# On the PropertyN-1

Authors: Stanisław Kowalczyk; Małgorzata Turowska
Journal: Abstract and Applied Analysis
 (2016)

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## Abstract

We construct a continuous functionf:[0,1]→R such that f possesses N-1-property, but f does not have approximate derivative on a set of full Lebesgue measure. This shows that Banach’s Theorem concerning differentiability of continuous functions with Lusin’s property (N) does not hold for N-1-property. Some relevant properties are presented.


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## Body

## 1. Introduction

First we will specify some basic notations. ByE we denote the Lebesgue measure of E⊂R. For any f:I→R, where I is an interval, by f↾E we denote the restriction of f to E⊂I and the symbol fap′(x) stands for approximate derivative of f at x.Definition 1 (see [1]).
LetD⊂R be measurable. We say that f:D→R has Lusin’s property (N), if the image f(E) of every set E⊂D of Lebesgue measure 0 has Lebesgue measure 0.This condition was studied exhaustively; some of results can be found in [1]. For the present paper the most important is the following.Theorem 2 (Third Banach Theorem, [1] Theorem 7.3).
Iff:[0,1]→R is continuous and has Lusin’s property (N), then f is differentiable on a set of positive Lebesgue measure.In the present paper we will study a similar property.Definition 3 (see [2, 3]).
We say thatf:D→R, defined on a measurable set D⊂R, has N-1-property, if the inverse image f-1(E) of every set E⊂R of Lebesgue measure 0 has Lebesgue measure 0.Some of results concerningN-1-property are presented in [2, 3]. In [2] a systematic study of N-1-property for smooth and almost everywhere differentiable functions can be found. Some applications of N-1-property in functional equation and geometric function theory can be found in [4–6].


## 2. Main Results

Our goal is to construct a continuous functionf:[0,1]→[0,1] with N-1-property which is not approximately differentiable on a set of full measure. We start with the basic theorem.Theorem 4.
LetB1={(2k-1)/2n:k∈{1,2,…,2n-1},n∈N}, B2={2k-1/2n+1/(3·2n):k∈{1,2,…,2n-1},n∈N}, and A=(0,1)∖(B1∪B2). There exists a homeomorphism f:A→A such that (a1)
f=f-1,(a2)
f has Lusin’s property (N) and N-1-property,(a3)
f has no approximate derivative (finite or not) at any x∈A.Proof.
Letx=0.i1i2⋯in⋯¯ denote a binary decomposition of x∈(0,1). It is easily seen that (2k-1)/2n+1/(3·2n)=0.i1i2⋯in-1101010⋯¯. Therefore, x∈(0,1) and x=0.i1i2⋯in⋯¯ belongs to A if and only if it has a binary decomposition x=0.i1i2⋯in⋯¯ such that(1)n∈N:i2n=0¯¯=ℵ0=n∈N:i2n=1¯¯orn∈N:i2n-1=0¯¯=ℵ0=n∈N:i2n-1=1¯¯(in other words x has infinitely many 0s and infinitely many 1s at even places or infinitely many 0s and infinitely many 1s at odd places). Let Δnk=(k/2n,(k+1)/2n) for k∈{0,1,…,2n-1} and n∈N. Obviously, A⊂⋃k=02n-1‍Δnk for every n∈N. Moreover,(2)A∩Δnk=x∈A:x=0.i1i2⋯in⋯¯where∑j=1n2n-jij=k.
Definef:A→A by(3)fx=0.i1i2′i3i4′⋯i2n-1i2n′⋯¯,where x=0.i1i2⋯in⋯¯ and ij′=1-ij. In other words, f(x)=0.m1m2⋯mn⋯¯, where mj=ij for odd j and mj=1-ij for even j. By (1), f(x)∈A for x∈A and f is well-defined. Moreover, directly from the definition of f, it follows that f is a bijection and the composition f∘f is the identity function, whence f-1=f. Moreover, by (2), for each n∈N and k∈{0,1,…,2n-1}, k=∑j=1n2n-jij, ij∈{0,1}, we have(4)fA∩Δnk=A∩Δnk′,where k′=∑j=1n2n-jmj.
We claim thatf is continuous. Fix x0∈A and ɛ>0. Choose n0∈N such that 1/2n0<ɛ. There exists k0≤2n0-1 for which x0∈Δn0k0. By (4), f(A∩Δn0k0)=A∩Δn0k0′. Since A∩Δn0k0 is a neighborhood of x0 and |y1-y2|≤1/2n0<ɛ for all y1,y2∈Δn0k0′, we conclude that f is continuous at x0. Thus, f is continuous, because x0 was arbitrary. By the equality f=f-1, f is a homeomorphism.
Now we will show condition (a2). LetH⊂A be any set of Lebesgue measure zero. Fix any ɛ>0. There exists an open in A set U⊂A such that H⊂U and |U|<ɛ. Let (5)B=Δnk∩A:k∈0,1,…,2n-1,n∈N.Clearly, B is a base of the natural topology in A. Since either any two sets from B are disjoint or one of them is contained in the other, it is easy to see that any open subset of A can be represented as a union of some subfamily of pairwise disjoint sets from B. Thus, U=⋃j∈J‍(Δnjkj∩A), where J is at most countable and Δnj1kj1∩Δnj2kj1=∅ for j1,j2∈J, j1≠j2. Then, by (4), ⋃j∈J‍f(Δnjkj∩A)=⋃j∈J‍(Δnjkj′∩A) is an open in A set containing f(H) and (6)∑j∈JΔnjkj′∩A=∑j∈JΔnjkj∩A=U<ɛ.Since ɛ>0 was arbitrary, |f(H)|=0 and f has Lusin’s property (N). Since f=f-1, f has also N-1-property.
Finally, we will show thatf has no approximate derivative at any x∈A. Fix x∈A and an even n∈N. Then x∈Δnk for some k≤2n-1. Let i1,i2,…,in∈{0,1} be such that k=∑j=1n‍2n-jij. Moreover, let k′ be understood as before. It is clear that (7)A∩Δnk=A∩Δn+24k∪A∩Δn+24k+1∪A∩Δn+24k+2∪A∩Δn+24k+3,fA∩Δn+24k=A∩Δn+24k′+1,fA∩Δn+24k+1=A∩Δn+24k′,fA∩Δn+24k+2=A∩Δn+24k′+3,fA∩Δn+24k+3=A∩Δn+24k′+2(remember that n is even).
Note that(8)fy-fxy-x<0if x∈Δn+24k and y∈A∩Δn+24k+1 or x∈Δn+24k+2 and y∈A∩Δn+24k+3. Moreover, (9)fy-fxy-x>2n3·2n=13if x∈Δn+24k and y∈A∩Δn+24k+2 or x∈Δn+24k+1 and y∈A∩Δn+24k+3.
Thus, ifn∈N is even and x∈A∩Δnk, we can find B,C⊂Δnk∩A such that(10)B=C=14Δnk,(11)fx-fyx-y<0∀y∈B,(12)fx-fyx-y>13∀y∈C.Since this is true for every even n and |Δnk|=1/2n, we conclude that f has no approximate derivative (finite or not) at x. The proof is completed.From Banach’s Theorem2, we easily get the following.Corollary 5.
Any functionf, defined on an interval, which possesses Lusin’s condition (N) such that the set of discontinuity points of f is finite, is derivable at every point of some set of positive Lebesgue measure.Meanwhile, by Theorem4, we have the following.Theorem 6.
There exists a bijectiong:[0,1]→[0,1] such that (b1)
g has Lusin’s property (N) and N-1-property,(b2)
the set of discontinuity points ofg is countable,(b3)
g has no approximate derivative at any point.Proof.
LetB1, B2, A, and f be the same as in Theorem 4. It is easily seen that every member of B2 is of the form 0.i1⋯in001010⋯¯ or 0.i1⋯in1101010⋯¯ for some n∈N, except 1/6, 1/3, 2/3, 5/6. Define φ:{0,1}∪B1∪B2→{0,1}∪B1∪B2 by(13)
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,where j2m-1=i2m-1 and j2m=1-i2m. It is easy to see that φ is a bijection. Let g:[0,1]→[0,1] be defined by (14)gx=fxforx∈A,φxforx∈0,1∪B1∪B2.Fix x∈(0,1)∖(B1∪B2), x=0.i1i2⋯in⋯¯, and ɛ>0. Let m be a positive integer such that 1/2m<ɛ. The set (15)C=0,1,16,13,23,56∪2k-12n:k∈1,2,…,2n-1,n≤m∪0.l1⋯ln001010⋯¯:n≤m∪0.l1⋯ln1101010⋯¯:n≤mis finite and C⊂{0,1}∪B1∪B2. Hence, we can find δ∈(0,1/2m+3) for which (x-δ,x+δ)∩C=∅. Take any y∈(x-δ,x+δ)∩(B1∪B2). Since |x-y|<1/2m+3 and y∈B1∪B2, we conclude y=0.i1i2⋯im⋯001010⋯¯ or y=0.i1i2⋯im⋯110101⋯¯ or y=0.i1i2⋯im⋯1¯. Hence, g(x)=f(x)=0.j1j2⋯jm⋯¯ and g(y)=φ(y)=0.j1j2⋯jm⋯¯ or g(y)=φ(y)=0.j1j2⋯jm⋯1¯. Therefore, |g(x)-g(y)|<1/2m<ɛ. Since f is continuous, g is continuous at x. Thus, we have proved that the set of all discontinuity points of g is contained in {0,1}∪B1∪B2. Therefore, g satisfies (b1), (b2), and (b3).Theorem 7.
For eachɛ∈(0,1) there exist a closed nowhere dense set F⊂(0,1) and a homeomorphism h:F→F such that (c1)
|F|>1-ɛ,(c2)
h=h-1,(c3)
h has Lusin’s property (N) and N-1-property,(c4)
h has no approximate derivative (finite or not) at any x∈F (more precisely, if h~:[0,1]→[0,1] is any extension of h then h~ has no approximate derivative (finite or not) at any x∈F).Proof.
LetB1, B2, A, and f be the same as in Theorem 4. Let {xn}n=1∞=B1∪B2. Fix ɛ>0 and choose a sequence (mn)n≥0 of even natural numbers satisfying(16)4∑n=0∞12mn<ɛ,(17)4∑j=n+1∞12mj<18·12mn∀n≥0.For each n≥1 there exists kn∈{1,…,2mn-1} such that xn∈(kn-1)/2mn,(kn+1)/2mn. Let (18)B=Δm00∪0∪Δm02m0-1∪1∪⋃n=1∞Δmnkn-1∪kn2mn∪Δmnkn.Since(19)Δmnkn-1∪kn2mn∪Δmnkn=kn-12mn,kn+12mn,B is an open subset of [0,1]. Moreover, B1∪B2∪{0,1}⊂B and, by (16),(20)B≤22m0+∑n=1∞22mn<ɛ2.
By (4), in the proof of Theorem 4, for each n≥1 there exist un,vn∈{1,…,2mn} such that (21)fΔmnkn-1∩A=Δmnun∩A,fΔmnkn∩A=Δmnvn∩A.Moreover,(22)fΔm00∩A=Δm0i0∩A,fΔm02m0-1∩A=Δm0i1∩Afor some i0,i1∈{0,1,…,2m0}. Hence, (23)C=fB∩A=fΔm00∩A∪Δm02m0-1∩A∪⋃n=1∞Δmnkn-1∩A∪Δmnkn∩A=Δm0i0∩A∪Δm0i1∩A∪⋃n=1∞Δmnun∩A∪Δmnvn∩A=Δm0i0∪Δm0i1∪⋃n=1∞Δmnun∪Δmnvn∩A.Again, applying (16), we have |C|=|B|<ɛ/2. Moreover, since 0,1∖A⊂IntB, the set (24)B∪C=0,1∖A∪B∪Δm0i0∪Δm0i1∪⋃n=1∞Δmnun∪Δmnvnis open in [0,1].
Finally, putH=[0,1]∖(B∪C). It is clear that H⊂A, H is a closed subset of [0,1], and |H|>1-2(ɛ/2)=1-ɛ. Since f is a bijection and f=f-1, we have (25)fB∪C∩A=fB∩A∪fC∩A=C∪B∩A=B∪C∩A.It follows that f(H)=H and h=f↾H is a homeomorphism.
Fixx0∈H and n∈N. There exists k∈{0,1,…,2n-1} such that x0∈Δmnk. Certainly, Δmnk⊄B∪C. Therefore, by (17), |(B∪C)∩Δmnk|<(1/8)|Δmnk|. By (10), (11), and (12), (26)x∈Δmnk:fx-fx0x-x0<0>14·18Δmnk,x∈Δmnk:fx-fx0x-x0>13>14·18Δmnk.Therefore, any extension h~:[0,1]→[0,1] of h has no approximate derivative at x0.Lemma 8.
Leta,b,c,d∈R, a<b, and c<d. For every ɛ∈(0,1) there exist a closed nowhere dense set H⊂(a,b) and a continuous injection g:H→[c,d] such that (d1)
|H|>(1/2)(b-a),(d2)
g-1:g(H)→H is continuous,(d3)
g has Lusin’s property (N) and N-1-property,(d4)
ifg~:[a,b]→[c,d] is any extension of g, then g~ has no approximate derivative (finite or not) at any x∈H,(d5)
|g(min⁡H)-c|<ɛ, |d-g(max⁡H)|<ɛ, and |g(bn)-g(an)|<ɛ for all n∈N, where {(an,bn):n∈N} is the set of all connected components of (a,b)∖H.Proof.
Fixɛ>0 and choose n∈N such that 1/(n+1)<ɛ/2(d-c). Let a=y0<x1<y1<x2<⋯<xn<yn<xn+1=b be a partition of [a,b] such that yi-xi=(1/(n+1))(b-a) for i∈{1,…,n} and xj-yj-1=(1/(n+1)2)(b-a) for j∈{1,…,n,n+1}. Let ψ:[a,b]→[c,d] be a linear homeomorphism, ψ(x)=((d-c)/(b-a))(x-a)+c. By Theorem 7, there exist a closed nowhere dense set F⊂(0,1) and a homeomorphism h:F→F satisfying conditions (c2)–(c4) such that |F|>(n+1)/2n. For each k∈{1,…,n} define linear homeomorphisms ψk:[xk,yk]→[0,1], (27)ψkx=1yk-xkx-xk,and ϕk:[0,1]→[ψ(xk),ψ(yk)],(28)ϕkx=ψyk-ψxkx+ψxk.Moreover, let Fk=ψk-1(F) for k≤n. Obviously, each Fk is a closed nowhere dense subset of (xk,yk). Besides, |Fk|=|F|·(yk-xk)=|F|·(b-a)/(n+1). For each k∈{1,…,n} define hk:Fk→[ψ(xk),ψ(yk)] by hk=ϕk∘h∘ψk. It is easy to see that each hk is a continuous injection, hk has Lusin’s property (N) and N-1-property, and, moreover, any extension of hk to [xk,yk] is not approximately differentiable at any point x∈Fk. Finally, let H=⋃k=1n‍Fk and define g:H→[c,d] by g(x)=hk(x) for x∈Fk, k∈{1,…,n}.
It is clear thatH and g satisfy conditions (d1)–(d4). Let (α,β) be any connected component of (a,b)∖H. If (α,β)⊂[xk,yk] for some k∈{1,…,n} then (29)gβ-gα≤ψyk-ψxk=b-an+1·d-cb-a=d-cn+1<ɛ.If (α,β)⊃[yk-1,xk] for some k∈{2,…,n} then (30)gβ-gα≤ψyk-ψxk-1=2d-cn+1<ɛ.Similarly, (31)gmin⁡F-c≤ψy1-c=d-cn+1+d-cn+12=n+2n+12d-c<2d-cn+1<ɛ.Analogously, |d-g(max⁡F)|<ɛ. This completes the proof.Now, we can prove the main theorem of the present paper.Theorem 9.
There exists a continuous functionf:[0,1]→[0,1] such that f has N-1-property, but fap′ exists almost nowhere.Proof.
We will construct inductively a sequence(Fn)n∈N of closed subsets of [0,1] and a sequence (fn)n∈N of continuous functions fn:[0,1]→[0,1] such that (1)
Fn⊂Fn+1 and |Fn|>1-1/2n for all n≥1,(2)
fn↾Fk=fk↾Fk for all n>k,(3)
|fn(x)-fn+1(x)|<1/2n for n∈{1,2,…} and x∈[0,1],(4)
everyfn restricted to Fn has N-1-property,(5)
every extension offn↾Fn has no approximate derivative at any x∈Fn.First, we give a useful definition. If E⊂(0,1) is closed and φ:E→(0,1), then by the linear extension of f we mean ψ:[0,1]→[0,1] such that ψ↾E=φ, ψ(0)=0, ψ(1)=1, and ψ is linear on every closed interval contiguous to E∪{0,1}. It is clear that ψ is continuous if and only if φ is continuous.
By Theorem7, there exist a closed set F⊂(0,1), |F|>1/2, and a bijection g1:F→F satisfying conditions (c1)–(c4). Let F1=F and f1:[0,1]→[0,1] be the linear extension of g1. Then f1 is continuous, f1 has N-1-property, and every extension of f1↾F1=g1 has no approximate derivative at any x∈F1.
Let((ak1,bk1))k≥1 be the family of all connected components of [0,1]∖(F1∪{0,1}). Moreover, for every k∈N, let Jk1 be an open interval with endpoints f1(ak1) and f1(bk1). By Lemma 8, for each k∈N there exist closed Fk1⊂(ak1,bk1) and gk1:Fk1→Jk1 satisfying conditions (d1)–(d5) with ɛ=1/2. Let F2=F1∪⋃k=1∞‍Fk1 and let g2:F2→F1∪⋃k=1∞‍Jk1 be defined by g2(x)=f1(x) for x∈F1 and g2(x)=gk1(x) for x∈Fk1, k∈{1,2,…}. We claim that g2 is continuous. The continuity of g2 at each point of ⋃k=1∞‍Fk1 is obvious. Fix x0∈F1 and ɛ>0. If x0 is not isolated from the right in F2, then there exist δ>0 such that |g2(x)-g2(x0)|=|g1(x)-g1(x0)|<ɛ for x∈F1∩(x0,x0+δ] and F2∩(x0,x0+δ)=(F1∩(x0,x0+δ))∪⋃k∈K‍Fk1 for some K⊂N. Since (32)g2x-g2x0<max⁡g2ak1-g2x0,g2bk1-g2x0for x∈Jk1, we have |g2(x)-g2(x0)|<ɛ for x∈F2∩(x0,x0+δ). Hence, g2 is continuous from the right at x0. Similarly, we can show that g2 is continuous from the left at x0. Since x0 was arbitrary, g2 is continuous.
Letf2 be the linear extension of g2. It is clear that F1⊂F2, |F2|>1-1/4, f2↾F1=f1↾F1, f2 restricted to F2 has N-1-property, and every extension of f2↾F2=g2 has no approximate derivative at any x∈F2. Moreover, |f2(x)-f1(x)|<1/2 for x∈[0,1].
Assume that closed setsF1,…,Fn⊂(0,1), F1⊂⋯⊂Fn, and continuous functions fr:[0,1]→[0,1], r∈{1,…,n}, are chosen. Moreover, assume that for every r∈{2,…,n} we have |Fr|>1-1/2r, fr restricted to Fr has N-1-property, every extension of fr↾Fr has no approximate derivative at any x∈Fr, |fr(x)-fr-1(x)|<1/2r-1 for each x∈[0,1], and fr↾Fs=fs↾Fs for every s∈{1,…,r-1}.
Let((akn,bkn))k≥1 be the family of all connected components of [0,1]∖(Fn∪{0,1}). Moreover, for every k∈N let Jkn be an open interval with endpoints fn(akn) and fn(bkn). By Lemma 8, for each k∈N there exist closed Fkn⊂(akn,bkn) and gkn:Fkn→Jkn satisfying conditions (d1)–(d5) with ɛ=1/2n. Let Fn+1=Fn∪⋃k=1∞‍Fkn and let gn+1:Fn+1→F1∪⋃k=1∞‍Jkn be defined by gn+1(x)=fn(x) for x∈Fn and gn+1(x)=gkn(x) for x∈Fkn, k∈{1,2,…}. Similarly, as in the case of g2, we can check that gn+1 is continuous.
Letfn+1 be the linear extension of gn+1. It is clear that Fn⊂Fn+1, |Fn+1|>1-1/2n+1, fn+1↾Fn=fn↾Fn, fn+1 restricted to Fn+1 has N-1-property, and every extension of fn+1↾Fn+1 has no approximate derivative at any x∈Fn+1. Moreover, |fn+1(x)-fn(x)|<1/2n for x∈[0,1]. Thus, we have proved inductively that there exist a sequence (Fn)n∈N of closed subsets of [0,1] and a sequence (fn)n∈N of continuous functions fn:[0,1]→[0,1] satisfying conditions (1)–(5).
Since|fn+1(x)-fn(x)|<1/2n for x∈[0,1] and n∈N, the sequence (fn)n∈N is uniformly convergent to some continuous function f:[0,1]→[0,1]. Moreover, f↾Fn=fn↾Fn for all n∈N. Therefore, by (5), f has no approximate derivative at any point from ⋃n=1∞‍Fn. Since, by (1), |⋃n=1∞‍Fn|=1, fap′ exists almost nowhere.
It remains to prove thatf has N-1-property. Take any E⊂[0,1] of the Lebesgue measure zero. Then, by (2), (33)f-1E⊂⋃n=1∞Fn∩f-1E∪0,1∖⋃n=1∞Fn=⋃n=1∞Fn∩fn-1E∪0,1∖⋃n=1∞Fn.Applying (1) and (4), we conclude that |f-1(E)|=0. Thus, f has N-1-property.



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