Coursehero >> California >> UCSD >> MATH 240a

Course Hero has millions of student submitted documents similar to the one below including study guides, homework solutions, papers, exam answer keys and textbook solutions.

functional ear on S satisfying property (D) is called a Daniell integral on S. We will also write S as D(I) the domain of I. 348 31 The Daniell Stone Construction of Integration and Measures Lemma 31.6. Let I be a non-negative linear functional on a lattice S. Then property (D) is equivalent to either of the following two properties: D1 If , n S satisfy; n n+1 for all n and limn n , then I( ) limn I( n ). D2 If uj S+ and S is such that j=1 uj then I( ) j=1 I(uj ). Proof. (D) = (D1 ) Let , n S be as in D1 . Then n and ( n ) 0 which implies I( ) I( n ) = I( ( n )) 0. Hence I( ) = lim I( n ) lim I( n ). n n 0 I(fn ) CK fn 0 as n . For example if X = R and F is an increasing function on R, then I (f ) := f dF is a Daniell integral on Cc (R, R), see Lemma 28.36. However it is not R generally true in this case that I(fn ) 0 for all fn S (S is the collection of compactly supported step functions on R) such that fn 0. The next example and proposition addresses this question. Example 31.8. Suppose F : R R is an increasing function which is not right continuous at x0 R. Then, letting fn = 1(x0 ,x0 +n 1 ] S, we have fn 0 as n but fn dF = F x0 + n 1 F (x0 ) F (x0 +) F (x0 ) = 0. R (D1 ) = (D2 ) Apply (D1 ) with n = n j=1 uj . (D2 ) = (D) Suppose n S N n=1 with n 0 and let un = n n+1 . Then hence N un = 1 N +1 1 and Proposition 31.9. Let (A, , S = Sf (A, ), I = I ) be as in De nition 28.34. If is a premeasure (De nition 30.1) on A, then fn S with fn 0 = I(fn ) 0 as n . Hence I is a Daniell integral on S. Proof. Let > 0 be given. Then fn = fn 1fn > f1 + fn 1fn f1 f1 1fn > f1 + f1 , I(fn ) I (f1 1fn > f1 ) + I(f1 ) = a>0 I( 1 ) n=1 I(un ) = lim N N I(un ) n=1 N (31.3) = lim I( 1 N +1 ) = I( 1 ) lim I( N +1 ) from which it follows that limN I( N +1 ) 0. Since I( N +1 ) 0 for all N we conclude that limN I( N +1 ) = 0. 31.0.1 Examples of Daniell Integrals Proposition 31.7. Suppose that (X, ) is locally compact Hausdor space and I is a positive linear functional on S := Cc (X, R). Then for each compact subset K X there is a constant CK < such that |I(f )| CK f for all f Cc (X, R) with supp(f ) K. Moreover, if fn Cc (X, [0, )) and fn 0 (pointwise) as n , then I(fn ) 0 as n and in particular I is necessarily a Daniell integral on S. Proof. Let f Cc (X, R) with supp(f ) K. By Lemma 15.8 there exists K X such that K = 1 on K. Since f K f 0, 0 I( f a (f1 = a, fn > a) + I(f1 ), and hence lim sup I(fn ) n a>0 a lim sup (f1 = a, fn > a) + I(f1 ). n (31.4) Because, for a > 0, A {f1 = a, fn > a} as n K f ) = f I( K ) I(f ) and (f1 = a) < , lim supn (f1 = a, fn > a) = 0. Combining this with Eq. (31.4) and making use of the fact that > 0 is arbitrary we learn lim supn I(fn ) = 0. from which it follows that |I(f )| I( K ) f . So the rst assertion holds with CK = I( K ) < . Now suppose that fn Cc (X, [0, )) and fn as 0 n . Let K = supp(f1 ) and notice that supp(fn ) K for all n. By Dini s Theorem (see Exercise 14.3), fn 0 as n and hence Page: 348 job: anal macro: svmonob.cls date/time: 26-Apr-2004/14:01 31.1 Extending a Daniell Integral 349 31.1 Extending a Daniell Integral In the remainder of this chapter we x a lattice, S, of bounded functions, f : X R, and a positive linear functional I : S R satisfying Property (D) of De nition 31.5. Lemma 31.10. Suppose that {fn } , {gn } S. 1. If fn f and gn g with f, g : X ( , ] such that f g, then n I (g) = lim I (gn ) if S n gn g. If f S S , then there exists fn , gn S such that fn f and gn f. Hence S (gn fn ) 0 and hence by the continuity property (D), I (f ) I (f ) = lim [I (gn ) I (fn )] = lim I (gn fn ) = 0. n n Therefore I = I on S S . (31.5) Notation 31.12 Using the above comments we may now simply write I (f ) for I (f ) or I (f ) when f S or f S . Henceforth we will now view I as a function on S S . Again because of Lemma 31.10, let I := I|S or I := I|S are positive functionals; i.e. if f g then I(f ) I(g). Exercise 31.1. Show S = S and for f S S that I( f ) = I(f ) R. Proposition 31.13. The set S and the extension of I to S in De nition 31.11 satis es: 1. (Monotonicity) I(f ) I(g) if f, g S with f g. 2. S is closed under the lattice operations, i.e. if f, g S then f g S and f g S . Moreover, if I(f ) < and I(g) < , then I(f g) < and I(f g) < . 3. (Positive Linearity) I (f + g) = I(f ) + I(g) for all f, g S and 0. 4. f S+ i there exists n S+ such that f = n=1 n . Moreover, I(f ) = m=1 I( m ). 5. If fn S+ , then n=1 fn =: f S+ and I(f ) = n=1 I(fn ). Remark 31.14. Similar results hold for the extension of I to S in De nition 31.11. Proof. lim I(fn ) lim I(gn ). n 2. If fn f and gn g with f, g : X [ , ) such that f g, then Eq. (31.5) still holds. In particular, in either case if f = g, then n lim I(fn ) = lim I(gn ). n Proof. 1. Fix n N, then gk fn fn as k and gk fn gk and hence I(fn ) = lim I(gk fn ) lim I(gk ). k k Passing to the limit n in this equation proves Eq. (31.5). 2. Since fn ( f ) and gn ( g) and g ( f ), what we just proved shows lim I(gn ) = lim I( gn ) lim I( fn ) = lim I(fn ) n n n n which is equivalent to Eq. (31.5). De nition 31.11. Let S = {f : X ( , ] : fn S such that fn f } and S = {f : X [ , ) : fn S such that fn f } . Because of Lemma 31.10, for f S and g S we may de ne I (f ) = lim I (fn ) if S n 1. Monotonicity follows directly from Lemma 31.10. 2. If fn , gn S are chosen so that fn f and gn g, then fn gn f g and fn gn f g. If we further assume that I(g) < , then f g g and hence I(f g) I(g) < . In particular it follows that I(f 0) ( , 0] for all f S . Combining this with the identity, I(f ) = I (f 0 + f 0) = I (f 0) + I(f

Find millions of documents here - Study Guides, Homework Solutions, Papers, Exam Answer Keys and more. Course Hero has millions of course related materials that will enable you to learn better, faster and get an A in all your courses.
Below is a small sample set of documents: