1.2. Lebesgue measure 19 X, which is an assigment E → m∗(E) of element of [0, +∞] to arbitrary subsets E of a space X that obeys the above three axioms of the empty set, monotonicity, and countable subadditivity thus Lebesgue outer measure is a model example of an abstract outer measure. On the other hand (and somewhat confusingly), Jordan outer measure will not be an abstract outer measure (even after adopting the convention that unbounded sets have Jor- dan outer measure +∞): it obeys the empty set and monotonicity axioms, but is only finitely subadditive rather than countably subadditive. (For in- stance, the rationals Q have infinite Jordan outer measure, despite being the countable union of points, each of which have a Jordan outer measure of zero.) Thus we already see a major benefit of allowing countable unions of boxes in the definition of Lebesgue outer measure, in contrast to the fi- nite unions of boxes in the definition of Jordan outer measure, in that finite subadditivity is upgraded to countable subadditivity. Of course, one cannot hope to upgrade countable subadditivity to un- countable subadditivity: Rd is an uncountable union of points, each of which has Lebesgue outer measure zero, but (as we shall shortly see), Rd has infi- nite Lebesgue outer measure. It is natural to ask whether Lebesgue outer measure has the finite addi- tivity property, that is to say that m∗(E ∪ F ) = m∗(E) + m∗(F ) whenever E, F ⊂ Rd are disjoint. The answer to this question is somewhat subtle: as we shall see later, we have finite additivity (and even countable additivity) when all sets involved are Lebesgue measurable, but that finite additivity (and hence also countable additivity) can break down in the non-measurable case. The diﬃculty here (which, incidentally, also appears in the theory of Jordan outer measure) is that if E and F are suﬃciently “entangled” with each other, it is not always possible to take a countable cover of E ∪ F by boxes and split the total volume of that cover into separate covers of E and F without some duplication. However, we can at least recover finite additivity if the sets E, F are separated by some positive distance: Lemma 1.2.5 (Finite additivity for separated sets). Let E, F ⊂ Rd be such that dist(E, F ) 0, where dist(E, F ) := inf{|x − y| : x ∈ E, y ∈ F } is the distance10 between E and F . Then m∗(E ∪ F ) = m∗(E) + m∗(F ). Proof. From subadditivity one has m∗(E ∪ F ) ≤ m∗(E) + m∗(F ), so it suﬃces to prove the other direction m∗(E) + m∗(F ) ≤ m∗(E ∪ F ). This is trivial if E ∪ F has infinite Lebesgue outer measure, so we may assume that 10Recall from the preface that we use the usual Euclidean metric |(x1, . . . , xd)| := x2 1 + . . . + x2 d on Rd.

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