Perfect Measures and Maps

Transcription

1 Preprint Ser. No. 26, 1991, Math. Inst. Aarhus Perfect Measures and Maps GOAN PESKI This paper is designed to provide a solid introduction to the theory of nonmeasurable calculus. In this context basic concepts are reviewed and fundamental facts are presented. Perfect measures and maps are shown to of a vital importance to support the theory, and therefore these concepts are treated separately as well. In particular, many equivalent statements to the property of being perfect are given. The last part of the paper deals with a setting arising in the investigation on extension of measures in the case of perturbations of their -algebras. 1. Introduction This paper is designed to provide a solid introduction to the theory of non-measurable calculus. The content is divided into three parts. Section 2 deals with the non-measurable calculus. Together with some new definitions of concepts which are shown useful, it contains several new results which are mainly generalizations or completions of the corresponding old ones. For additional information in this context we refer to [2]. Section 3 deals with perfect measures and maps. From the historical point of view the concept of perfectness relies upon the following two well-known facts: (1.1) If is a finite, finitely additive measure defined on the algebra A of subsets of a set X such that for each decreasing sequence f A n j n 1 g in A with empty intersection we have lim n!1 (A n ) = 0, then is countably additive and admits a unique extension to the -algebra (A) generated by A. (1.2) If f K n j n 1 g is a decreasing sequence of compact sets in a topological space X with empty intersection, then there is N 1 such that T N n=1 K n is empty. These two facts in a certain way form a base of the general axiomatic theory of Kolmogorov, see [10]. Marczewski in [12] generalized this idea and introduced compact measures. In [4] Gnedenko and Kolmogorov were defined perfect measures, while yll-nardzewski in [20] introduced and studied the same concept of quasi-compact measures. Since that time the properties of compactness and perfectness have been subject of a number of studies, see [6], [11], [13], [14], [15], [16], [19], [21], [22], [23]. It is shown that by the aid of these properties we avoid many pathological phenomena that arise within the framework of Kolmogorov s probability theory (independence, see [21]; the existence of the regular conditional probability, see [15], [16], [21]; countable additivity of the product measures, see [12], [13], [15], [20] ). Also, using mainly this idea Alexandroff in [1] has given an extensive development of the theory of bounded regular finitely additive measures in topological spaces, which is at the background of many later studies. In this section we present many equivalent statements to the property of being perfect, as well as many general properties AMS 1980 subject classifications. Primary 28A12, 28A25. Secondary 28A05, 28A20. (Second edition) Key words and phrases: The non-measurable calculus, completion, the kernel (hull), restriction, the outer (inner) trace, extension, the upper (lower) integral (envelope), Segal s localization principle, finitely founded, image (coimage) measure, perfect measure (map), Blackwell space, Luzin measurable, the (outer, inner) approximating paving, (dual) functional, the Hahn-Banach theorem, perturbation. goran@imf.au.dk 1

2 of perfect measures and closely related perfect maps. Many of these facts have its own origin in Høffmann-Jørgensen s unpublished notes [7]. For simplicity, most of the proofs are usually omitted or briefly sketched. Section 4 deals with an algebraic setting arising in the investigation on extension of measures in the case of perturbations of their -algebras, see [17] and [18]. The approach presented could be suitable for investigations carried out in the framework of the general vector space theory, with the given examples in mind. 2. Non-measurable calculus 2.1. Completions. Let (X; A; ) be a finite measure space. Then 3 and 3 denote the outer and the inner -measure, A denotes the -algebra of -measurable subsets of X, i.e. A = f B 2 2 X j 9A; A 0 2 A ; A B A 0 such that (A 0 n A) = 0 g and A 3 denotes the -algebra of universally measurable subsets of X, i.e. A 3 = T A where ranges over the family of all finite measures on (X; A). Let us recall that the -algebra A is called the completion of A with respect to, that A 2 A if and only if 3 (A) = 3 (A), and that the restriction of 3 or 3 on A is the only measure on A that agrees on A with. This measure is called the completion of and is denoted by. In particular, for each finite measure on (X; A) there is a unique measure on (X; A 3 ) that agrees with on A. Let us recall that the given unique finite measure space (X; A ; ) is called the completion of (X; A; ), and that a finite measure space (X; A; ) ( or a measure ) is called complete if A = A and = Kernels and hulls. Let (X; A; ) be a finite measure space. If C 2 2 X is an arbitrary subset of X, then C 3 denotes the -hull of C, and C 3 denotes the -kernel of C, i.e. C 3 ; C 3 2 A and (C 3 ) = 3 (C), (C 3 ) = 3 (C). It is easily seen that each subset of X has the -hull and the -kernel, uniquely determined up to a -nullset, and we will without loss of generality always assume that C 3 C C 3 for all C 2 2 X estrictions, traces and extensions. Let (X; A; ) be a finite measure space, let A 0 be a sub--algebra of A, and let C 2 2 X be an arbitrary set. The restriction of on the -algebra A 0 is a measure r(; A 0 ) : A 0! [0; 1> defined by: r(; A 0 )(A 0 ) = (A 0 ) for all A 0 2 A 0. The trace of A on the set C is a -algebra defined by: tr(a; C) = fa \ C j A 2 Ag. If C 2 A, then the trace of on the set C is a measure tr(; C) : A! [0; 1> defined by: tr(; C)(A) = (A \ C) 2

3 for all A 2 A. If C is not necessarily from A, then we distinguish the following two possibilities: the outer trace of on the set C is a measure tr 3 (; C) : (A[fCg)! [0; 1> defined by: tr 3 (; C)([A; B]) = (A \ C 3 ) where [A; B] = (A \ C) [ (B \ C c ), with A; B 2 A, is a general representation of an element from (A [ fcg) ; the inner trace of on the set C is a measure tr 3 (; C) : (A[fCg)! [0; 1) defined by: where [A; B] is as above. tr 3 (; C)([A; B]) = (A\C 3 ) It is easily seen that the definitions of the outer and inner traces are good, i.e. they do not depend on the choice of the -hull and the -kernel of the set C, as well of the representation of elements from the -algebra (A [ fcg). Moreover, the following properties are satisfied: (2.1) tr 3 (; C)([A; B]) = 3 (A \ C) ; 8[A; B] 2 (A [ fcg) (2.2) tr 3 (; C)([A; B]) = 3 (A \ C) ; 8[A; B] 2 (A [ fcg) (2.3) tr 3 (; C)(A) + tr 3 (; C c )(A) = (A), 8A 2 A (2.4) tr 3 (; C)(C) = 3 (C) (2.5) tr 3 (; C)(C) = 3 (C) (2.6) tr 3 (; C)(C c ) = tr 3 (; C)(C c ) = 0 (2.7) If C 2 A, then tr 3 (; C) = tr 3 (; C) = tr(; C) (2.8) If C 2 A, then tr 3 (; C) = tr 3 (; C) = r 0 tr(; C); (A [ fcg)1. Let (X; A; ) be a finite measure space, and let ^X X be a set. The extension of A on the set ^X is a -algebra defined by: ^A = f ^A 2 2 ^X j ^A\X 2 A g. The extension of the measure on the set ^X is a measure ext(; ^X) : ^A! [0; 1) defined by: ext(; ^X)( ^A) = ( ^A\X) for all ^A 2 ^A. It is easily checked that the following properties are satisfied: (2.9) A ^A and 0 r ext(; ^X); A 1 = (2.10) ext(; ^X)( ^A) = 0, whenever ^A ^X n X (2.11) trfext(; ^X); Xg = ext(; ^X) (2.12) 0 r ext (tr(; C); ^X 1 ; A = tr(; C), whenever C 2 A (2.13) fext(; ^X)g 3 ( ^C) = 3 ( ^C \ X), whenever ^C 2 2 ^X 3

4 (2.14) fext(; ^X)g 3 ( ^C) = 3 ( ^C \ X), whenever ^C 2 2 ^X Lower and upper integrals. Let (X; A; ) be a finite measure space and let L 1 () be the set of all -integrable -valued functions on X. Then by: Z 3 f d = inf f g d j g 2 L 1 (); f g g Z 3 f d = sup f g d j g 2 L 1 (); g f g the upper and lower -integral of an arbitrary -function f on X are defined. It is well-known that the upper and lower -integrals satisfy the following properties: (2.15) 3 f d + 3 g d 3 (f + g) d 3 f d + 3 g d (2.16) 3 f d + 3 g d 3(f + g) d 3 f d + 3 g d (2.17) 3(af ) d = a 3 f d ; 8a 2 + (2.18) 3 (af ) d = a 3 f d ; 8a 2 + (2.19) 3(af ) d = a 3 f d ; 8a 2 0 (2.20) 3 (af ) d = a 3 f d ; 8a 2 0 (2.21) If f g, then 3 f d 3 g d (2.22) If f g, then 3 f d 3 g d (2.23) If f g, then 3 f d 3 g d (2.24) If f n " f and 3 f1 d > 01, then 3 f d = lim 3 fn d (2.25) If f n # f and 3 f 1 d < +1, then 3 f d = lim 3 f n d 3( (2.26) If inf fn 3 ) d > 01, then ( lim inf fn 3 ) d lim inf fn d (2.27) If 3 ( sup f n ) d < +1, then 3 ( lim sup f n) d lim sup 3 f n d (2.28) 3 1C d = 3 (C) ; 8C 2 2 X (2.29) 3 1 C d = 3(C) ; 8C 2 2 X (2.30) f d exists if and only if 3 3 f d = f d, and in this case we have f d = 3 fd = 3 fd provided that all relations are well-defined. All preceding statements remain valid for an arbitrary measure space (X; A; ), where in the definition of the upper and inner -integral one can use: L() = f f 2 X j f d exists in g instead of L 1 (), but with the same meaning if is finitely founded, see below. emark 2.1 Let H be a Hamel basis for as a vector space over Q, and let f H n j n S 1g be a strictly increasing sequence of subsets of H such that 1 n=1 H n = H. Let L n be the 4

5 smallest subspace of over Q which contains H n for all n 1. Then L n " and L n 6= for all n 1. If B 2 B() and B L n, then (B) > 0 implies <0" ; " > di(b) = f x 0 y j x; y 2 B g L n, for some " > 0 and hence the smallest subspace of over Q which contains <0" ; " > is contained in L n. But this subspace is actually equal to. Thus 3 (L n ) = 0 and hence 3 ([0; 1] \ L c n ) = 1 for all n 1, but A n = [0; 1] \ L c n # ;. Now putting f n = 1 An or 1 A c n on the finite measure space 0 ([0; 1]; tr B(); [0;1]1 ; ), it is easily verified that (2.24)-(2.27) do not hold if we replace the upper integral 3 with the lower 3, or reversely Lower and upper envelopes. The upper and lower -integral can be much easier handled by using so-called upper and lower -envelope of the integrand. These envelopes are built in order to establish a simple connection with the ordinary -integral. The central point in the construction of the envelopes plays the following well-known property of a finite measure. Lemma 2.2 (Segal s localization principle) Let (X; A; ) be a finite measure space, and let F be an arbitrary family of A-measurable -valued functions on X. Then there exists a sequence f f n j n 1 g F such that the A measurable function S : X!, defined by: S(x) = sup f n (x) n1 for x 2 X, satisfies the following properties: (2.31) f S -a.a. for all f 2 F (2.32) If S 0 : X! is an another A-measurable function such that f S 0 -a.a. for all f 2 F, then fs 0 < Sg = 0, i.e. S S 0 -a.s. Proof. Since x 7! (2=) arctan x is a strictly increasing function from onto [01; 1], it is no restriction to assume that: jf (x)j 1 ; 8x 2 X and 8f 2 F. Define G = f g 2 X j g = sup n1 f n, for some ff n j n 1g F g and put: G = sup f g d j g 2 G g. By assumption, we see that G 2, and by definition of G, we can choose a sequence f gn j n 1 g G such that G = sup n1 gn d. Put: S = sup gn n1 and note that S 2 G actually. Since S gn for all n 1, hence it follows S d = G. Furthermore, since F G we have S _ f 2 G. This implies (S _ f ) d = G = S d for all f 2 F, i.e. S = S _ f -a.a. for all f 2 F which implies (i), while (ii) is an easy consequence of (i) and definition of S. Such a function S, which satisfies (2.31) and (2.32) and which is uniquely determined up to 5

6 a -nullset actually, is called the -essential supremum of F and is denoted by S = -ess sup F. Similarly, the -essential infimum of F is a function T : X! denoted and defined by T = -ess inf F = 0f -ess sup (0F ) g, where 0F = f0f j f 2 Fg. If (X; A; ) is a finite measure space, then M(A) denotes the set of all A-measurable real valued function on X, and M(A) denotes the set of all A-measurable -valued functions on X. Then by: f 3 = -ess inf f g 2 M(A) j f g g f 3 = -ess sup f g 2 M(A) j g f g the upper and lower -envelope of an arbitrary -valued function f on X are defined. The following properties show that f 3 and f 3 are the smallest and the largest A-measurable -valued function on X which is greater or less than f, respectively: (2.33) f 3 and f 3 are A-measurable -valued functions (2.34) f 3 (x) f (x) f 3 (x) ; 8x 2 X (2.35) g f 3 f 3 h -a.s. for all g; h 2 M(A) such that g f h -a.s. (2.36) 3 ff 3 < g fg = 3 ff g < f 3 g = 0 ; 8g 2 M(A) (2.37) f is -measurable if and only if f 3 = f 3 -a.s. For (2.36) suppose that 3 f f 3 < g f g > 0 for some g 2 M(A). Then there exists A 2 A ; A f f 3 < g f g such that (A) > 0. Put: f+ = g 1 1A + f 3 1 1A c. Then we have f+ 2 M(A) ; f+ f and f f 3 < f+ g = (A) > 0 which contradicts the definition of the lower -envelope f 3 of f and proves (5.4). Actually, by (2.33)-(2.35) we see that the upper and lower -envelope f 3 and f 3 of a function f are uniquely determined up to a -nullset, while (2.36) is equivalent to (2.37). More generally, let (X; A; ) be an arbitrary measure space and let M() be the set of all -measurable -valued functions on X. Then M() is a lattice ordered by -a.s., and we say that satisfies Segal s localization principle if ( M() ; -a.s.) is a complete lattice. We obviously have: (2.38) If satisfies Segal s localization principle, then f 3 and f 3 exist for all functions f : X!. Using the preceding result on a finite measure and standard extension arguments we easily find: (2.39) Every -finite measure satisfies Segal s localization principle. Moreover, if there exists a disjoint family C A such that: (1) (C) < 1, for all C 2 C (2) If A \ C 2 A for all C 2 C, then A 2 A (3) If A 2 A and (A \ C) = 0 for all C 2 C, then (A) = 0 6

7 then satisfies Segal s localization principle. Let N () = f N 2 2 X j 3 (N ) = 0 g be the family of all -null sets in X, and let a measure on (X; A). Then we obviously have: be (2.40) If N () = N () and satisfies Segal s localization principle, then so does. The measure is said to be 6-finite, if there is a sequence of finite measures f n j n 1 g on (X; A) such that = P 1 n=1 n. It is clear that each -finite measure is 6-finite. Moreover: (2.41) Every 6-finite measure satisfies Segal s localization principle. For this, let = P 1 n=1 n be a 6-finite measure. It is no restriction to assume n 6= 0 for all n 1. Then: (A) = 1X n=1 1 2 n n(x) n(a) is a finite measure on (X; A) such that N () = N (). Thus (2.41) follows from (2.40). Using the definition properties of the -envelopes (2.33)-(2.35) one can easily check that the following statements are satisfied: (2.42) f 3 + g 3 (f + g) 3 f 3 + g 3 (f + g) 3 f 3 + g 3 (2.43) (f ) 3 = f 3 ; (f ) 3 = f 3 ; 8 0 (2.44) (f ) 3 = f 3 ; (f ) 3 = f 3 ; 8 < 0 (2.45) 0jfj 3 f 3 f 3 jfj 3 (2.46) jf 3 j jfj 3 ; jf 3 j jfj 3 (2.47) jf 3 0 g 3 j jf 0 g j 3 (2.48) jf 3 0 g 3 j jf 0 g j 3 (2.49) f 3 g 3 (f g) 3 f 3 g 3 (f g) 3 f 3 g 3, if f; g 0 provided that all relations are well-defined. We shall now state the basic connection between upper and lower -integrals, upper and lower -envelopes, and outer and inner -measures. Namely, if (X; A; ) is a finite measure space, then for each -valued function f on X, and for each subset C of X, we have: Z 3 (2.50) (2.51) f d = ( f 3 d ; if f 3 2 L() +1 ; otherwise Z ( f f d = 3 3 d ; if f 3 2 L() 01 ; otherwise (2.52) (1C) 3 = 1C 3 and (1 C) 3 = 1C 3 (2.53) 3 (C) = (C 3 ) = 3 1C d (2.54) 3 (C) = (C 3 ) = 3 1 C d. 7

8 More generally, if an arbitrary measure space (X; A; ) satisfies Segal s localization principle, then (2.52)-(2.54) hold. We shall now characterize those measures for which (2.50) and (2.51) remain valid. Let us recall that a measure is called finitely founded if for every A 2 A with (A) = 1, there is B 2 A, B A such that 0 < (B) < 1, i.e. if has no infinite atoms. It is easily verified that any finite, -finite or 6-finite measure is finitely founded, resp. Theorem 2.3 Let (X; A; ) be a measure space. Then the following three statements are equivalent: (2.55) is finitely founded (2.56) (A) = sup f (B) j B 2 A; B A, (B) < 1 g for all A 2 A (2.57) For each -measurable function f : X! + such that f d = 1, there exists a sequence f f n j n 1 g in L 1 () such that 0 f n f for all n 1 and fn d! 1 as n! 1. Moreover, if satisfies Segal s localization principle, then (2.55)-(2.57) is equivalent to each of the following two equivalent statements: Z 3 (2.58) (2.59) Z 3 f d =( f 3 d ; if f 3 2 L() +1 ; otherwise f d =( f 3 d ; if f 3 2 L() 01 ; otherwise, for all f 2 X, for all f 2 X. Proof. (2.55) ) (2.56): Take A 2 A with (A) = 1 and suppose that S = sup f (B) j B 2 C g < 1, where C = f C 2 A j C A ; (C) < 1 g. Then there is a sequence B n 2 C, n 1 such that (B n ) > S 0 1 n ; 8n 1. Hence B = S 1 n=1 B n 2 C with (B) = S. Since (A) = 1, we have (A n B) = 1, and by (2.55) there is C A n B such that 0 < (C) < 1. Then B [ C 2 C, and hence S (B [ C) = (B) + (C) > (B) = S, which is a contradiction. This completes the proof of (2.56). (2.56) ) (2.57): Let f : X! + be a -measurable function such that f d = 1. Let: f n Xmn = i=1 x n i 1 Ai n ; n 1 be an increasing sequence of simple -measurable + -valued functions on X such that f n " f as n! 1. For fixed n 2 N, and each i = 1; 2;... ; m n such that (A n i ) = 1, by (2.56) we can find an increasing sequence Bp;i n 2 A ; p 1 such that Bn p;i A i ; 0 < (Bp;i n ) < 1 and (Bp;i n )! 1 for p! 1. For each i = 1; 2;... ; m n such that (A n i ) < 1 put Bp;i n = An i ; 8p 1. Define: X g n p = m n i=1 x n i 1 B n p;i Then by definition of g n p we easily find: ; p 1. 8

9 Z 0 g n p f n, g n p d < 1, 8n; p 1 and Z g n p d! Z f n d as p! 1, for all n 1. Since by the monotone convergence theorem f n d! f d = 1, it is easily checked that the sequence f g n n j n 1 g satisfies: Z 0 g n n f, g n n d < 1, 8n 1 and Z as n! 1, which completes the proof of (2.57). g n n d! 1 (2.57) ) (2.55): Let A 2 A with (A) = 1. Then by (2.57) there is a sequence f f n j n 1 g in L 1 () such that 0 f n 1 A ; fn d < 1 ; 8n 1 and fn d! 1 as n! 1. It is no restriction to assume that each f n ; n 1 is A-measurable with 0 f n 1 A -a.s. Then each B n = ff n > 0g is in A and B n A -a.s. Moreover fn d = Bn f n d (B n ), which shows (B n ) < 1 ; n 1 and (B n )! 1 for n! 1. This completes the proof of (2.55). Since (0f ) 3 = 0f 3 and (0f ) 3 = 0f 3, the equivalence of (2.58) and (2.59) is obvious. Hence it is enough to show that (2.57) and (2.59) are equivalent. (2.57) ) (2.59): Since f 3 f, we have 3 f 3 d 3 f d. If there exists g 2 L1 () such that g f, then g f 3 -a.s., and hence g d f3 d, which implies 3 f d f 3 d = 3 f 3 d. If there is no such g 2 L 1 (), then by definition we have f d = 01 which together with the preceding statements in both cases implies: 3 Z Z (2.60) 3 f d = 3 f 3 d. Now we shall show that 3 f 3 d is equal to the right side in (2.59). If f 3 2 L 1 () this is obvious, and if f3 d = 01 then this holds by definition of the lower -integral. In the case where f 3 62 L(), by (2.60) and definition of the lower -integral we get f d = 01. So, 3 let us suppose that f3 d = 1. Then (f 3 ) + d = 1 and by (2.57) we can find a sequence f f n j n 1 g in L 1 () such that 0 f n (f 3 ) + ; n 1 and f n d! 1 as n! 1. Since (f 3 ) 0 d < 1, by g n = f n 0 (f 3 ) 0 ; n 1 a sequence from L 1 () is defined such that g n = f n 0 (f 3 ) 0 (f 3 ) + 0 (f 3 ) 0 = f 3 f and such that gn d = fn d 0 (f 3 ) 0 d! 1 as n! 1. This shows f d = 1 and completes the proof of (2.59). 3 (2.59) ) (2.57): Let f : X! + be a given -measurable function such that f d = 1. Since f 3 3 = f -a.s. hence by (2.59) it follows f d = 1. Then by definition of the upper -integral there is a sequence f f n j n 1 g in L 1 () satisfying f n f ; n 1 and fn d! 1 as n! 1. Since it is no restriction to assume f n 0 ; n 1, the proof of (2.57) is complete. Next we extend some well-known results for the ordinary integral to the upper integral case. In this direction the following proposition (see [7]) is going to play a central role. Proposition 2.4 Let (X; A; ) be a measure space such that satisfies Segal s localization principle, and let f : X! and G :! be given functions. Then we have: 9

10 (2.61) If G is increasing and left continuous, then G(f ) 3 = G(f 3 ). (2.62) If G is increasing and right continuous, then G(f ) 3 = G(f 3 ). (2.63) If G is decreasing and left continuous, then G(f ) 3 = G(f 3 ). (2.64) If G is decreasing and right continuous, then G(f ) 3 = G(f 3 ). Proof. (2.61): Since G is increasing we have G(f ) G(f 3 ), and hence G(f ) 3 G(f 3 ). Conversely, take g 2 M(A) such that G(f ) g, and define: G (01) (y) = sup f x 2 j G(x) y g for all y 2 with sup ; = 01. Put G (01) (01) = 01 and G (01) (+1) = +1. Then G (01) :! is an increasing function and hence Borel measurable. Since G is increasing and left continuous we have: (2.65) G(x) y if and only if G (01) (y) x for all x; y 2. Hence we find G (01) (g) f, and since G (01) belongs to M(A), it follows G (01) (g) f 3. Moreover, by (2.65) we get G(f 3 ) g. This implies G(f 3 ) G(f ) 3, and the proof of (2.61) is complete. The proof of (2.62)-(2.64) is quite similar. Theorem 2.5 Let (X; A; ) Suppose that be a measure space, and let us for every 1 p < 1 define: L <p> () = 8 f 2 X j 3 jfj p d < 1 9. satisfies Segal s localization principle. Then we have: (2.66) (Markov s inequality) If f : X! is an arbitrary function, then: 3 f jfj > " g 1 " p Z 3 jfj p d for all p > 0 and all " > 0. (2.67) (Holder s inequality) If f 2 L <p> () and g 2 L <q> () for 1 < p; q < 1 with 1 p + 1 q = 1, then f 1 g 2 L <1> () and we have: Z 3 nz 3 1 jf 1gj d jfj do 1nZ 3 o 1 p p jgj q q d. (2.68) (Minkowski s inequality) If f; g 2 L <p> () for 1 p < 1, then f + g 2 L <p> () and we have: nz 3 1 jf +gj do nz 3 o 1 p p jfj p p d +nz 3 o 1 jgj p p d. (2.69) (Jensen s inequality) If (X) = 1 and C :! is convex, then for every f 2 L <1> () we have: 10

11 CZ 3 Z 3 f d C f d. Proof. The statement (2.66) follows by (2.50) and (***) below from the classic Markov s inequality. The statement (2.67) follows by (2.49), (2.50) and (2.61) from the classic Holder s inequality. The statement (2.68) follows by (2.50) and (2.61) from the classic Minkovski s inequality. For (2.69) note if C is decreasing, then C(f 3 ) C(f ) and hence measurability of C implies C(f 3 ) C(f ) 3. If C is increasing, then by (2.61) we have C(f 3 ) = C(f ) 3. Combining these two facts, using (i) of lemma 4.4 in [19] and piecewise monotonicity of a convex function, the statement (2.69) follows by (2.50) from the classic Jensen s inequality. Let us remark that all relations (equalities, inequalities,... ) concerning the hulls and kernels of sets and envelopes of functions should be understood in the a.s.-sense, i.e. like ordinary relations with properly chosen representant from the given equivalence classes. Also, to avoid technical difficulties and to be concerned with main ideas only, we shall restrict ourselves to the finite case, i.e. we shall mainly deal with finite measure spaces and real valued functions in question. We will now show that some known results on the envelopes (see [2]) remain valid if we replace Euclidean topology by Sorgenfrey topology on the real line. For this, if X is a set and F a family of subsets of X, then (F ) denotes the smallest topology on X which contains F. Let us further introduce the following notation for families of intervals on the real line: I l () = f<a; b] j a; b 2 g, I r () = f[a; b > j a; b 2 g, and I() = f<a; b > j a; b 2 g. Then by L() = (I l ()), () = (I r ()) and E() = (I()) left, right Sorgenfrey and Euclidean topology on the real line are defined, resp. It is well-known that (; L()) and (; ()) are separable perfectly normal non-metrizable topological spaces which do not allow countable basis for their topology. Families I l () and I r () are the basis of the corresponding topologies, and their elements are open and closed at the same time. The corresponding weights of equals continuum, i.e. wf(; L())g = wf(; ())g = c. Note that E() = L()\(). Although topologies L() and () do not allow countable basis, let us note if L 2 L() is an open set in the left Sorgenfrey topology, then S there exists a countable family of disjoint intervals f I n j n 1 g I l () such that 1 L = n=1 I n. Of course, the analogous conclusion holds for right Sorgenfrey topology () as well. Indeed, if L is L()-open, then there exists a family f I j 2 A g I l () such that L = S 2A I. Put L Q = L \ Q and for each q 2 L Q look at the family I(g) = f I j q 2 I ; 2 A g. Then it is easily checked that I q := S I2I(q) I 2 fi l() [ I()g and L = S q2lq I q. Since each I 2 I() is a countable union of elements from I l (), hence we find that there exists a countable subfamily f J n j n 1g S S 1 of I l () such that L = n=1 J n. Now take n01 I1 = J1 ; I2 = J2 n J1... I n = J n n ( i=1 J i) for n 2, and then one can S easily verify that f I n j n 1 g is a family of disjoint elements 1 from I l () such that L = n=1 I n. This completes the claim. Proposition 2.6 Let (X; A; ) be a finite measure space, and let f be an arbitrary real valued function on X. (2.70) If G 2 L(), then ff 3 2 Gg = 3 ff 3 2 G ; f 2 G g. (2.71) If G 2 (), then ff 3 2 Gg = 3 ff 3 2 G ; f 2 G g. 11

12 Proof. (2.70): First note that it is enough to show: 3f f 3 2 G ; f 62 G g = 0. For this, take A 2 A such that A f 3 f 2 G ; f 62 G g. Since G 2 L(), we can find a family of disjoint intervals f <a n ; b n ] j n 1 g such that S 1 G = n=1 <a n; b n ]. Put A n = A \ f 3 f 2<a n ; b n ] g for n 1, and 3 f + = f 1 P 1 1 A c + n=1 a n1 An. Then f + 2 M() and since A f f 62 G g, by definition of 3 f we find f f + 3 f and A = f 3 f + < f g. This implies (A) = 0, and the proof of (2.70) is complete. The proof of (268) follows similarly if we take on the right place f + = f3 1 P 1 1 A c + n=1 b n1 An. Corollary 2.7 Let (X; A; ) be a finite measure space, and let f be an arbitrary real valued function on X. (2.72) If F 2 2 is L()-closed, then f f 3 2 F g = 3(f f 3 2 F g [ f f 2 F g). (2.73) If F 2 2 is ()-closed, then f f3 2 F g = 3(f f3 2 F g [ f f 2 F g). (2.74) If F 2 2 is E()-closed, then 3f f 2 F g = 1 implies f f 3 2 F ; f3 2 F g = 1. Proof. Since for an arbitrary subset C of X we have: 3 (C) + 3(C c ) = (X) then by Proposition 2.6 we find 3(f f 3 2 F g [ f f 2 F g) = (X) 0 3 f f 3 2 F c ; f 2 F c g = (X) 0 f f 3 2 F c g = f f 3 2 F g which proves (2.72), and 3(f f3 2 F g [ f f 2 F g) = (X) 0 3 f f3 2 F c ; f 2 F c g = (X) 0 f f3 2 F c g = f f3 2 F g which implies (2.73), while (2.74) follows directly by (2.72) and (2.73). Since Euclidean topology E() equals to L() \ (), all the preceding statements from proposition 2.6 and Corollary 2.7 hold in the case when G is E()-open and F is E()- closed. In particular, we find that for an arbitrary real valued function f on a finite measure space (X; A; ), and for each t 2, the following relations are satisfied: (2.75) f f 3 > t g = 3 f f > t g (2.76) f f3 < t g = 3 f f < t g (2.77) f f 3 t g = 3f f t g (2.78) f f3 t g = 3f f t g. p 1 0 (2.79) Since for each f 2 M(A) with f 0, we have 1 X f p d = p t p01 ff tg dt = 0 t p01 ff > tg dt, then from (2.75) and (2.78) together with (2.50) and (2.51) we obtain: Corollary 2.8 Z 3 f p d = pz 1 If f : X! + is an arbitrary function, then for all p 1 we have: 0 t p01 3 ff tg dt = p 12 Z 1 t p01 3 ff > tg dt 0

13 (2.80) Z 3 Z 1 Z 1 f p d = p t p01 3ff tg dt = p t p01 3ff > tg dt. 0 0 Let us remark that (2.75)-(2.80) remain valid if is an arbitrary measure on (X; A) which satisfies Segal s localization principle Image and coimage measures. Let (X; A; ) be a finite measure space, let Y be a set, and let f : X! Y be a function. Then the image measure f of under f is defined on the -algebra B = f 1 (A) = f B 2 2 Y j f 01 (B) 2 A g as follows: f (B) = ff 01 (B)g for all B 2 B. Let X be a set, let (Y; B; ) be a finite measure space, and let g : X! Y be a function. Then the outer and inner coimage measure g 01 and g 01 of under g are defined on the -algebra A = g 01 (B) as follows: g 01(A) = 3 fg(a)g and g01 (A) = 3fg(A)g for all A 2 A, resp. It is easily verified that f, g 01 and g01 are finite measures, and that the following statements are satisfied: (2.81) ( f )3 3 f 01 3 f 01 ( f ) 3 on 2 Y (2.82) g01 g 01 on A = g 01 (B) (2.83) g01 (A) = g 01(A) if and only if g(a) 2 B, where A 2 A = g 01 (B) (2.84) ( g 01) 3 (C) = 3 fg(c)g ; 8C 2 2 X (2.85) ( g 01) 3 (C) 3 fg(c)g ; 8C 2 2 X (2.86) ( g01 ) 3 (C) 3 fg(c)g ; 8C 2 2 X (2.87) ( g 01 ) 3(C) 3 fg(c)g ; 8C 2 2 X, with equality if g is injective (2.88) ( f ) f 01 = on A = f 01 (B) (2.89) ( f ) f01 on A = f 01 (B) (2.90) ( f ) f01 (A) = (A) if and only if f (A) 2 B f, where A 2 A = f 01 (B) (2.91) ( g 01) g = if and only if 3 fy n g(x)g = 0 (2.92) ( g01 ) g = if and only if 3 fy n g(x)g = 0. In this context note that for each A 2 A = f 01 (B), there is B 2 B such that A = g 01 (B). Moreover, it is very useful to observe that we can replace this general set B by the -hull f (A) 3 of f (A), since without loss of generality we can assume that f (A) 3 B. Proposition 2.9 Let X be a set, let (Y; B; ) be a finite measure space, let f : X! Y be a function, and let A = f 01 (B) be the -algebra of subsets of X. For given C 2 2 X let C 3 and C 3 denote the ( g 01)-hull and kernel of C, and for given D 2 2 Y let D 3 and D 3 denote the -hull and 13

14 kernel of D, resp. Then we have: (2.93) g(c 3 ) 3 = g(c) 3 ; 8C 2 2 X (2.94) 3 fg(c 3 ) n g(c)g = 0 ; 8C 2 2 X (2.95) 3 fg(c) n g(c 3 )g = 0 if and only if g(c 3 ) 3 = g(c) 3 (2.96) If g is injective, then 3 fg(c) n g(c 3 )g = 0, and hence g(c 3 ) 3 = g(c) 3 (2.97) 3 fg(c 3 ) n g(c)g = 0 if and only if g(c 3 ) 3 = g(c) 3 if and only if g(c) 2 B (2.98) 3 fg(c) n g(c 3 )g = 0 if and only if g(c 3 ) 3 = g(c 3 ) 3 if and only if C 2 A g01. Proof. (2.93): Since g(c) g(c 3 ), it follows g(c) 3 g(c 3 ) 3. Further, without loss of generality we can assume C 3 = g 01 fg(c 3 ) 3 g. Thus g 01 fg(c) 3 g C 3, and hence it is no restriction assume that C 3 = g 01 fg(c) 3 g. This implies g(c 3 ) g(c) 3 and hence g(c 3 ) 3 g(c) 3, which together with the preceding inclusion implies (2.93). (2.94): Take B 2 B with B g(c 3 ) n g(c). Since C 3 = g 01 fg(c 3 ) 3 g it follows g 01 (B) C 3 n C, and since B g(c 3 ) by definition of f g 01g-hull C 3 we find 0 = g 01fg 01 (B)g = 3 fg(g 01 (B))g = (B). This implies (2.94) and completes the proof. (Note that (2.94) follows directly by (2.93) as well.) (2.95): First suppose that g(c 3 ) 3 = g(c) 3. Then 3 fg(c) n g(c 3 )g 3 fg(c) n g(c 3 ) 3 g = 3 fg(c) n g(c) 3 g = 0, and hence 3 fg(c) n g(c 3 )g = 0. Conversely, without loss of generality suppose C 3 = g 01 fg(c 3 ) 3 g. Hence we have g(c 3 ) = g(c)\g(c 3 ) 3. Then g(c) 3 = g(c 3 ) 3 [B, with B 2 B ; B g(c) \ fg(c 3 ) 3 g c. But, then 3 fg(c) n g(c 3 )g = 0 implies (B) = 0. Thus g(c) 3 = g(c 3 ) 3, and the proof of (2.95) is complete. (2.96): If g is injective, then g 01 (B) = A C n C 3, where g(c) 3 = g(c 3 ) 3 [ B with B 2 B ; B g(c) \ fg(c 3 ) 3 g c. Hence g 01(A) = 0 by definition of ( g 01)-kernel C 3. But g 01(A) = 3 fg(a)g = (B), since B g(c), and thus (B) = 0, which implies g(c) 3 = g(c 3 ) 3. Now (2.96) follows by (2.95). (2.97): First suppose that 3 fg(c 3 ) n g(c)g = 0. Since g(c) g(c 3 ), then it follows 3 fg(c 3 ) n g(c)g = (fg(c 3 ) n g(c)g 3 ) = fg(c 3 ) 3 n g(c) 3 g. Hence g(c 3 ) 3 = g(c) 3. If g(c 3 ) 3 = g(c) 3, then by (2.93) we find g(c) 3 = g(c) 3 and hence g(c) 2 B. If g(c) 2 B, then g(c) 3 = g(c) 3 and hence by (2.93) we have 3 fg(c 3 ) n g(c)g = (fg(c 3 ) n g(c)g 3 ) = fg(c 3 ) 3 n g(c) 3 g = fg(c) 3 n g(c) 3 g = 0. This completes the proof. (2.98): It is no restriction to assume g(c) 3 = g(c 3 ) 3 [ fg(c) n g(c 3 )g 3, and now if 3 fg(c) n g(c 3 )g = (fg(c) n g(c 3 )g 3 ) = 0, then by (1) we have g(c 3 ) 3 = g(c) 3 = g(c 3 ) 3. If g(c 3 ) 3 = g(c 3 ) 3, then C 3 = g 01 fg(c 3 ) 3 g = g 01 fg(c 3 ) 3 g = C 3 which implies C 2 A g01. If C 2 A g01, then g 01fC 3 n C 3 g = 0. Since C 3 = g 01 fg(c 3 ) 3 g we have fg(c) n g(c 3 )g g(c 3 n C 3 ) g(c 3 ) 3 n g(c 3 ) 3 which implies g 01(C 3 n C 3 ) = 3 fg(c 3 n C 3 )g = 3 fg(c)ng(c 3 )g and hence 3 fg(c) n g(c 3 )g = 0. Proposition 2.10 Let X be a set, let (Y; B; ) be a finite measure space, let f : X! Y be a function, and 14

15 let A = f 01 (B) be the -algebra of subsets of X. For given C 2 2 X let C 3 and C 3 denote the ( )-hull and kernel of g C, and for given D Y let D 3 and D 3 denote the -hull and kernel of D, resp. Then we have: (2.99) g(c 3 ) 3 = g(c) 3 ; 8C 2 2 X (2.100) 3 fg(c 3 ) n g(c)g = 0 ; 8C 2 2 X (2.101) 3 fg(c) n g(c 3 )g = 0 if and only if g(c 3 ) 3 = g(c) 3 (2.102) If g is injective, then 3 fg(c) n g(c 3 )g = 0, and hence g(c 3 ) 3 = g(c) 3 (2.103) 3 fg(c 3 ) n g(c)g = 0 if and only if g(c 3 ) 3 = g(c) 3 if and only if g(c) 2 B (2.104) 3 fg(c) n g(c 3 )g = 0 if and only if g(c 3 ) 3 = g(c 3 ) 3 (2.105) 3 fg(c) n g(c 3 )g = 0 ) C 2 A g 01 ) 3 fg(c) n g(c 3 )g = 0 (2.106) C 2 A g 01 if and only if fg(c) n g(c 3 )g = 0. Proof. The proofs of (2.99)-(2.104) are very similar to the proofs of the facts from the preceding proposition. For (2.105) it is no restriction assume that C 3 = g 01 fg(c 3 ) 3 g ; g(c) 3 = g(c 3 ) 3 [ fg(c) n g(c 3 )g 3 and fg(c) n g(c) 3 g 3 [g(c 3 ) 3 = g(c) 3. Note that g(c)ng(c 3 ) g(c 3 nc 3 ) g(c) 3 n g(c 3 ) 3, which implies 3 fg(c) n g(c 3 )g 3 fg(c 3 n C 3 )g fg(c) 3 n g(c 3 ) 3 g. Since 3 fg(c 3 n C 3 )g = g 01 (C3 n C 3 ) and fg(c) 3 n g(c 3 ) 3 g = 3 fg(c) n g(c 3 )g, this completes the proof of (2.105). Finally, it remains to note that (2.106) is an easy consequence of (2.105). We will conclude this section by noting if (X; A; ) is a finite measure space and C 2 2 X, then for the map g : C! X defined by g(x) = x for x 2 C we have: g 01 = rftr 3 (; C); tr(a; C)g g 01 = rftr 3(; C); tr(a; C)g. We remark that all of the preceding results can be applied to this particular case. 3. Perfect measures and maps 3.1. Perfect measures. A finite measure on a measurable space (X; A) is said to be perfect, if f (X) belongs to B() f whenever f is an A-measurable real valued function on X. A finite measure space (X; A; ) is said to be perfect, if the measure is perfect. A measurable space (X; A) is said to be perfect, if every finite measure on (X; A) is perfect. Proposition 3.1 Let (X; A; ) (3.1) is perfect (3.2) is perfect be a finite measure space. Then the following statements are equivalent: (3.3) 8f 2 M(A); 9B f 2 B(); B f f (X) such that ff 01 ( n B f )g = 0 15

16 (3.4) ff 01 ( n f (X) 3 )g = 0, 8f 2 M(A), where f (X) 3 is the f -kernel of f (X) (3.5) ( f ) 3 ff (X)g = (X), 8f 2 M(A) (3.6) 8f 2 M(A); 9N f 2 A; (N f ) = 0 such that f (X n N f ) 2 B() (3.7) 8f 2 M(A); 9g 2 M(A); g = f -a.s. such that g(x) 2 B() (3.8) 8f 2 M(A); 9N f 2 A; (N f ) = 0 such that f (X n N f ) 2 B() f (3.9) 8f 2 M(A); 9g 2 M(A); g = f -a.s. such that g(x) 2 B() f. Proof. For (3.8) suppose first that is perfect. Then there is B 2 B() such that B f (X) and f ( n B) = 0. Put N f = f 01 ( n B), then (N f ) = 0 and since B f (X), we have f (X n N f ) = B 2 B() B() f. Conversely, for given f 2 M(A), let N f be a set from A such that (N f ) = 0 and f (X n N f ) 2 B() f. Let M = f (N f ) n f (X n N f ), then f (X) = f (X n N f )[M. Since f ff (X n N f ) 3 n f (X n N f ) 3 g = 0, it follows f (X) 3 = f (X n N f ) 3 [ M 3 and hence f ff (X) 3 g = f ff (X n N f ) 3 g + f (M 3 ). But f (M 3 ) = ff 01 (M 3 )g = 0, since f 01 (M 3 ) f 01 (M ) N f. Similarly, we find that f (X) 3 = f (XnN f ) 3 [fm 3 nf (XnN f ) 3 g and hence f ff (X) 3 g = f ff (X n N f ) 3 g + f fm 3 n f (X n N f ) 3 g. But f fm 3 n f (X n N f ) 3 g = 0, since f 01 fm 3 nf (XnN f ) 3 g f 01 (M ) N f. Now, since f ff (XnN f ) 3 g = f ff (XnN f ) 3 g, it follows f ff (X) 3 g = f ff (X) 3 g, i.e. f (X) 2 B() f and the proof is complete. Let us recall that a class K of subsets of X is said to be compact, if for each sequence T 1 T n=1 K n = ;, there is N 1 such that N n=1 K n = ;. f K n j n 1 g in K such that It is well-known (see [12]) that we have: (3.10) Every subclass of a compact class is compact. (3.11) The class consisting of countable intersections of the sets in some compact class is itself compact. (3.12) The class formed by taking finite unions of the sets in some compact class is itself compact. A finite measure on a measurable space (X; A) is said to be compact, if there is a compact class K of subsets of X such that: (A) = sup 3 (K). K2K;KA The compact class K which approximates the given compact measure in the sense of the above relation can always be chosen such that it consists of sets in A, see [12]. If (X; A; ) and (Y; B; ) are finite measure spaces, then the product of and is a finitely additive measure defined on the smallest algebra (A B) generated by the family A B = f A 2 B j A 2 A ; B 2 B g which satisfies: (A 2 Y ) = (A) and (X 2 B) = (B) for all A 2 A and B 2 B. Let us recall that a finite measure on a topological space X is said to be adon, if for 16

17 each Borel set B 2 B(X) we have: (B) = sup 3 (K) K2K(X );KB where K(X) is the family of all compact sets in X. It is well-known that the following statements (see [16], [20], [21]) are also equivalent to (3.1)-(3.9): (3.13) 8" > 0 ; 8f A n j n 1 g A ; 9A 2 A such that (A) (X)0" and f A\A n j n 1 g is compact. (3.14) Any product of with any finite measure is countably additive. (3.15) If f : X! S is a Borel measurable function into a separable metric space S, then the image measure f is adon. (3.16) If (Y; B; ) is a finite measure space and is a measure on the product -algebra (A B) such that its restriction to the algebra (A B) is the product of and, then 3 (X 2 C) = 3 (C) for every C subset of Y. (3.17) For each countably generated -algebra B contained in the -algebra A, the restriction r(; B) of on B is compact. Furthermore, if we extend the definition of the product of two finite measures to an arbitrary family of finite measures naturally, then we have (see [20]): (3.18) Any product of an arbitrary family of perfect measures is countably additive and its countably additive extension to the product -algebra is perfect. In [20] we can also find that: (3.19) Every compact measure is perfect. Combining this result with (3.17) we get: (3.20) If the -algebra A is countably generated, then a finite measure on A is perfect, if and only if it is compact. emark 3.2 Let us take C 2 2 [0;1] such that 3 (C) = 0 and 3 (C) = 1, where as usual denotes the Lebesgue measure on the Borel -algebra B([0; 1]) = trfb(); [0; 1]g. Put X = C ; A = trfb([0; 1]); Cg and = tr 3 (; C). Then (X; A; ) is a finite measure space, but the measure is not perfect. Indeed, for f 2 M(A) defined by f (x) = x ; 8x 2 X, and for B 2 B() such that B f (X) = C we have f (B) = ff 01 (B)g = (B \ C) = (B) = tr 3 (; C)(B) = (B \ C 3 ) = (B) = 0. Hence we can conclude ( f ) 3 ff (X)g = 0 6= 1 = (X) which by (3.5) shows that is not perfect Blackwell spaces. Last example is dealing with a very pathological measurable space as we shall see now introducing a very wide and powerful class of perfect measurable spaces. A measurable space (X; A) is called Blackwell space, if f (X ) is an analytic subset of the real line, whenever f is a measurable real valued function on X. Blackwell spaces have a lot 17

18 of nice properties. Let us recall some of them. For this, if P is a given paving on a set X, let S(P) denote the paving of all P-Souslin sets in X. Then we have: (The projection theorem) Let (X; A) be a Blackwell space, and let (Y; B) be a measurable space. If p Y of X 2 Y onto Y, then p Y (A) 2 S(B) for all A 2 S(A 2 B). (The image theorem) is the projection Let (X; A) be a Blackwell space, and let (Y; B) be a countably separated measurable space. If f : X! Y is a measurable function, then f (A) 2 S(B) for all A 2 S(A). (The first separation theorem) Let (X; A) be a Blackwell space. If A; B 2 S(A) are disjoint sets, then there are disjoint sets A 0 ; B 0 2 A such that A A 0 and B B 0. Let us recall that an analytic space is a Hausdorff space which is a continuous image of some Polish space. It is well-known that if X is an analytic space and P = F (X) is the paving of all closed subsets of X, then S(P) = SfF (X)g = A(X) is the paving of all analytic subsets of X. Furthermore, the main point in our considerations have the following well-known statements: (3.21) Every analytic space together with its Borel -algebra is a Blackwell space. (3.22) Every Blackwell space is perfect. (3.23) If X is analytic, then (X; B(X)) is perfect. Note that (3.23) follows straightforward by (3.21) and (3.22), and that (3.22) is a trivial consequence of our definition of the Blackwell space. This shows that there is a lot of nice Blackwell spaces, and that the class of Blackwell spaces (and thus of the perfect ones as well) is very wide. Moreover, let us recall that a Borel or Baire measure on a topological space X is said to be tight, if we have: (X) = It is well-known (see [9]) that we have: sup 3 (K) : K2K(X);KX (3.24) Every tight Borel or Baire measure is perfect. For separable metric spaces the converse statement is also true, see [21]: (3.25) Every perfect Borel measure on a separable metric space is tight. The following proposition answers the question when the image and coimage measures of a perfect measure are perfect. Proposition 3.3 (3.26) Let (X; A; ) be a perfect measure space, let Y be a set, let f : X! Y be a function, and let B = f 1 (A) = f B 2 2 Y j f 01 (B) 2 A g be the -algebra of subsets of Y. Then 18

19 (Y; B; f ) is perfect. (3.27) Let X be a set, let (Y; B; ) be a perfect measure space, let g : X! Y be a function, and let A = g 01 (B) be the -algebra of subsets of X. If g is injective and g(a) B, then (X; A; g 01) and (X; A; g 01 ) are perfect. Proof. (3.26): Let g 2 M(B), then g f 2 M(A) and since is perfect, we have f( f ) g g 3 fg(y )g = ( gf ) 3 fg(y )g ( gf ) 3 f(g f )(X)g = (X). On the other hand, for each B 2 B() we have f( f ) g g(b) = gf (B) = f(g f ) 01 (B)g (X), and hence together with the preceding relation it follows that f( f ) g g 3 fg(y )g = (X) = f (Y ). This by (3.5) implies that f is perfect. (3.27): Let f 2 M(A). Since g is injective and g(a) B, for given B 2 B() we find that ( g 01) f (B) = g 01ff 01 (B)g = 3 fg(f 01 (B))g = fg f 01 (B)g = gf01(b). Thus f( g 01) f g 3 ff (X)g = ( 01) f g 3ff (X)g ( 01) f g 3ff g 01 (Y )g. Since is perfect, then by (3.2) we have that is also perfect. Then f g 01 2 M(B) implies ( 01) f g 3f(f g 01 )(Y )g = (Y ) = (Y ) 3 fg(x)g = g 01(X). On the other hand we have f( g 01) f g(b) = g 01ff 01 (B)g g 01(X), and hence together with the preceding relations it follows f( g 01) f g 3 ff (X)g = g 01(X). This by (3.5) implies that g 01 is perfect is similar. g01 Corollary 3.4 is a perfect measure. The proof that the inner coimage measure (3.28) If (X; A; ) is a perfect measure space and A 0 is a sub--algebra of A, then the restriction measure space 0 X; A 0 ; r(; A 0 )1 is perfect. (3.29) If (X; A; ) is a perfect measure space, and if A 2 A, then the trace measure space 0 A; tr(a ; A); r(tr(; A); tr(a ; A)) 1 is perfect. (3.30) If (X; A) is a perfect measurable space, and if A 2 A 3, then the trace measurable space (A; tr(a 3 ; A)) is perfect. (3.31) If (X; A; ) is a perfect measure space and ^X X is a set, then the extension measure space ( ^X; ^A; ext(; ^X)) is perfect. (3.32) If (X; A) is a perfect measurable space and ^X X is a set, then the extension measurable space ( ^X; ^A) is perfect. Proof. (3.28): It follows directly by definition of perfectness. (3.29): If (X; A; ) is perfect, then by (3.2) (X; A ; ) is perfect. For given A 2 A, define g : A! X by g(x) = x, 8x 2 A. Then g is injective, g 01 (A ) = tr(a ; A) and gfg 01 (A )g = gftr(a ; A)g = tr(a ; A) A. Hence we see that (3.27) can be applied, and since g 01 = g01 = r(tr(; A); tr(a ; A)), it follows that the trace measure space 0 A; tr(a ; A); r(tr(; A); tr(a ; A))1 is perfect. (3.30): Suppose that (X; A) is perfect. Let A 2 A 3 be given, and consider any finite measure 0 on (A; tr(a 3 ; A)). Let = r(ext( 0 ; X); A), i.e. let (B) = 0 (B \ A) for all B 2 A. Then is a finite measure on (X; A), and hence is perfect. Since A 2 A 3 A, by (3.29) it follows that the trace measure space 0 A; tr(a ; A); r(tr(; A); tr(a ; A))1 is perfect, 19

20 while by (3.28) the restriction measure space 0 A; tr(a 3 ; A); r(tr(; A); tr(a 3 ; A))1 is perfect. But, since r( 0 ; tr(a; A)) = r(tr(; A); tr(a; A)), it follows r(tr(; A); tr(a 3 ; A)) = 0, and the proof of (3.30) is complete. (3.31): Let f 2 M( ^A), then f 01 fb()g ^A and hence f 01 (B) \ X 2 A, whenever B 2 B(). This shows that the restriction g of f to X is an A-measurable function on X, and since (X; A; ) is perfect, there is B g 2 B() such that fg 01 ( n B)g = 0. Since g(x) = f (X) f ( ^X) and g 01 (B) f 01 (B), it follows ext(; ^X)ff 01 ( n B g )g = 0 and hence ( ^X; ^A; ext(; ^X)) is perfect by definition. This proves (3.31). (3.32): It is a direct consequence of (3.31) Perfect maps. Perfectness of a given finite measure is defined globally, that is, by using the whole space, and one can be interested to see how much has this property local influence to measurability properties. For this, let (X; A; ) be a finite measure space, let (Y; B) be a measurable space, and let f : X! Y be a measurable map. Then f is said to be -perfect, if 8A 2 A; 9B 2 B; B f (A) such that fa n f 01 (B)g = 0. Proposition 3.5 Under the hypotheses stated above, we have: (2.33) f is -perfect, if and only if fa n f 01 ff (A) 3 gg = 0 for all A 2 A, where f (A) 3 is the f -kernel of f (A) (2.34) f is -perfect, if and only if 8A 2 A; 9A 0 2 A; A 0 A such that (A n A 0 ) = 0 and f (A 0 ) 2 B (2.35) If f (A) 2 B f for all A 2 A, then f is -perfect (2.36) If f : X! Y is -perfect and injective, then f (A) 2 B f for all A 2 A. Proof. It follows easily by definition. The set of all real valued -perfect maps on (X; A; ) is denoted by M p (A). If M() is the set of all -measurable ( i.e. A -measurable ) real valued functions on X, then we clearly have: Moreover, it is easily seen that: M p (A) M(A) M(). (3.37) The measure is perfect, if and only if every A-measurable real valued function on X is -perfect, i.e. if and only if M p (A) = M(A) (3.38) The measure is perfect and complete, if and only if M p (A) = M(). Theorem 3.6 Let (X; A; ) be a finite measure space, let (Y; B) be a measurable space, let f : X! Y be a measurable function. Then for any function : Y! we have: (3.39) 3 f ( f ) 3 ( f ) 3 3 f (3.40) 3 d f 3 f d 3 f d 3 df. 20

21 Moreover, the following statements are equivalent: (3.41) f is -perfect (3.42) ( f ) 3 = 3 f 01 (3.43) ( f ) 3 = 3 f (3.44) df = f d, for all functions : Y! (3.45) d 3 f = f d, for all functions : Y! 3 (3.46) 3 f = ( f ) 3, for all functions : Y! (3.47) 3 f = ( f ) 3, for all functions : Y! (3.48) 3 ff 01 (C)g = 1 ; 8C 2 2 Y such that ( f ) 3 (C) = 1 (3.49) 3 ff 01 (C)g = 0 ; 8C 2 2 Y such that ( f ) 3 (C) = 0 (3.50) f A j f = y g = 0 ; 8y 2 Y n f (A) ; A 2 A (3.51) f A j f = y g = 1 ; 8y 2 Y n f (X n A) ; A 2 A (3.52) Ef ( f ) 3 j f g = 3, for all functions : Y! (3.53) Ef ( f ) 3 j f g = 3, for all functions : Y!. Furthermore, for a given map g : X! define: M (f; g)(y) = m(f; g)(y) = sup g(x) ; x2f 01 (y) y 2 Y inf g(x) ; y 2 Y. x2f 01 (y) Then the following statements are also equivalent to (3.41)-(3.53): (3.54) Ef g3 j f g M (f; g), for all functions g : X! (3.55) Ef g 3 j f g m(f; g), for all functions g : X! (3.56) 8h : X! ; 9g 2 M() ; g = h3 -a.s. such that M (f; g) is A-measurable (3.57) 8h : X! ; 9g 2 M() ; g = h 3 -a.s. such that m(f; g) is A-measurable (3.58) 3 g d 3 M (f; d g) f, for all functions g : X! 3 3 (3.59) g d m(f; g) d f, for all functions g : X! (3.60) g 3 M (f; g)3 f, for all functions g : X! (3.61) g 3 m(f; g) 3 f, for all functions g : X!. Proof. The proof of (3.39) and (3.40) follows by definitions of the lower and inner -envelopes and integrals. The rest of the proof can be reconstructed from [2] and [7]. Let (X; A; ) be a finite measure space, and let N () be the set of all -null sets, i.e. 21

22 N () = f C 2 2 X j 3 (C) = 0 g. Then f 01 fn ( f )g N () and f 01 (B f ) A, and if g 2 M( f ) is an arbitrary function, then g f 2 M(). Furthermore, if f is -perfect, then we have: (3.62) N ( f ) = f 1 (N ()) = f B 2 2 Y j f 01 (B) 2 N () g (3.63) B f = f 1 (A ) = f B 2 2 Y j f 01 (B) 2 A g (3.64) g 2 M( f ) if and only if g f 2 M(), where g : Y! is an arbitrary function. Let (X; A; ) be a finite measure space, let (Y; B) and (Z; C) be measurable spaces, and let f : X! Y ; g : Y! Z be measurable functions. It is well-known (see [2]) that we have: (3.65) If f is -perfect and g is ( f )-perfect, then g f is -perfect (3.66) If g f is -perfect, then g is ( f )-perfect (3.67) If g f is -perfect, and g is injective, then f is -perfect. We shall now see that a certain classes of -measurable functions are closely related to perfect ones, see [6]. Let (X; A; ) be a finite measure space, let (Y; B) be a measurable space, and let f : X! Y be a function. Then f is said to be Lusin -measurable, if f is - measurable (measurable with respect A and B) and 8A 2 A; 9A 0 2 A ; A 0 A such that (A n A 0 ) = 0 and f (A 0 ) 2 B f. A finite measure space (X; A; ) is called a Lusin space, if all -measurable real valued functions on X are Lusin -measurable. Let us first note that Lusin -measurability can be expressed, up to -measurability, in terms which do not involve completions of the corresponding -algebras. Namely, it is easily verified that: (3.68) f is Lusin -measurable, if and only if f is -measurable and 8A 2 A; 9A 0 2 A; A 0 A such that (A n A 0 ) = 0 and f (A 0 ) 2 B: Now, one can easily check the following statements: (3.69) If f is -perfect, then f is Lusin -measurable. (3.70) If f is Lusin -measurable and g is an A-measurable function such that f = g -a.s., then g is -perfect. (3.71) A Lusin -measurable function f is -perfect, if and only if f is A-measurable. (3.72) If f is Lusin -measurable and g = f -a.s., then g is Lusin -measurable. From (3.69), (3.70) and (3.71) we get: (3.73) A finite measure space (X; A; ) is a Lusin space, if and only if is perfect. According to (3.70) and the well-known fact that for every -measurable function f A-measurable function g such that f = g -a.s., we easily find: there is an (3.74) A -measurable function f is Lusin -measurable, if and only if either of the 22