Homepage Solution manuals Terence Tao An Introduction to Measure Theory Exercise 1.2.26 (Lebesgue outer measure is not additive)

Exercise 1.2.26 (Lebesgue outer measure is not additive)

Show that there exist disjoint bounded subsets E,F of the real line such that

m(E F)m(E) + m(F)

Answers

Consider the sets E and X from the proof of Proposition 1.2.18. From the countable sub-additivity and translation invariance we have

m(X) q[1,1]m(q+E) = q[1,1]m(E) = { 0 if m(E) = 0 if m(E) > 0

We have demonstrated that 1 m(X) 3 which contradicts to the first case m(E) = 0. Thus, m(E) > 0 (maybe small). Pick an n large enough so that 1n < m(E). If m were finitely additive, then for a subset J [1,1] such that #J = 3n we would have the result

m ( qJq + E) = qJm(q + E) = qJm(E) = 3n m(E) > 3

But qJq + E is a subset of X which has outer measure of at most 3. This contradicts monotonicity, and we conclude that m is not (finitely) additive.

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2020-05-30 00:00
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