 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2009/d/d1/solution

m^{*} (E) = in f {\sum_{j = 1}^{\infty} ∣ I_{j} ∣ : E \subseteq ⋃_{j = 1}^{\infty} I_{j}, I_{j} open intervals} .

m^{*} (E) = 0

, equivalently it admits such a cover of total length below every positive

ε

The usual Cantor set is the intersection of the sets obtained from

[0, 1]

by repeatedly removing every open middle third. At stage

n

there are

2^{n}

closed intervals of length

3^{- n}

. Enlarge them slightly to open intervals whose total added length is below

ε

. Thus

m^{*} (C) \leq (2/3)^{n} + ε

for every

n

and

ε > 0

, proving

m^{*} (C) = 0

For the different deletion rule in this problem, let

L_{n}

be the total length of the stage-

n

intervals. Removing the stated fraction from each component gives

L_{n + 1} = L_{n} \frac{2 ^{n} + 1}{2 ^{n} + 2}, L_{0} = 1.

For

n \geq 1

L_{n} = \prod_{j = 0}^{n - 1} \frac{2 ^{j} + 1}{2 ^{j} + 2} = \frac{2}{3} \prod_{j = 1}^{n - 1} \frac{2 ^{j} + 1}{2 ( 2 ^{j - 1} + 1 )} = \frac{2 ^{n - 1} + 1}{3 2 ^{n - 1}} .

Because

K \subseteq K_{n}

, the same finite interval-enlargement argument gives

m^{*} (K) \leq L_{n}

. Taking the limit proves

m^{*} (K) \leq \frac{1}{3} .

Only interval covers and their lengths were used; no continuity theorem for integrals is needed.

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