By Tao T.

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8. We are now able to explain why not every bounded open set or compact set is Jordan measurable. Consider the countable set Q ∩ [0, 1], which we enumerate as {q1 , q2 , q3 , . }, let ε > 0 be a small number, and consider the set ∞ U := (qn − ε/2n , qn + ε/2n ). n=1 This is the union of open sets and is thus open. On the other hand, by countable subadditivity, one has ∞ m∗ (U ) ≤ 2ε/2n = 2ε. e. U contains [0, 1]), we have m∗,(J) (U ) = m∗,(J) (U ) ≥ m∗,(J) ([0, 1]) = 1. g. ε := 1/3), we see that the Lebesgue outer measure and Jordan outer measure of U disagree.

From the above exercises, we see that when describing what it means for a set to be Lebesgue measurable, there is a tradeoff between the type of approximation one is willing to bear, and the type of things one can say about the approximation. If one is only willing to approximate to within a null set, then one can only say that a measurable set is approximated by a Gδ or a Fσ set, which is a fairly weak amount of structure. If one is willing to add on an epsilon of error (as measured in the Lebesgue outer measure), one can make a measurable set open; dually, if one is willing to take away an epsilon of error, one can make a measurable set closed.

3) m(B1 ∪ . . ∪ Bk ) = |B1 | + . . + |Bk | holds for almost disjoint boxes B1 , . . , Bk , and not just for disjoint boxes. 9 (Outer measure of countable unions of almost disjoint ∞ boxes). Let E = n=1 Bn be a countable union of almost disjoint boxes B1 , B2 , . .. Then ∞ m∗ (E) = |Bn |. n=1 Thus, for instance, Rd itself has an infinite outer measure. Proof. 6 we have ∞ m∗ (E) ≤ ∞ m∗ (Bn ) = n=1 |Bn |, n=1 so it suffices to show that ∞ |Bn | ≤ m∗ (E). n=1 But for each natural number N , E contains the elementary set B1 ∪ .