Spring 2021 - Problem 1

Solution.

First suppose $\mu \ll m$ so that $\diff\mu = f \,\diff{x}$ with $f \in L^1\p{\br{0,1}}$. For $t \geq 0$, let $F_t = \set{f\p{x} > t}$ and set

so that $m\p{F_\lambda} \leq \frac{1}{2}$. Observe that

Set $E_\lambda = \set{f\p{x} = \lambda}$, and notice that $x \mapsto m\p{E_\lambda \cap \br{0, x}}$ is a continuous function by continuity of measures. In particular, by the intermediate value theorem and our estimate above, we see that there exists $A \subseteq E_\lambda$ such that $m\p{A} = \frac{1}{2} - m\p{F_\lambda}$. Thus, if we let $F = F_\lambda \cup A$, then $m\p{F} = \frac{1}{2}$. We claim that $\mu\p{F} = C$ as well. Indeed, suppose $E \subseteq \br{0, 1}$ satisfies $m\p{E} = \frac{1}{2}$. Observe that

so because $m\p{E} = m\p{F}$, we have $m\p{E \setminus F} = m\p{F \setminus E}$. But on $E \setminus F$, we have $f \leq \lambda$, and on $F \setminus E$, we have $f \geq \lambda$. Thus,

so because $E$ was arbitrary, it follows that $\mu\p{F} = C$.

Finally, in the most general case, we may apply the Lebesgue-Radon-Nikodym theorem to write $\diff\mu = f \,\diff{x} + \diff\lambda$, where $\lambda$ is singular with respect to Lebesgue measure. Then by the special case, there exists $F$ such that $m\p{F} = \frac{1}{2}$ and $\int_F f \,\diff{x} = C$. Let $S$ be such that $m\p{S} = \lambda\p{S^\comp} = 0$, and let $F' = F \cap S^\comp$. Then $\lambda\p{F'} = 0$ and so

which was what we wanted to show.