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Spring 2017 - Problem 6

measure theory

Let $\cl{\D}$ be the closed unit disc in the complex plane, let $\set{p_n}_n$ be distinct points in the open disc $\D$ and let $r_n > 0$ be such that the discs $D_n = \set{z \mid \abs{z - p_n} \leq r_n}$ satisfy

$D_n \subseteq \D$ ,
$D_n \cap D_m = \emptyset$ if $n \neq m$ , and
$\sum r_n < \infty$ .

Prove that $X = \cl{\D} \setminus \bigcup_n D_n$ has positive area.

Hint: For $-1 < x < 1$ consider $\#\set{n \mid D_n \cap \set{\Re{z} = x} \neq \emptyset}$ .

Solution.

Let $f\p{x} = \sum_n \chi_{\pi\p{D_n}}\p{x}$ , where $\func{\pi}{\C}{\R}$ , $z \mapsto \Re{z}$ . Thus, $m\p{\pi\p{D_n}} = 2r_n$ , where $m$ denotes the Lebesgue measure, and because each term in the sum is non-negative, the monotone convergence gives us

\int_{-1}^1 f\p{x} \,\diff{x} = \sum_n 2r_n < \infty,

so $f\p{x} < \infty$ for almost every $x \in \p{-1, 1}$ . Fix such an $x$ , so that

f\p{x} = \abs{\set{n \mid D_n \cap \set{\Re{z} = 0} \neq \emptyset}} < \infty.

Since the $D_n$ are disjoint, we see that $D_n \cap \set{\Re{z} = x}$ comprises of finitely many closed segments, which cannot be all of $\D \cap \set{\Re{z} = x}$ , a non-closed segment. Since $\D \cap \set{\Re{z} = x}$ is an open segment, $E_x = \p{\D \cap \set{\Re{z} = x}} \setminus \bigcup_n D_n$ contains an open segment, hence of positive one-dimensional Lebesgue measure.

Thus, the function $\chi_{E_x}\p{y}$ has positive integral for almost every $x \in \p{-1, 1}$ and so

\begin{aligned} m\p{X} &= \int_{\cl{\D}} \chi_{\cl{D} \setminus \bigcup_n D_n}\p{x, y} \,\diff{A} \\ &= \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \chi_{\p{\cl{\D} \setminus \bigcup_n D_n} \cap \set{\Re{z} = x}}\p{y} \,\diff{y} \,\diff{x} \\ &= \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \chi_{E_x}\p{y} \,\diff{y} \,\diff{x} \\ &> 0, \end{aligned}

as required.