SOOCHOW JOURNAL OF MATHEMATICS
Volume 33, No. 4, pp. 543-551, October 2007
FUNCTIONS WHOSE COMPOSITION WITH BAIRE CLASS ONE FUNCTIONS ARE BAIRE CLASS ONE BY DONGSHENG ZHAO
Abstract. We study the functions whose composition with Baire class one functions are Baire class one functions. We first prove some characterizations of such functions, then investigate a subclass of such functions which are defined in a natural way.
1. Introduction Given a Polish space X, a real valued function f : X → R is called Baire class one if there is a sequence {fn } of continuous functions fn : X → R such that limn→∞ fn (x) = f (x) for every x ∈ X. Various different equivalent definitions of Baire class one functions have been obtained since Baire first defined and studied such functions over a hundred years ago. One of such definition is: f : X → R is Baire class one if and only if, for every open subset U of R, f −1 (U ) is an Fσ set. In [4] a new characterization of Baire class one functions is proved, which is similar to the ǫ-δ-definition of continuous functions. By [4], a function f : R → R is Baire class one if for any positive number ǫ there is a positive valued function δ(·) defined on R such that for any x, y ∈ R, |x − y| < min{δ(x), δ(y)} implies |f (x) − f (y)| < ǫ. This new characterization has been further investigated by several other authors (see [2], [3] and [5]). Received June 9, 2005; revised December 15, 2006; April 16, 2007. AMS Subject Classification. 26A21, 26A15. Key words. Baire class one function, k-continuous function, right B1 compositor. 543
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If f : R → R and g : R → R are continuous functions then f ◦ g is continuous. The composition f ◦g is Baire class one if one of f and g is continuous and another is Baire class one. However the composition of two Baire class one functions is not necessarily Baire class one. Thus it is natural to ask which functions have the property that their composition with any Baire class one function is Baire class one. A function g : R → R is called a right B1 compositor if f ◦ g is Baire class one whenever f is Baire class one. In this paper we shall prove some characterizations of right B1 compositors. More specifically, the main result of this paper is the following: Theorem 1. Let g : R → R be a function. Then the following statements are equivalent: (1) For any closed subset A of R, g−1 (A) is an Fσ set. (2) For any Fσ set A, g−1 (A) is an Fσ set. (3) For every positive Baire class one function ǫ, there is a positive function δ on R such that |x − y| < δ(x) ∧ δ(y) implies |g(x) − g(y)| < ǫ(g(x)) ∧ ǫ(g(y)). (4) g is a right B1 compositor. We also study a class of right B1 compositors which are defined in a very natural way. Although some results proved here are also true for functions on more general spaces, we shall only consider functions defined on the real line. 2. D-Continuous Functions In the following, R will denote the set of all real numbers as well as the ordinary topological space of real numbers. To simplify the statements and expressions below, we shall denote min{a, b} by a ∧ b for any two real numbers a and b. Definition 1. Let D be a class of positive real valued functions defined on R. A function f : R → R is called D-continuous if for any ǫ(·) ∈ D there is a positive real valued function δ(·) such that for any x, y ∈ R, |x − y| < δ(x) ∧ δ(y) implies |f (x) − f (y)| < ǫ(f (x)) ∧ ǫ(f (y)).
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Remark 1. (a) If for each α ∈ D1 there exists β ∈ D2 with β ≤ α, then every D2 -continuous function is D1 -continuous. (b) Let D be a class of positive real valued function. Put D ∗ = {α1 ∧α2 ∧· · ·∧αk : αi ∈ D}. Then a function f is D-continuous iff it is D ∗ -continuous. Example 1. (a) Let D0 be the set of all positive constant functions. Then D0 -continuous functions are exactly the Baire class one functions ([4]). (b) Let D be the set of all positive valued functions α with α(0) > 1, α(1) > 1. Then the Dirichlet function, which is the characteristic function of the set Q of all rational numbers, is D-continuous. Note that the Dirichlet function is not Baire class one. Let f : R → R be the Riemann function defined by 1 p q , if x = q , p and q are co-prime intergers and 0 < q, f (x) = 1, if x = 0, 0, otherwise. Define g : R → R by
g(x) =
(
1,
if x = n1 ,
0,
otherwise.
n ∈ N − {0},
Then both f and g are Baire class one, but their composition g ◦ f is the Dirichlet function. Hence the composition of two Baire class one functions need not be Baire class one. (c) Take D1 to be the set of all positive continuous functions. Then every continuous function is D1 -continuous, and every D1 -continuous function is Baire class one. (d) Define D∞ to be the set of all positive functions. Thus D∞ -continuous functions are D-continuous for every D. A D∞ -continuous function will be called a k-continuous function. We shall prove more properties of such functions in Section 4. We now prove that D1 -continuous functions are the same as Baire class one functions. Note that a positive continuous function ǫ defined on the whole real line may not attain its infimum, so the result of Proposition 1 is not very obvious.
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Lemma 1.([1]) If R = ∪∞ n=1 En with each En an Fσ set, then there are disjoint Fσ sets Fn (n = 1, 2, . . .) such that Fn ⊆ En and R = ∪∞ n=1 Fn . Lemma 2. Let R = ∪∞ n=1 Fn where Fn ’s are disjoint Fσ sets. Then there is a positive function δ(·) on R such that x ∈ Fn , y ∈ Fm and n 6= m imply |x − y| ≥ δ(x) ∧ δ(y). Proof. For any n, we assume that Fn = ∪∞ i=1 Fn,i where each Fn,i is a closed set such that Fn,i ⊆ Fn,i+1 for all i. For each n and x ∈ Fn , let Fn,i(x) be the closed set containing x and Fn,i(x)−1 does not contain x, i.e. i(x) is the minimal index i S for which x ∈ Fn,i . Choose δ(x) > 0 so that k 0, by Corollary 33 of [3], there is a positive Baire class one function δ such that |x − y| < δ(x) ∧ δ(y) implies |f (x) − f (y)| < ǫ. By (3), there is a positive function λ such that |x − y| < λ(x) ∧ λ(y) implies |g(x) − g(y)| < δ(g(x)) ∧ δ(g(y)). Then obviously |x − y| < λ(x) ∧ λ(y) implies |f (g(x)) − f (g(y))| < ǫ. So f ◦ g is Baire class one. (4) implies (1). Suppose that g is a right B1 compositor and A is a closed set. Let h : R → R be the characteristic function of A. Then because A is both Fσ and Gδ , h is Baire class one. Thus the composition h ◦ g is Baire class one. Now g−1 (A) = (h ◦ g)−1 ( 12 , 32 ), which is an Fσ set. At last we prove that (2) implies (3). First g is Baire class one because every open set U of R is an Fσ set. Given any positive Baire class one function ǫ, we have −1 −n R = ∪∞ , +∞), n=1 ǫ (2
where each ǫ−1 (2−n , +∞) is an Fσ set. For each n, let En = g−1 (ǫ−1 (2−n , +∞)). By the assumption on g, En is an Fσ set for every n. We now just have to apply Lemma 3 to conclude the proof.
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The class of right B1 compositors contains all the continuous functions and is closed under composition. If g is a right B1 compositor, then for each constant c, c+ g and cg are right B1 compositors as well because they are the compositions of g and the function h(x) = c + x and k(x) = cx, respectively. However it is still not clear whether the addition (product) of a continuous function and a right B1 compositor is a right B1 compositor. 4. k-Continuous Functions A function f is called k-continuous if for every positive function ǫ there is a positive function δ such that for any x, y ∈ R, |x − y| < δ(x) ∧ δ(y) implies |f (x) − f (y)| < ǫ(f (x)) ∧ ǫ(f (y)). By Theorem 1, every k-continuous function is a right B1 compositor. We still do not have a counterexample of a right B1 compositor which is not k-continuous. In this section we shall prove some properties of k-continuous functions. Proposition 2. (1) The composition of two k-continuous functions is k-continuous. (2) If f and g are injective k-continuous functions, then so is f + g. (3) If f and g are injective k-continuous functions, then so is the product function f g. Proof. (1) Follows directly from the definition of k-continuous functions. To prove (2), given any positive function ǫ, define two positive functions ǫ1 and ǫ2 on R such that ǫ1 (f (x)) = 12 ǫ(f (x) + g(x)), ǫ2 (g(x)) = 12 ǫ(f (x) + g(x)), and ǫ1 (y) = 1, ǫ2 (y) = 1 for other y. Since f and g are k-continuous, there are positive functions δ1 and δ2 such that |x − y| < δ1 (x) ∧ δ1 (y) implies |f (x) − f (y)| < ǫ1 (f (x)) ∧ ǫ1 (f (y)), and |x − y| < δ2 (x) ∧ δ2 (y) implies |g(x) − g(y)| < ǫ2 (g(x)) ∧ ǫ2 (g(y)). Let δ(x) = δ1 (x) ∧ δ2 (x), x ∈ R. Then |x − y| < δ(x) ∧ δ(y) implies |f (x) + g(x) − (f (y) + g(y))| ≤ |f (x) − f (y)| + |g(x) − g(y)| < min{ǫ1 (f (x)), ǫ1 (f (y))}+min{ǫ2 (g(x)), ǫ2 (g(y))} = 12 min{ǫ(f (x)+g(x)), ǫ(f (y)+ g(y))} + 12 min{ǫ(f (x) + g(x)), ǫ(f (y) + g(y))} = ǫ(f (x) + g(x)) ∧ ǫ(f (y) + g(y)). Thus f + g is k-continuous. S To prove (3), let ǫ be a given positive function. Then R = ∞ i=1 Ei , where Ei = {x : |f (x)| ≤ i, |g(x)| ≤ i}(i ∈ N). Since f and g are right B1 compositors,
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Ei is an Fσ set. By Lemma 1 there are disjoint Fσ sets Fi ⊆ Ei (i ∈ N) with R = S∞ i=1 Fi . By Lemma 2 there is a positive function δ1 so that |x− y| ≥ δ1 (x)∧ δ1 (y) for any x ∈ Fi , y ∈ Fj with i 6= j. For any n ∈ N, define two new positive 1 1 functions ǫnf and ǫng such that ǫnf (f (x)) = 2n ǫ(f (x)g(x)), ǫng (g(x)) = 2n ǫ(f (x)g(x)) n n and ǫf (y) = 1, ǫg (y) = 1 for other y. Since f and g are k-continuous, there exist positive functions λn and µn such that |x − y| < λn (x) ∧ λn (y) implies |f (x) − f (y)| < ǫnf (f (x)) ∧ ǫnf (f (y)), |x − y| < µn (x) ∧ µn (y) implies |g(x) − g(y)| < ǫng (g(x)) ∧ ǫng (g(y)). To define the function δ, for each x there is a unique n with x ∈ Fn , then let δ(x) = δ1 (x) ∧ λn (x) ∧ µn (x). Now if |x − y| < δ(x) ∧ δ(y), then by the definition of δ1 there is n such that x, y ∈ Fn . So |f (x)| ≤ n and |g(y)| ≤ n. In addition |f (x)g(x) − f (y)g(y)| ≤ |f (x)||g(x) − g(y)| + |g(y)||f (x) − f (y)| ≤ n(|g(x) − g(y)| + |f (x) − f (y)|) < n(ǫng (g(x)) ∧ ǫng (g(y)) + ǫnf (f (x)) ∧ ǫnf (f (y))) = ǫ(f (x)g(x)) ∧ ǫ(f (y)g(y)). All these show that f g is k-continuous. Corollary 1. (1) If f : R → R is k-continuous, then for every constant c, c + f and cf are k-continuous functions. (2) If f : R → R is k-continuous, then so is |f |. The class of continuous functions and the class of Baire class one functions are closed under uniform convergence. The following example shows that the class of k-continuous functions is not closed under uniform convergence. Example 2. Let Q = {rn : n ∈ N} be the set of all rational numbers. For each n, let gn be the function defined by ( 1 , if x = rm , m ≤ n, gn (x) = m+1 0, otherwise. Then each gn is k-continuous and {gn } converges uniformly to the function h : R → R, where ( 1 , if x = rn , h(x) = n+1 0, otherwise.
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In the same way as we prove that the Riemann function is not a right B1 compositor, we can show that h is not a right B1 compositor. Hence h is not k-continuous. Since each gn in the above example is also a right B1 compositor, we can also conclude that the class of right B1 compositors is not closed under uniform convergence.
(1)
(2) (3) (4)
Remark 2. The following conclusions can be easily verified. If the range f (R) is a finite set, then f is k-continuous iff it is Baire class one. In particular, the characteristic function χA of a set A is k-continuous iff A is both Fσ and Gδ . If f and g have different values at only a finite number of points, then f is k-continuous iff g is k-continuous. If the set of discontinuity points of f is a discrete set, then f is k-continuous. S Suppose that R = i Fi with each Fi an Fσ set, and that f is k-continuous on each Fi . Then f is k-continuous on R.
It is also natural to consider the left B1 compositors. Of course all continuous functions are such functions. As a matter of fact, if f : R → R has only a finite number of discontinuity points, then f is a left B1 compositor. It would be nice if one can find a necessary and sufficient condition for such functions as well.
Acknowledgments I must express my sincere thanks to the referees for their valuable suggestions that helped to improve the paper. I also wish to thank Professor Yixiang Chen of Shanghai Teachers University for the useful conversation with him during my visit to his university in the summer of 2004. The notion of k-continuous function was inspired by the earlier collaboration on the new characterization of Baire class one functions with my colleagues Peng Yee Lee and Wee Kee Tang, I also would like to thank them for their ideas. References [1] C. S. Ding, P. Y. Lee and P. Bullen, Introduction to Real Analysis, China Scientific Press, 2001.
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[2] J. Jachymski, M. Lindner and S. Lindner, On Cauchy type characterizations of continuity and Baire one functions, Real Anal. Exchange, 30:1(2004/05), 339-346. [3] D. Lecomte, How can we recover Baire class one functions? Mathematika, 50:1-2(2003), 171-198. [4] P. Y. Lee, W. K. Tang and D. Zhao, An equivalent definition of functions of the first Baire class, Proc. Amer. Math. Soc., 129:8(2001), 2273-2275. [5] D. N. Sarkhel, Baire one functions, Bull. Inst. Math. Acad. Sinica, 31:2(2003), 143-149. Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, Singapore. E-mail:
[email protected] 
