1.2. Lebesgue measure 21 In general, disjoint sets E, F need not have a positive separation from each other (e.g. E = [0, 1) and F = [1, 2]). But the situation improves when E, F are closed, and at least one of E, F is compact: Exercise 1.2.4. Let E, F Rd be disjoint closed sets, with at least one of E, F being compact. Show that dist(E, F ) 0. Give a counterexample to show that this claim fails when the compactness hypothesis is dropped. We already know that countable sets have Lebesgue outer measure zero. Now we start computing the outer measure of some other sets. We begin with elementary sets: Lemma 1.2.6 (Outer measure of elementary sets). Let E be an elementary set. Then the Lebesgue outer measure m∗(E) of E is equal to the elementary measure m(E) of E: m∗(E) = m(E). Remark 1.2.7. Since countable sets have zero outer measure, we note that we have managed to give a proof of Cantor’s theorem that Rd is uncountable. Of course, much quicker proofs of this theorem are available. However, this observation shows that the proof this lemma must somehow use some crucial fact about the real line which is not true for countable subfields of R, such as the rationals Q. In the proof we give here, the key fact about the real line we use is the Heine-Borel theorem, which ultimately exploits the important fact that the reals are complete. In the one-dimensional case d = 1, it is also possible to exploit the fact that the reals are connected as a substitute for completeness (note that proper subfields of the reals are neither connected nor complete). Proof. We already know that m∗(E) m∗,(J)(E) = m(E), so it suffices to show that m(E) m∗(E). We first establish this in the case when the elementary set E is closed. As the elementary set E is also bounded, this allows us to use the powerful Heine-Borel theorem, which asserts that every open cover of E has a finite subcover (or in other words, E is compact). Again, we use the epsilon of room strategy. Let ε 0 be arbitrary, then we can find a countable family B1,B2,... of boxes that cover E, E n=1 Bn, and such that n=1 |Bn| m∗(E) + ε. We would like to use the Heine-Borel theorem, but the boxes Bn need not be open. But this is not a serious problem, as one can spend another epsilon
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