<Home>Qualifying Exams><Analysis>Fall 2016 - Problem 2

Fall 2016 - Problem 2

Lebesgue differentiation theorem

Let $\mu$ be a finite positive Borel measure on $\R$ that is singular to Lebesgue measure. Show that

\lim_{r\to0^+}\frac{\mu\p{\br{x - r, x+ r}}}{2r} = \infty

for $\mu$ -almost every $x \in \R$ .

Solution.

Since $\mu$ is singular with respect to Lebesgue measure (which we denote $m$ ), there exists $S \subseteq \R$ Borel such that $\mu\p{S} = 0$ and $m\p{S^\comp} = 0$ . For $k \in \N$ , let

E_k = \set{x \in S^\comp \st \limsup_{r\to0^+} \frac{\mu\p{B\p{x, r}}}{m\p{B\p{x, r}}} < \frac{1}{k}}.

Let $\epsilon > 0$ . By regularity of the Lebesgue measure, there exists an open set $U \supseteq S^\comp$ such that

m\p{U} \leq m\p{U \setminus S^\comp} < \epsilon.

For any $x \in E_k$ , there exists an $r_x > 0$ so that $B\p{x, r_x} \subseteq U$ (since $U$ is open) and

\frac{\mu\p{B\p{x, 5r_x}}}{m\p{B\p{x, 5r_x}}} < \frac{1}{k},

by definition of $E_k$ . By the Vitali covering lemma, there exists a subset $\set{x_n}_n \subseteq E_k$ such that

\bigcup_{x \in S^\comp} B\p{x, r_x} \subseteq \bigcup_{n=1}^\infty B\p{x_n, 5r_n},

where $r_n = r_{x_n}$ and $\set{B\p{x_n, r_n}}_n$ are pairwise disjoint. Hence,

\begin{aligned} \mu\p{S^\comp} &\leq \mu\p{\bigcup_{x \in S^\comp} B\p{x, r_x}} \\ &\leq \mu\p{\bigcup_{n=1}^\infty B\p{x, 5r_n}} \\ &\leq \sum_{n=1}^\infty \mu\p{B\p{x_n, 5r_n}} \\ &\leq \frac{1}{k}\sum_{n=1}^\infty m\p{B\p{x_n, 5r_n}} && \p{x_n \in E_k} \\ &= \frac{5}{k} m\p{\bigcup_{n=1}^\infty B\p{x_n, r_n}} && \p{B\p{x_n, r_n} \text{ are disjoint}} \\ &\leq \frac{5}{k} m\p{U} \\ &\leq \frac{5\epsilon}{k}. \end{aligned}

Hence, $\mu\p{E_k} = 0$ for all $k$ , so on $S^\comp$ , the limit holds $\mu$ -almost everywhere, hence $\mu$ -almost everywhere on $\R$ since $\mu\p{S} = 0$ .