GAUGES OF BAIRE CLASS ONE FUNCTIONS

GAUGES OF BAIRE CLASS ONE FUNCTIONS ZULIJANTO ATOK, WEE-KEE TANG, AND DONGSHENG ZHAO Abstract. Let K be a compact metric space and f : K R be a bounded Baire class one function. We proved that for any ε > 0 there exists an upper semicontinuous positive function δ of f with finite oscillation index and f(x) f(y) < ε whenever d(x, y) < min{δ(x), δ(y)}. Throughout this paper, let K denote a compact metric space. A real-valued function f : K R on K is said to be of Baire class one, or simply Baire-1, if there exists a sequence (f n ) of real-valued continuous functions that converges pointwise to f. The Baire characterization theorem [?] states that a function f is of Baire-1 if and only if its restriction to every closed subset F of K has a point of continuity. This leads naturally to the definition of oscillation index β of a function f. The oscillation indices of bounded Baire-1 functions have been studied by many authors (see, e.g., [?], [?], [?]). The study has also been extended to unbounded Baire-1 functions [?]. In this paper, we only consider bounded Baire-1 functions. Let C denote the collection of all closed subsets of K. A derivation on C is a map D : C C such that D(H) H for all H C. The oscillation index β is associated with a family of derivations. Let ε > 0 and a function f : K R be given. For any H C, let D 0 (f, ε, H) = H and D 1 (f, ε, H) be the set of those x H such that for every open set U containing x there are two points x 1 and x 2 in U H with f(x 1 ) f(x 2 ) ε. The derivation D 1 (f, ε,.) may be iterated in the usual manner. For α < ω 1, let D α+1 (f, ε, H) = D 1 (f, ε, D α (f, ε, H)). If α is a countable limit ordinal, let D α (f, ε, H) = α <α D α (f, ε, H). Define β H (f, ε) to be the least countable ordinal α with D α (f, ε, H) = if such an α exists, and ω 1 otherwise. The oscillation index of f on the set H is defined by β H (f) = sup{β H (f, ε) : ε > 0}. We shall write β(f, ε) and β(f) for β K (f, ε) and β K (f) respectively. Note that each set D α (f, ε, H) is closed. By [?, Proposition 1.2] a function f : K R is Baire-1 if and only if β(f) < ω 1. In [?], Lee, Tang and Zhao gave an equivalent condition for a function to be of Baire class one. This condition is analogous to the ε-δ definition of continuous functions. Theorem 1 ([?, Theorem 1]). Suppose that f : X R is a real-valued function on a complete separable metric space (X, d). Then the following statements are 2000 Mathematics Subject Classification. Primary 26A21; Secondary 03E15, 54C30. Key words and phrases. Baire class one functions, oscillation index, gauges. 1

2 Z.ATOK, W.K.TANG, AND D.ZHAO equivalent: (1) For any ε > 0, there exists a positive function δ on X such that f(x) f(y) < ε whenever d(x, y) < min{δ(x), δ(y)}. (2) The function f is of Baire class one. We call the positive function δ in (1) an ε-gauge of f. The ε-gauge δ originates from the study of Kurzweil-Henstock integrals. It was shown that the gauge for Kurzweil-Henstock integrals can always be taken to be measurable (see, e.g., [?] and [?]). In this paper we show that the ε-gauge δ for bounded Baire-1 functions can be chosen to be well-behaved. More precisely, for any bounded Baire-1 function f : K R and for any ε > 0, there exists an upper semicontinuous ε-gauge δ of f with finite oscillation index. The main result of this paper is the following theorem: Theorem 2. Let f : K R be a bounded Baire-1 function. For any ε > 0, there exists an upper semicontinuous ε-gauge δ of f such that β(δ) < ω. We recall that if a function has a finite oscillation index then it is a difference of two bounded semicontinuous functions [?]. On the other hand every semicontinuous function has oscillation index at most ω [?]. We start by constructing desired ε-gauges for characteristic functions. Then, using the fact that every bounded Baire-1 function may be approximated uniformly by simple Baire-1 functions, we are able to deduce Theorem 2. For any 0 < ε < 1 and any A K with β(χ A ) < ω 1 we construct an upper semicontinuous ε-gauge δ of χ A such that β(δ) = 2. We omit the easy proof of the following lemma. Lemma 1. For any 0 < ε < 1, A K and α < ω 1 we have D α+1 (χ A, ε, K) = D α (χ A, ε, K) A D α (χ A, ε, K) A c. Set N r (x) = {y K : d(x, y) < r} and let A α denote D α (χ A, ε, K) for any α < ω 1. It follows from the compactness of K that β(f, ε) is a successor for any ε > 0. Obviously β(χ A, ε 1 ) = β(χ A, ε 2 ) for any 0 < ε 1, ε 2 < 1, therefore β(χ A ) = β(χ A, 1 2 ). Hence β(χ A) must be a successor. Let A K and β(χ A ) = ξ + n with 1 n < ω and ξ = 0 or ξ is a limit ordinal. Define δ : K R + by 1 2 d(x, A α A) if x A α \ A α A, α < ξ + n, 1 δ(x) = 2 d(x, A α A c ) if x A α A \ A α A c, α < ξ + n 1, 1 if x A ξ+n 1 A. The function δ is well defined because for each x K, there is a unique α such that x A α \ A α+1, and A α \ A α+1 is a disjoint union of (A α \ A α A) and (A α A \ A α A c ). Indeed, by Lemma 1, A α \ A α+1 = A α \ (A α A A α A c ) = (A α \ A α A) (A α A \ A α A c ). We now prove that δ is the desired ε-gauge for χ A. Lemma 2. δ is an ε-gauge of χ A. Proof. Let x, y K with d(x, y) < min{δ(x), δ(y)}. If x A α \ A α+1 for some α < ξ + n 1 then y A α \ A α+1. Indeed, if y A α+1 then d(x, y) d(x, A α A) whenever x A α \A α A, otherwise d(x, y) d(x, A α A c ). In both case we have d(x, y) > δ(x). Thus y A β \ A β+1 for some β α. By symmetry β = α.

GAUGES OF BAIRE CLASS ONE FUNCTIONS 3 Recall that A α \A α+1 = (A α \A α A) (A α A\A α A c ). If x A α \A α A, then it follows from N δ(x) (x) A α A = and y N δ(x) (x) that y A α \ A α A. Similarly, if x A α A \ A α A c then y A α A \ A α A c. If x A ξ+n 1 then y A ξ+n 1. Indeed, if y A α \A α+1 for some α < ξ +n 1, by the above arguments we have x A α \A α+1 as well, contradicting the assumption that x A ξ+n 1. Therefore, if x A ξ+n 1 A c = A ξ+n 1 \ A ξ+n 1 A, then as N δ(x) (x) A ξ+n 1 A = and y N δ(x) (x), we have y A ξ+n 1 A c. It follows by symmetry that, if x A ξ+n 1 A then y A ξ+n 1 A. Thus for any x, y K with d(x, y) < min{δ(x), δ(y)}, either both x and y are in A or both of them are in A c. Therefore Lemma 3. β(δ) = 2. Proof. For any α < ξ + n and η > 0, let χ A (x) χ A (y) = 0 < ε. B α,η = {x A α \ A α A : d(x, A α A) η}, C α,η = {x A α A \ A α A c : d(x, A α A c ) η}. For any η > 0 we have (1) D 1 (δ, η, K) ( B α,η ) ( C α,η ) (A ξ+n 1 A). Indeed, suppose that x ( α<ξ+n α<ξ+n B α,η ) ( C α,η ) (A ξ+n 1 A). If x A α \ A α A for some α < ξ + n, then d(x, A α A) < η. If y N δ(x) (x), then y A α A. Also for all β > α, as A β A α A, we have y A β. Therefore for any y N δ(x) (x), there are only three possibilities: (i) y A α \ A α A; (ii) y A β \ A β A for some β < α; and (iii) y A β A \ A β A c, for some β < α. In each of the three cases, we can prove that δ(y) δ(x) < η. Since the proofs for the three cases are similar, we shall only give the proof for case (iii). Note that as β < α, A α A β+1 A β A c. If δ(y) > δ(x), then δ(y) δ(x) = δ(y) δ(x) = 1 2 d(y, A β A c ) 1 2 d(x, A α A) 1 2 d(y, A α A) 1 2 d(x, A α A) 1 2 d(y, x) < 1 2 δ(x) < η 4. If δ(y) δ(x) then δ(y) δ(x) < δ(x) < η 2. Therefore for any y N δ(x)(x), δ(y) δ(x) < η 2. Hence x D1 (δ, η, K). Similarly, if x A α A \ A α A c for some α < ξ + n 1 then x D 1 (δ, η, K). Claim (a) d(b α,η, B β,η ) η for α < β < ξ + n. (b) d(c α,η, C β,η ) η for α < β < ξ + n 1.

4 Z.ATOK, W.K.TANG, AND D.ZHAO (c) d(b α,η, C β,η ) η for α < ξ + n, 0 β < ξ + n 1. (d) d(a ξ+n 1 A, B α,η ) η for α < ξ + n. (e) d(a ξ+n 1 A, C α,η ) η for α < ξ + n 1. We only prove (c) and (d). Let x B α,η and y C α,η for some α < ξ + n 1. Since C α,η A α A, we have d(x, y) d(x, A α A) η. Therefore d(b α,η, C α,η ) η. Now suppose that β < α. Take any x B β,η and y C α,η. It follows from C α,η A α A A β A that d(x, y) d(x, A β A) η. Therefore d(b β,η, C α,η ) η. Take any x B α,η and y C β,η. Since B α,η A α A β A c, we see that d(y, x) d(y, A β A c ) η. Therefore d(b α,η, C β,η ) η. So (c) is proved. To prove (d), take any x A ξ+n 1 A and y B α,η, where α < ξ + n. It follows from A ξ+n 1 A A α A that d(y, x) d(y, A α A) η, hence d(a ξ+n 1 A, B α,η ) η. So (d) is valid. Now for any x D 1 (δ, η, K), by (1) we have x ( B α,η ) ( C α,η ) (A ξ+n 1 A). α<ξ+n Suppose that x B γ,η for some 0 γ < ξ + n. By Claims (a), (c) and (d) we have (x) (( B α,η ) ( C α,η ) (A ξ+n 1 A)) =. N η 2 α<ξ+n, α γ Then for any x 1, x 2 N η 2 (x) D1 (δ, η, K) we have x 1, x 2 B γ,η. Therefore δ(x 1 ) δ(x 2 ) = 1 2 d(x 1, A γ A) 1 2 d(x 2, A γ A) 1 2 d(x 1, x 2 ) 1 2 (d(x 1, x) + d(x, x 2 )) < η 2. Thus, x D 2 (δ, η, K). Similarly, if x C α,η for some α < ξ + n 1, then by Claims (b), (c) and (e) we deduce that x D 2 (δ, η, K). Now if x A ξ+n 1 A, then for any x 1, x 2 N η 2 (x) D1 (δ, η, K), by Claims (d) and (e) we have x 1, x 2 A ξ+n 1 A. Therefore δ(x 1 ) δ(x 2 ) = 1 1 = 0 < η, hence again x D 2 (δ, η, K). It follows that D 2 (δ, η, K) =. Since this is true for arbitrary η > 0 we conclude that β(δ) = 2. Lemma 4. δ is upper semicontinuous. Proof. Let x K and (x n ) be a sequence in K that converges to x. By taking a subsequence if necessary we can assume that (δ(x n )) converges to a point, say to l. To prove that δ is upper semicontinuous, it is enough to show that l δ(x). There are 3 possible cases: (i) x A α \ A α A for some α < ξ + n; (ii) x A α A \ A α A c for some α < ξ + n 1; and (iii) x A ξ+n 1 A. We only consider (i), the proofs for the rest of the cases are similar. Suppose that x A α \ A α A for some α < ξ + n. Since (x n ) converges to x we may assume (x n ) N δ(x) (x). As N δ(x) (x) A α A = we have (x n ) K \A α A. Since A α+1 A α A, we consider two cases. First, there is a subsequence (x nj ) j 1

GAUGES OF BAIRE CLASS ONE FUNCTIONS 5 of (x n ) such that x nj A α \ A α A for all j 1. Second, there is k N such that for all n k, x n A α \ A α A. For the first case, since δ Aα\A α A is continuous at x, we have l = lim j δ(x n j ) = δ(x). For the second case, for each n k, x n A β for some β < α as x n A α. Therefore δ(x n ) is either equal to 1 2 d(x n, A β A) or 1 2 d(x n, A β A c ). In any case, δ(x n ) 1 2 d(x n, A β+1 ) 1 2 d(x n, A α ) 1 2 d(x n, x). It follows that Thus δ is upper semicontinuous. l = lim δ(x 1 n) lim sup n n 2 d(x, x n) = 0 < δ(x). By Lemma 2, Lemma 3 and Lemma 4 we have the following. Proposition 1. Let A K with β(χ A ) < ω 1. Then for any 0 < ε < 1 there exists an upper semicontinuous ε-gauge δ of χ A such that β(δ) = 2. Now we are ready to prove our main result. Proof of Theorem 2. For any ε > 0, there exist a i R and subsets A i of K, i = 1, 2,..., N with β(χ Ai ) < ω 1 for all i such that f(x) a i χ Ai (x) < ε 3 for all x K(see, e.g., [?] and [?]). By Proposition 1, for each i = 1, 2,..., N, there exists an upper semicontinuous ε 3N -gauge δ i of a i χ Ai such that β(δ i ) = 2. Let δ : K R + be defined by δ(x) = min{δ i (x) : 1 i N}. Then δ is upper semicontinuous. Also by [?, Theorem 1.3] we have β(δ) < ω. Let x, y K with d(x, y) < min{δ(x), δ(y)}. Then d(x, y) < min{δ i (x), δ i (y)} for all i = 1, 2,..., N. Therefore ε a i χ Ai (x) a i χ Ai (y) a i χ Ai (x) a i χ Ai (y) < 3N = ε 3. It follows that f(x) f(y) f(x) a i χ Ai (x) + a i χ Ai (x) a i χ Ai (y) + a i χ Ai (y) f(y) < ε. Acknowledgement. We would like to express our sincere thanks to Professor Peng-Yee Lee and the referee for their invaluable comments and suggestions.

6 Z.ATOK, W.K.TANG, AND D.ZHAO References [1] R. Baire, Sur les fonctions des variables réeles, Ann. Mat. Pura ed Appl. 3(1899), 1-122. [2] F. Chaatit, V. Mascioni and H. Rosenthal, On functions of finite Baire index, J. Funct. Anal. 142(1996), 277-295. [3] R. A. Gordon, The integral of Lebesgue, Denjoy, Perron, and Henstock. AMS, New York and Amsterdam, 1991. [4] R. Haydon, E. Odell and H. P. Rosenthal, On certain classes of Baire-1 functions with applications to Banach space theory, in: Functional Analysis ( Austin, TX, 1987-89), Lecture Notes in Math. 1470, Springer, New York, 1991, 1-35. [5] A. S. Kechris and A. Louveau, A classification of Baire class 1 functions, Trans. Amer. Math. Soc. 318 (1990), 209-236. [6] P. Kiriakouli, A classification of Baire-1 functions, Trans. Amer. Math. Soc. 351 (1999), 4599-4609. [7] K. Kuratowski, Topology, Academic Press, London, 1966. [8] P.-Y. Lee, W.-K. Tang and D. Zhao, An equivalent definition of functions of the first Baire class, Proc. Amer. Math. Soc. 129 (2000), 2273-2275. [9] T.-Y. Lee, The Henstock variational measure, Baire functions and a problem of Henstock, Rocky Mountain J. Math. 35 (2005), no. 6, 1981-1997. [10] D. H. Leung and W.-K. Tang, Functions of Baire class one, Fund. Math. 179 (2003), 225-247. (Zulijanto Atok) Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, 1 Nanyang Walk, Singapore 637616 ; and Department of mathematics, Gadjah Mada University, Sekip Utara, Yogyakarta, Indonesia 55281. E-mail address: atokzulijanto@yahoo.com (Wee-Kee Tang) Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, 1 Nanyang Walk, Singapore 637616 E-mail address: weekee.tang@nie.edu.sg (Dongsheng Zhao) Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, 1 Nanyang Walk, Singapore 637616 E-mail address: dongsheng.zhao@nie.edu.sg