22 1. Measure theory to enlarge the boxes to be open. More precisely, for each box Bn one can find an open box Bn containing Bn such that |Bn| ≤ |Bn| + ε/2n (say). The Bn still cover E, and we have ∞ n=1 |Bn| ≤ ∞ n=1 (|Bn| + ε/2n) = ( ∞ n=1 |Bn|) + ε ≤ m∗(E) + 2ε. As the Bn are open, we may apply the Heine-Borel theorem and conclude that E ⊂ N n=1 Bn for some finite N. Using the finite subadditivity of elementary measure, we conclude that m(E) ≤ N n=1 |Bn| and thus m(E) ≤ m∗(E) + 2ε. Since ε 0 was arbitrary, the claim follows. Now we consider the case when the elementary E is not closed. Then we can write E as the finite union Q1 ∪ . . . ∪ Qk of disjoint boxes, which need not be closed. But, similarly to before, we can use the epsilon of room strategy: for every ε 0 and every 1 ≤ j ≤ k, one can find a closed sub-box Qj of Qj such that |Qj| ≥ |Qj|− ε/k (say) then E contains the finite union of Q1 ∪ . . . ∪ Qk disjoint closed boxes, which is a closed elementary set. By the previous discussion and the finite additivity of elementary measure, we have m∗(Q1 ∪ . . . ∪ Qk) = m(Q1 ∪ . . . ∪ Qk) = m(Q1) + . . . + m(Qk) ≥ m(Q1) + . . . + m(Qk) − ε = m(E) − ε. Applying by monotonicity of Lebesgue outer measure, we conclude that m∗(E) ≥ m(E) − ε for every ε 0. Since ε 0 was arbitrary, the claim follows. The above lemma allows us to compute the Lebesgue outer measure of a finite union of boxes. From this and monotonicity we conclude that the Lebesgue outer measure of any set is bounded below by its Jordan inner measure. As it is also bounded above by the Jordan outer measure, we have (1.2) m∗,(J)(E) ≤ m∗(E) ≤ m∗,(J)(E) for every E ⊂ Rd.

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