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20 1. Measure theory it has finite Lebesgue outer measure (and then the same is true for E and F , by monotonicity). We use the standard “give yourself an epsilon of room” trick (see Section 2.7 of An epsilon of room, Vol I.). Let ε 0. By definition of Lebesgue outer measure, we can cover E ∪F by a countable family B1,B2,... of boxes such that ∞ n=1 |Bn| ≤ m∗(E ∪ F ) + ε. Suppose it was the case that each box intersected at most one of E and F . Then we could divide this family into two subfamilies B1,B2,... and B1 , B2 , B3 , . . ., the first of which covered E, and the second of which covered F . From definition of Lebesgue outer measure, we have m∗(E) ≤ ∞ n=1 |Bn| and m∗(F ) ≤ ∞ n=1 |Bn| summing, we obtain m∗(E) + m∗(F ) ≤ ∞ n=1 |Bn| and thus m∗(E) + m∗(F ) ≤ m∗(E ∪ F ) + ε. Since ε was arbitrary, this gives m∗(E) + m∗(F ) ≤ m∗(E ∪ F ) as required. Of course, it is quite possible for some of the boxes Bn to intersect both E and F , particularly if the boxes are big, in which case the above argument does not work because that box would be double-counted. However, observe that given any r 0, one can always partition a large box Bn into a finite number of smaller boxes, each of which has diameter11 at most r, with the total volume of these sub-boxes equal to the volume of the original box. Applying this observation to each of the boxes Bn, we see that given any r 0, we may assume without loss of generality that the boxes B1,B2,... covering E ∪ F have diameter at most r. In particular, we may assume that all such boxes have diameter strictly less than dist(E, F ). Once we do this, then it is no longer possible for any box to intersect both E and F , and then the previous argument now applies. 11The diameter of a set B is defined as sup{|x − y| : x, y ∈ B}.