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Suppose Proof. n=1 I (fn ) < , for otherwise the result is trivial. Let > 0 be given and choose gn S+ such that fn gn and I(gn ) = I (fn ) + n where n=1 n . (For example take n 2 n .) Then n=1 gn =: G S+ , F G and so I (F ) I (G) = I(G) = n=1 I(gn ) = n=1 (I (fn ) + n ) n=1 I (fn ) + . Since > 0 is arbitrary, the proof is complete. Proposition 31.20. The space L1 (I) is an extended lattice and I : L1 (I) R is linear in the sense of De nition 31.3. macro: svmonob.cls date/time: 26-Apr-2004/14:01 352 31 The Daniell Stone Construction of Integration and Measures Proof. Let us begin by showing that L1 (I) is a vector space. Suppose that g1 , g2 L1 (I), and g (g1 + g2 ). Given > 0 there exists fi S L1 (I) and hi S L1 (I) such that fi gi hi and I(hi fi ) < /2. Let us now show f1 (x) + f2 (x) g(x) h1 (x) + h2 (x) x X. (31.8) 0 h1 h2 f1 f2 h1 f1 + h2 f2 , because, for example, if h1 h2 = h1 and f1 f2 = f2 then h1 h2 f1 f2 = h1 f2 h1 f1 . Therefore, I (h1 h2 f1 f2 ) I (h1 f1 + h2 f2 ) < and hence by Remark 31.17, g1 g2 L1 (I). Theorem 31.21 (Monotone convergence theorem). If fn L1 (I) and fn f, then f L1 (I) i limn I(fn ) = supn I(fn ) < in which case ) = limn I(fn ). I(f Proof. If f L1 (I), then by monotonicity I(fn ) I(f ) for all n and n ) I(f ) < . Conversely, suppose := limn I(fn ) < therefore limn I(f and let g := n=1 (fn+1 fn )0 . The reader should check that f (f1 +g) (f1 + g) . So by Lemma 31.19, I (f ) I ((f1 + g) ) I (f1 ) + I (g) This is clear at points x X where g1 (x) + g2 (x) is well de ned. The other case to consider is where g1 (x) = = g2 (x) in which case h1 (x) = and f2 (x) = while , h2 (x) > and f1 (x) < because h2 S and f1 S . Therefore, f1 (x)+f2 (x) = and h1 (x)+h2 (x) = so that Eq. (31.8) is valid no matter how g(x) is chosen. Since f1 + f2 S L1 (I), h1 + h2 S L1 (I) and I(gi ) I(fi ) + /2 and /2 + I(hi ) I(gi ), we nd I(g1 ) + I(g2 ) I(f1 ) + I(f2 ) = I(f1 + f2 ) I (g) I (g) I(h1 + h2 ) = I(h1 ) + I(h2 ) I(g1 ) + I(g2 ) + . Because > 0 is arbitrary, we have shown that g L1 (I) and I(g1 ) + I(g2 ) = I(g), i.e. I(g1 + g2 ) = I(g1 ) + I(g2 ). It is a simple matter to show g L1 (I) and I( g) = I(g) for all g L1 (I) and R. For example if = 1 (the most interesting case), choose f S L1 (I) and h S L1 (I) such that f g h and I(h f ) < . Therefore, S L1 (I) h g f S L1 (I) I (f1 ) + n=1 I ((fn+1 fn )0 ) = I(f1 ) + n=1 I (fn+1 fn ) (31.9) = I(f1 ) + n=1 I(fn+1 ) I(fn ) = I(f1 ) + I(f1 ) = . with I( f ( h)) = I(h f ) < and this shows that g L1 (I) and I( g) = I(g). We have now shown that L1 (I) is a vector space of extended real valued functions and I : L1 (I) R is linear. To show L1 (I) is a lattice, let g1 , g2 L1 (I) and fi S (I) L1 and hi S L1 (I) such that fi gi hi and I(hi fi ) < /2 as above. Then using Proposition 31.13 and Remark 31.14, S L1 (I) Moreover, 0 h1 h2 f1 f2 h1 f1 + h2 f2 , because, for example, if h1 h2 = h1 and f1 f2 = f2 then h1 h2 f1 f2 = h1 f2 h2 f2 . Therefore, I (h1 h2 f1 f2 ) I (h1 f1 + h2 f2 ) < and hence by Remark 31.17, g1 g2 L1 (I). Similarly Page: 352 job: anal Because fn f, it follows that I(fn ) = I (fn ) I (f ) which upon passing to limit implies I (f ). This inequality and the one in Eq. (31.9) shows I (f ) I (f ) and therefore, f L1 (I) and I(f ) = = limn I(fn ). Lemma 31.22 (Fatou s Lemma). Suppose {fn } L1 (I) , then inf fn L1 (I). If lim inf n I(fn ) < , then lim inf n fn L1 (I) and in this case I(lim inf fn ) lim inf I(fn ). n n + f1 f2 g1 g2 h1 h2 S L1 (I). Proof. Let gk := f1 fk L1 (I), then gk g := inf n fn . Since gk g, gk L1 (I) for all k and I( gk ) I(0) = 0, it follow from Theorem 31.21 1 that g L (I) and hence so is inf n fn = g L1 (I). By what we have just proved, uk := inf n k fn L1 (I) for all k. Notice that uk lim inf n fn , and by monotonicity that I(uk ) I(fk ) for all k. Therefore, k lim I(uk ) = lim inf I(uk ) lim inf I(fn ) < k k and by the monotone convergence Theorem 31.21, lim inf n fn limk uk L1 (I) and macro: svmonob.cls = date/time: 26-Apr-2004/14:01 31.2 The Structure of L1 (I) 353 I(lim inf fn ) = lim I(uk ) lim inf I(fn ). n k n 6. By Proposition 31.20, |f | L1 (I) and so by Chebyshev s inequality (Item 2 of Proposition 31.18), {|f | = } is a null set. Before stating the dominated convergence theorem, it is helpful to remove some of the annoyances of dealing with extended real valued functions. As we have done when studying integrals associated to a measure, we can do this by modifying integrable functions by a null function. De nition 31.23. A function n : X R is a null function if I (|n|) = 0. A subset E X is said to be a null set if 1E is a null function. Given two functions f, g : X R we will write f = g a.e. if {f = g} is a null set. Here are some basic properties of null functions and null sets. Proposition 31.24. Suppose that n : X R is a null function and f : X R is an arbitrary function. Then 1. n L1 (I) and I(n) = 0. 2. The function n f is a null function. 3. The set {x X : n(x) = 0} is a null set. 4. If E is a null set and f L1 (I), then 1E c f L1 (I) and I(f ) = I(1E c f ). 1 1 ) = I(g). 5. If g L (I) and f = g a.e. then f L (I) and I(f 6. If f L1 (I), then E := {|f | = } is a null set. Proof. 1. If n is null, using n |n| we nd I ( n) I (|n|) = 0, i.e. I (n) 0 and I (n) = I ( n) 0. Thus it follows that I (n) 0 I (n) and therefore n L1 (I) with I (n) = 0. 2. Since |n f | |n| , I (|n f |) I ( |n|) . For k N, k |n| L1 (I) and I(k |n|) = kI (|n|) = 0, so k |n| is a null function. By the monotone convergence Theorem 31.21 and the fa

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