Multivariable Advanced Calculus

192 THE ABSTRACT LEBESGUE INTEGRALProof: Why is the integral linear? Let {s n } and {t n } be sequences of simplefunctions attached to f and g respectively according to the definition.∫∫(af + bg) dµ ≡ lim (as n + bt n ) dµn→∞( ∫ ∫ )= lim a s n dµ + b t n dµn→∞∫∫= a lim s n dµ + b lim t n dµn→∞n→∞∫ ∫= a fdµ + b gdµ.The fact that ∫ is linear makes the triangle inequality easy to verify. Let f ∈ L 1 (Ω)and let θ ∈ C such that |θ| = 1 and θ ∫ fdµ = ∣ ∫ fdµ ∣ ∣. Then∫∫ ∫∫∣ fdµ∣ = θfdµ = Re (θf) dµ = Re (θf) + − Re (θf) − dµ∫∫∫≤ Re (θf) + dµ ≤ |Re (θf)| dµ ≤ |f| dµNow the last assertion follows from the definition. There exists a sequence of simplefunctions {s n } converging pointwise to f such that for all m, n large enough,∫ε2 > |s n − s m | dµFix such an m and let n → ∞. By Fatou’s lemmaε > ε ∫∫≥ lim inf |s n − s m | dµ ≥ |f − s m | dµ.2 n→∞Let s = s m . Recall that it has been shown that in computing the integrals on the right in ??,then for g one of those integrands,∫ ∫ ∞∞∑gdµ = µ ([g > t]) dt = sup µ ([g > kh]) h{∫= sup0h>0k=1}sdµ : s ≤ g, s a nonnegative simple functionOne of the major theorems in this theory is the dominated convergence theorem.Before presenting it, here is a technical lemma about lim sup and lim inf .Lemma 8.8.2 Let {a n } be a sequence in [−∞, ∞] . Then lim n→∞ a n exists if andonly iflim inf a n = lim sup a nn→∞ n→∞and in this case, the limit equals the common value of these two numbers.Proof: Suppose first lim n→∞ a n = a ∈ R. Then, letting ε > 0 be given, a n ∈(a − ε, a + ε) for all n large enough, say n ≥ N. Therefore, both inf {a k : k ≥ n} andsup {a k : k ≥ n} are contained in [a − ε, a + ε] whenever n ≥ N. It follows lim sup n→∞ a nand lim inf n→∞ a n are both in [a − ε, a + ε] , showing∣ ∣∣∣ ∣lim inf a n − lim sup a n < 2ε.n→∞n→∞