Multivariable Advanced Calculus

168 MEASURES AND MEASURABLE FUNCTIONSSince this holds for all n, you can take the limit as n → ∞ and conclude,∞∑µ(A i ) = µ(A)i=1which establishes 7.9.Consider part 7.10. Without loss of generality µ (F k ) < ∞ for all k since otherwisethere is nothing to show. Suppose {F k } is an increasing sequence of sets of S. Thenletting F 0 ≡ ∅, {F k+1 \ F k } ∞ k=0is a sequence of disjoint sets of S since it was shownabove that the difference of two sets of S is in S. Also note that from 7.12and so if µ (F k ) < ∞, thenTherefore, lettingwhich also equalsit follows from part 7.9 just shown thatµ (F ) =µ (F k+1 \ F k ) + µ (F k ) = µ (F k+1 )µ (F k+1 \ F k ) = µ (F k+1 ) − µ (F k ) .F ≡ ∪ ∞ k=1F k∪ ∞ k=1 (F k+1 \ F k ) ,∞∑µ (F k+1 \ F k ) = limk=1= limn∑n→∞k=1n→∞k=1n∑µ (F k+1 \ F k )µ (F k+1 ) − µ (F k ) = limn→∞ µ (F n+1) .In order to establish 7.11, let the F n be as given there. Then, since (F 1 \ F n )increases to (F 1 \ F ), 7.10 implieslim (µ (F 1) − µ (F n )) = µ (F 1 \ F ) .n→∞Now µ (F 1 \ F ) + µ (F ) ≥ µ (F 1 ) and so µ (F 1 \ F ) ≥ µ (F 1 ) − µ (F ). Hencewhich impliesBut since F ⊆ F n ,lim (µ (F 1) − µ (F n )) = µ (F 1 \ F ) ≥ µ (F 1 ) − µ (F )n→∞lim µ (F n) ≤ µ (F ) .n→∞µ (F ) ≤ limn→∞ µ (F n)and this establishes 7.11. Note that it was assumed µ (F 1 ) < ∞ because µ (F 1 ) wassubtracted from both sides.It remains to show S is closed under countable unions. Recall that if A ∈ S, thenA C ∈ S and S is closed under finite unions. Let A i ∈ S, A = ∪ ∞ i=1 A i, B n = ∪ n i=1 A i.Thenµ(S) = µ(S ∩ B n ) + µ(S \ B n ) (7.13)= (µ⌊S)(B n ) + (µ⌊S)(B C n ).