1.1. Prologue: The problem of measure 5 length), thus for instance, an interval is a one-dimensional box. The volume |B| of such a box B is defined as |B| := |I1| × . . . × |Id|. An elementary set is any subset of Rd which is the union of a finite number of boxes. Exercise 1.1.1 (Boolean closure). Show that if E, F Rd are elementary sets, then the union E F , the intersection E F , and the set theoretic difference E\F := {x E : x F }, and the symmetric difference EΔF := (E\F ) (F \E) are also elementary. If x Rd, show that the translate E + x := {y + x : y E} is also an elementary set. We now give each elementary set a measure. Lemma 1.1.2 (Measure of an elementary set). Let E Rd be an elemen- tary set. (i) E can be expressed as the finite union of disjoint boxes. (ii) If E is partitioned as the finite union B1 ∪...∪Bk of disjoint boxes, then the quantity m(E) := |B1| + . . . + |Bk| is independent of the partition. In other words, given any other partition B1 . . . Bk of E, one has |B1| + . . . + |Bk| = |B1| + . . . + |Bk |. We refer to m(E) as the elementary measure of E. (We occasionally write m(E) as md(E) to emphasise the d-dimensional nature of the measure.) Thus, for example, the elementary measure of (1, 2)∪[3, 6] is (2−1)+(6−3) = 4. Proof. We first prove (i) in the one-dimensional case d = 1. Given any finite collection of intervals I1,...,Ik, one can place the 2k endpoints of these intervals in increasing order (discarding repetitions). Looking at the open intervals between these endpoints, together with the endpoints themselves (viewed as intervals of length zero), we see that there exists a finite collection of disjoint intervals J1,...,Jk such that each of the I1,...,Ik are a union of some subcollection of the J1,...,Jk . This already gives (i) when d = 1. To prove the higher dimensional case, we express E as the union B1,...,Bk of boxes Bi = Ii,1 × . . . × Ii,d. For each j = 1,...,d, we use the one- dimensional argument to express I1,j,...,Ik,j as the union of subcollections of a collection J1,j,...,Jk j ,j of disjoint intervals. Taking Cartesian products, we can express the B1,...,Bk as finite unions of boxes Ji1,1 × . . . × Jid,d, where 1 ij kj for all 1 j d. Such boxes are all disjoint, and the claim follows. To prove (ii) we use a discretisation argument. Observe (exercise!) that for any interval I, the length of I can be recovered by the limiting formula |I| = lim N→∞ 1 N #(I 1 N Z)
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