous with respect to Lebesgue's measure). Then, for every measurable set $E⊆\mathbb R^2$, with $m(E)<\infty$, there exists a measurable subset $A⊆E$ such that $m(A)=m(E)/2$.

Proof. For each interval $I\subseteq \mathbb R$, consider the vertical strip $$ V_I=I\times \mathbb R, $$ and let $$ E_I = V_I\cap E. $$

Fixing any real number $t_0$, notice that $$ \bigcap_{t>t_0} (-\infty, t] = (-\infty, t_0] \qquad \text { and } \qquad \bigcup_{t<t_0} (-\infty, t] = (-\infty, t_0). $$ Consequently $$ \bigcap_{t>t_0} E_{(-\infty, t]} = E_{(-\infty, t_0]} \qquad \text { and } \qquad \bigcup_{t<t_0} E_{(-\infty, t]} = E_{(-\infty, t_0)}. $$ Using that $E$ has finite measure we deduce that $$ \inf_{t>t_0} m (E_{(-\infty, t]}) = m (E_{(-\infty, t_0]}) \qquad \text { and } \qquad \sup_{t<t_0} m (E_{(-\infty, t]}) = m (E_{(-\infty, t_0)}). $$ Because positive measures are monotonic functions we also get $$ \lim_{t\to t_0^+} m (E_{(-\infty, t]}) = m (E_{(-\infty, t_0]}) \qquad \text { and } \qquad \lim_{t\to t_0^-} m (E_{(-\infty, t]}) = m (E_{(-\infty, t_0)}). $$ Regarding the function $ f(t):= m (E_{(-\infty, t]}), $ we conclude from the first identity above that $f$ is right-continuous. Moreover, referring to the degenerate interval $[t_0,t_0]$, observe that $$ m (E_{(-\infty, t_0]})= m (E_{(-\infty, t_0)}) + m (E_{[t_0 t_0]})= m (E_{(-\infty, t_0)}), $$ where we have used the hypothesis to conclude that $m (E_{[t_0 t_0]})= 0$. We therefore also have that $f$ is left-continuous, hence continuos.

Since $$ \lim_{t\to-\infty}f(t) = 0 \qquad \text { and } \qquad \lim_{t\to\infty}f(t) = m (E), $$ we can now employ the intermediate value Theorem to obtain some $t_1$ such that $f(t_1)=m (E)/2$. The desired set $A$ mentioned in the statement can then taken to be $A=E_{(-\infty,t_1]}$.