24 1. Measure theory so it suffices to show that n=1 |Bn| m∗(E). But for each natural number N, E contains the elementary set B1 ∪...∪BN , so by monotonicity and Lemma 1.2.6, m∗(E) m∗(B1 . . . BN ) = m(B1 . . . BN ) and thus by (1.3), one has N n=1 |Bn| m∗(E). Letting N we obtain the claim. Remark 1.2.10. The above lemma has the following immediate corollary: if E = n=1 Bn = n=1 Bn can be decomposed in two different ways as the countable union of almost disjoint boxes, then n=1 |Bn| = ∑∞ n=1 |Bn|. Although this statement is intuitively obvious and does not explicitly use the concepts of Lebesgue outer measure or Lebesgue measure, it is remarkably difficult to prove this statement rigorously without essentially using one of these two concepts. (Try it!) Exercise 1.2.5. Show that if a set E Rd is expressible as the countable union of almost disjoint boxes, then the Lebesgue outer measure of E is equal to the Jordan inner measure: m∗(E) = m∗,(J)(E), where we extend the definition of Jordan inner measure to unbounded sets in the obvious manner. Not every set can be expressed as the countable union of almost disjoint boxes (consider for instance the irrationals R\Q, which contain no boxes other than the singleton sets). However, there is an important class of sets of this form, namely the open sets: Lemma 1.2.11. Let E Rd be an open set. Then E can be expressed as the countable union of almost disjoint boxes (and, in fact, as the countable union of almost disjoint closed cubes). Proof. We will use the dyadic mesh structure of the Euclidean space Rd, which is a convenient tool for “discretising” certain aspects of real analysis. Define a closed dyadic cube to be a cube Q of the form Q = i1 2n , i1 + 1 2n × . . . × id 2n , id + 1 2n
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