By Tao T.
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Define a distance d : 2A / ∼ ×2A / ∼→ R+ between two equivalence classes [E], [E ] by defining d([E], [E ]) := m∗ (E∆E ). Show that this distance is well-defined (in the sense that m(E∆E ) = m(F ∆F ) whenever [E] = [F ] and [E ] = [F ]) and gives 2A / ∼ the structure of a complete metric space. (iii) Let E ⊂ 2A be the elementary subsets of A, and let L ⊂ 2A be the Lebesgue measurable subsets of A. Show that L/ ∼ is the closure of E/ ∼ with respect to the metric defined above. In particular, L/ ∼ is a complete metric space that contains E/ ∼ as a dense subset; in other words, L/ ∼ is a metric completion of E/ ∼.
Iii) (Countable subadditivity) If E1 , E2 , . . ⊂ Rd is a count∞ ∞ able sequence of sets, then m∗ ( n=1 En ) ≤ n=1 m∗ (En ). (Hint: Use the axiom of countable choice, Tonelli’s theorem 22 1. ) Note that countable subadditivity, when combined with the empty set axiom, gives as a corollary the finite subadditivity property m∗ (E1 ∪ . . ∪ Ek ) ≤ m∗ (E1 ) + . . + m∗ (Ek ) for any k ≥ 0. These subadditivity properties will be useful in establishing upper bounds on Lebesgue outer measure. Establishing lower bounds will often be a bit trickier.
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 ∪ . . 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.