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Additional info for An Introduction To Measure Theory (January 2011 Draft)

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Show that there exists disjoint bounded subsets E, F of the real line such that m∗ (E ∪ F ) = m∗ (E) + m∗ (F ). 27 (Projections of measurable sets need not be measurable). Let π : R2 → R be the coordinate projection π(x, y) := x. Show that there exists a measurable subset E of R2 such that π(E) is not measurable. 46 1. 19. The above discussion shows that, in the presence of the axiom of choice, one cannot hope to extend Lebesgue measure to arbitrary subsets of R while retaining both the countable additivity and the translation invariance properties.

It is convenient to introduce the following notion: two boxes are almost disjoint if their 28 1. Measure theory interiors are disjoint, thus for instance [0, 1] and [1, 2] are almost disjoint. 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.

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 ∪ . . 6, m∗ (E) ≥ m∗ (B1 ∪ . . ∪ BN ) = m(B1 ∪ . . 3), one has N |Bn | ≤ m∗ (E). n=1 Letting N → ∞ we obtain the claim. 10. 2. Lebesgue measure ∞ n=1 29 ∞ |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.