Folia Mathematica Acta Universitatis Lodziensis Vol. 11, No. 1, pp. 53–57 c 2004 for University ofL´od´zPress SOME PROPERTIES OF THE SET OF DISCONTINUITY POINTS OF A MONOTONIC FUNCTION OF SEVERAL VARIABLES WANDA LINDNER‡ Abstract. In the note [1] the neccesary and sufficient condition for a set E ⊂ R2 to be the set of discontinuity points for some nondecreasing function f : R2 → R has been proved. The problem of the similar characterization for the function of more than two variables is still open. This paper provides the neccesary condition for the set E ⊂ Rn to be the set of discontinuity points of some nondecreasing function. 1. Introduction n Let x = (x1,x2,...,xn), y = (y1,y2,...,yn) ∈ R . Definition 1. We say, that x ≺ y iff (xi <yi) for every i ∈ {1, 2,...,n}. Definition 2. We say, that x y iff (xi ≤ yi) for every i ∈ {1, 2,...,n}. If one of the following relations occurs: x y or y x we say, that this points are comparable. Definition 3. The function f : Rn → R is called nondecreasing iff for every x, y ∈ Rn x y ⇒ f(x) ≤ f(y). Definition 4. The function f : Rn → R is called increasing iff for every x, y ∈ Rn (x 6= y) ∧ (x y) ⇒ f(x) <f(y). Proposition 1. The function f : Rn → R is nondecreasing (increasing) iff it is nondecreasing (increasing) as the function of one variable, with the others fixed. ‡Technical University ofL´od´z, Institute of Mathematics, Al. Politechniki 11, 90–924 L´od´z. Poland.. E-mail: [email protected]. Key words and phrases: monotonic function, discontinuity points. AMS subject classifications: 26A48. 54 WANDA LINDNER We will consider the space Rn equipped with the metric d(x, y) = max{|xi − yi| : i = 1, 2,...,n}. 2. A dislocation of the set of discontinuity points For every function g : Rk → R we will denote by g(i) the function defined as follows: (i) g (u1, u2,...,ui−1, ui, ui+1,...,uk)= g(u1, u2,...,ui−1, −ui, ui+1,...,uk). Definition 5. The hypersurface n−1 Γ= ϕ(u) = (g1(u), g2(u),...,gn(u)) : u ∈ R is called semidecreasing iff all the functions (1) (2) (n−1) g1(u), g2 (u), g3 (u),..., gn (u) are continuous and nondecreasing. Definition 6. Let f : Rn → R be a nondecreasing function. Let t ∈ R. The set Pt(f)= {x : x ≺ y ⇒ f(y) ≥ t ∧ y ≺ x ⇒ f(y) <t} is called the level line of the function f on the level t. Lemma 1. Let f : Rn → R be a nondecreasing function. If x ≺ y then for every number t the conjunction x ∈ Pt(f) ∧ y ∈ Pt(f) is false. Proof. In the oposite case we have that x + y x + y x ≺ , so f( ) ≥ t 2 2 and x + y x + y ≺ y so f( ) < t, 2 2 the contradiction. Lemma 2. Each nonempty level line of the nondecreasing function f : Rn → R is a semidecreasing hypersurface. n−1 Proof. Let x = (x1,x2,...,xn) ∈ Pt(f). Fix u = (u1, u2,...,un−1) ∈ R . Let a = min{x1,x2 + u1,x3 + u2,...,xn + un−1} − 1, and b = max{x1,x2 + u1,x3 + u2,...,xn + un−1} + 1. Then f(a, a − u1,...,a − un−1) <t ≤ f(b, b − u1,...,b − un−1). Hence the number x(u) = inf{x : t ≤ f(x,x − u1,...,x − un−1)} is finite. Let us define: g1(u)= x(u), g2(u)= x(u) − u1, ..., gn(u)= x(u) − un−1. SOME PROPERTIES OF THE SET OF DICONTINUITY POINTS... 55 This is easy to see, that for every u ∈ Rn−1 the point (g1(u), g2(u),...,gn(u)) ∈ Pt(f). (1) (n−1) Now we show, that g1, g2 ,...,gn are nondecreasing functions of (n− 1) variables. Let u v. Then for every x ∈ R (x,x − u1,...,x − un−1) (x,x − v1,...,x − vn−1), so f(x,x − u1,...,x − un−1) ≥ f(x,x − v1,...,x − vn−1), and, consequently {x : t ≤ f(x,x − u1,...,x − un−1)} ⊃ {x : t ≤ f(x,x − v1,...,x − vn−1)} hence inf{x : t ≤ f(x,x−u1,...,x−un−1)}≤ inf{x : t ≤ f(x,x−v1,...,x−vn−1)}. As a result g1(u) ≤ g1(v). (i−1) Let us fix i ∈ {2, 3,...,n}. We will prove that the function gi is nondecreasing. By virtue of proposition 1 it suffices to check, that the function gi is nonincreasing with respect to the (i − 1)-th variable, and is nondecreasing with respect to others. At first we show, that gi(u1, u2,...,un−1) = g1(u1, u2,...,un−1) − ui−1 is nonincreasig with respect to the (i − 1)-th variable. In order to do this ∗ consider u = (u1, u2,...,un) and u = (u1, u2,...,ui−1 + h,...,un), where h> 0. We have ∗ g1(u ) = inf{x : t ≤ f(x,x − u1,...,x − ui−1 − h,...,x − un−1)} = = inf{x + h : t ≤ f(x + h, x − u1 + h,...,x − ui−1,...,x − un−1 + h)} = = inf{x : t ≤ f(x + h, x − u1 + h,...,x − ui−1,...,x − un−1 + h)} + h ≤ ≤ inf{x : t ≤ f(x,x − u1,...,x − ui−1,...,x − un−1)} + h = g1(u)+ h. So ∗ g1(u ) − h ≤ g1(u), ∗ g1(u ) − ui−1 − h ≤ g1(u), ∗ gi(u ) ≤ gi(u). The last inequality shows, that gi is nonincreasing with respect to the (i − 1)-th variable. Of course gi(u)= x(u) − ui−1 is nondecreasing with respect to any other variable, as g1 is. 56 WANDA LINDNER Now we check, that the function g1 is continuous. In order to do this take u, v ∈ Rn−1. Let ∗ u = (min(u1, v1), min(u2, v2),..., min(un, vn)) and ∗∗ u = (max(u1, v1), max(u2, v2),..., max(un, vn)). Note, that d[u∗, u∗∗]= d[u, v]. According to the previous part of the proof, by induction we have, that n−1 ∗ ∗∗ ∗ ∗∗ ∗ g1(u ) ≤ g1(u ) ≤ g1(u )+ |ui − ui |, Xi=1 hence n−1 ∗ ∗∗ ∗∗ ∗ ∗ ∗∗ |g1(u ) − g1(u )|≤ |ui − ui |≤ (n − 1) ∗ d[u , u ]. Xi=1 and, further |g1(u) − g1(v)|≤ (n − 1) ∗ d[u, v]. From this we obtain, that the function g1 satisfies the Lipschitz condi- tion, and, as a result, is continuous. Of course it implies the continuity of functions g2, g3,..., gn. Theorem 1. The set of discontinuity points of a given nondecreasing func- tion f : Rn → R is contained in a countable union of semidecreasing hy- persurfaces. Proof. The set E of discontinuity points of a given nondecreasing function n f : R → R can be expressed as E = k∈N Ek, where S 1 Ek = x : inf f(y) − sup f(y) ≥ . x≺y y≺x k Let W be a given countable and dense subset of R. We show, that Ek ⊂ Pj(f). j[∈W Let x ∈ Ek. There exists such a number j ∈ W , that sup f(y) < j ≤ inf f(y), y≺x x≺y so x ∈ Pj(f). Lemma 3. If x, y ∈ Pt(f) and x y, then the segment xy ⊂ Pt(f). The proof is obvious and hence ommited. SOME PROPERTIES OF THE SET OF DICONTINUITY POINTS... 57 n Theorem 2. Let f : R → R be a nondecreasing function. If x, y ∈ Pt(f) are comparable, and the function f is continuous in those points, then f is continuous in every point of the segment xy. Proof. Suppose, contrary to our claim, that there exists such a point z ∈ xy that the function f is not continuous in z. Hence there exists such ǫ > 0 that for any δ > 0 there exists a point z∗ such that d(z, z∗) < δ and |f(z) − f(z∗)| > ǫ. Since f is continuous in x, there exists such δx > 0 that the condition ∗ ∗ ǫ d(x, x ) < δx implies |t − f(x )| < 2 . Analogously, there exists such δy > 0 ∗ ∗ ǫ that the condition d(y, y ) < δy implies |t − f(y )| < 2 . 1 ∗ Assume, that x y. Let δ = 2 min(δx, δy), and let d(z, z ) < δ and |f(z) − f(z∗)| > ǫ. Then ∗ (x1 − δ, x2 − δ,...,xn − δ) ≺ z ≺ (y1 + δ, y2 + δ,...,yn + δ), but ∗ ∗ f(z ) <f(x1 − δ, x2 − δ,...,xn − δ) or f(z ) >f(y1 + δ, y2 + δ,...,yn + δ). This contradicts the monotonicity of the function f. References [1] W. Lindner, A characterization of the set of discontinuity points of a two variables monotone function, Scientific Bulletin ofL´od´zTechnical University 466 (1986). [2] S. Mazurkiewicz, Podstawy rachunku prawdopodobie´nstwa, Monografie Matematy- czne 32 (1956), PWN, Warszawa..