Analysis Problem of the Day 79

Today’s problem appeared as Problem 5 on the UCLA Fall 2009 Analysis Qual:

Problem 9. Construct a Borel subset $E$ of $\mathbb{R}$ such that $0<m(E \cap [a,b])<b-a$ for all $a<b, a, b \in \mathbb{R}.$

Solution: We are asked to construct a subset that “lives” everywhere on the real line, yet which is not the complement of a measure zero set. Recall that for every $0 < \lambda < 1,$ there exists a fat Cantor set $C_\lambda \subseteq [0,1]$ of measure $\lambda$ which is Borel (as it is closed). Note that for every natural $n, \mathbb{R}$ is a countable union of disjoint copies of half-open intervals of size $2^{-n}.$ Let $I_1,I_2,...$ be an enumeration of these intervals for all $n,$ which is countable as a countable union of countable sets. For each $I_j=[a_j,b_j),$ let $C_j:=(b_j-a_j)C_{\frac{b_j-a_j}{2^{j+8}}}+a_j$ be the Cantor set with measure $2^{-(j+8)}$ of the measure of $I_j$ placed into $I_j.$ I claim that $C = \bigcup_{n=1}^\infty C_n$ is the desired Borel set. Indeed, any interval $[a,b]$ will contain some subinterval $I_j$ of small enough length, and since $C_j \subset I_j \subset [a,b)$ and $m(C_j)>0,$ we have $m(C \cap [a,b]) >0.$ Moreover, note that any interval $[a,b]$ can intersect at most $N:=\lceil(b-a)2^{n} \rceil+1$ subintervals of length $2^{-n},$ which can contribute at most

$2^{-n} \sum_{j=1}^{N} 2^{-j} = 2^{-n} (2-2^{-N}) \leq 2^{1-n}$

to the measure of $C$ on $[a,b].$ Since $\sum_{n=1}^\infty 2^{-n}<\infty,$ the contributions of the fat Cantor sets for $2^{-n} \leq \frac{b-a}{8}$ are clearly bounded above by $\frac{b-a}{4}.$ Finally, for $2^{-n} \geq \frac{b-a}{8}$ there are at most $2+3+5+9 \leq 20$ intervals for $n=1,2,3,4$ intersecting $[a,b],$ which contribute less than

$32 \sum_{j=0}^{19} 2^{-(j+8)} \leq \frac{b-a}{2}$

to the measure of $C.$ Thus, $m(C \cap [a,b]) \leq \frac{b-a}{4}+\frac{b-a}{2} \leq \frac{3(b-a)}{4}<b-a.$ Thus, $C$ satisfies the desired properties.

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