1.2. Lebesgue measure 23 Remark 1.2.8. We are now able to explain why not every bounded open set or compact set is Jordan measurable. Consider the countable set Q ∩ [0, 1], which we enumerate as {q1,q2,q3,...}, let ε 0 be a small number, and consider the set U := ∞ n=1 (qn − ε/2n,qn + ε/2n). This is the union of open sets and is thus open. On the other hand, by countable subadditivity, one has m∗(U) ≤ ∞ n=1 2ε/2n = 2ε. Finally, as U is dense in [0, 1] (i.e. U contains [0, 1]), we have m∗,(J)(U) = m∗,(J)(U) ≥ m∗,(J)([0, 1]) = 1. For ε small enough (e.g. ε := 1/3), we see that the Lebesgue outer measure and Jordan outer measure of U disagree. Using (1.2), we conclude that the bounded open set U is not Jordan measurable. This in turn implies that the complement of U in, say, [−2, 2], is also not Jordan measurable, despite being a compact set. Now we turn to countable unions of boxes. It is convenient to introduce the following notion: two boxes are almost disjoint if their interiors are disjoint, thus, for instance, [0, 1] and [1, 2] are almost disjoint. As a box has the same elementary measure as its interior, we see that the finite additivity property (1.3) m(B1 ∪ . . . ∪ Bk) = |B1| + . . . + |Bk| holds for almost disjoint boxes B1,...,Bk, and not just for disjoint boxes. This (and Lemma 1.2.6) has the following consequence: Lemma 1.2.9 (Outer measure of countable unions of almost disjoint boxes). Let E = ∞ n=1 Bn be a countable union of almost disjoint boxes B1,B2,.... Then m∗(E) = ∞ n=1 |Bn|. Thus, for instance, Rd itself has an infinite outer measure. Proof. From countable subadditivity and Lemma 1.2.6 we have m∗(E) ≤ ∞ n=1 m∗(Bn) = ∞ n=1 |Bn|,

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