26 1. Measure theory so it suﬃces to show that inf E⊂U,U open m∗(U) ≤ m∗(E). This is trivial for m∗(E) infinite, so we may assume that m∗(E) is finite. Let ε 0. By definition of outer measure, there exists a countable family B1,B2,... of boxes covering E such that ∞ n=1 |Bn| ≤ m∗(E) + ε. We use the ε/2n trick again. We can enlarge each of these boxes Bn to an open box Bn such that |Bn| ≤ |Bn| + ε/2n. Then the set ∞ n=1 Bn, being a union of open sets, is itself open, and contains E and ∞ n=1 |Bn| ≤ m∗(E) + ε + ∞ n=1 ε/2n = m∗(E) + 2ε. By countable subadditivity, this implies that m∗( ∞ n=1 Bn) ≤ m∗(E) + 2ε and thus inf E⊂U,U open m∗(U) ≤ m∗(E) + 2ε. As ε 0 was arbitrary, we obtain the claim. Exercise 1.2.6. Give an example to show that the reverse statement m∗(E) = sup U⊂E,U open m∗(U) is false. (For the corrected version of this statement, see Exercise 1.2.15.) 1.2.2. Lebesgue measurability. We now define the notion of a Lebesgue measurable set as one which can be eﬃciently contained in open sets in the sense of Definition 1.2.2, and set out their basic properties. First, we show that there are plenty of Lebesgue measurable sets. Lemma 1.2.13 (Existence of Lebesgue measurable sets). (i) Every open set is Lebesgue measurable. (ii) Every closed set is Lebesgue measurable. (iii) Every set of Lebesgue outer measure zero is measurable. (Such sets are called null sets.) (iv) The empty set ∅ is Lebesgue measurable. (v) If E ⊂ Rd is Lebesgue measurable, then so is its complement Rd\E.

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