AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 10.1155/2016/1256906 1256906 Research Article On the Property N-1 Kowalczyk Stanisław 1 http://orcid.org/0000-0001-9802-7874 Turowska Małgorzata 1 Minhós Feliz Institute of Mathematics Pomeranian University in Słupsk Ulica Arciszewskiego 22d 76-200 Słupsk Poland apsl.edu.pl 2016 332016 2016 02 11 2015 07 02 2016 2016 Copyright © 2016 Stanisław Kowalczyk and Małgorzata Turowska. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We construct a continuous function f:[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.

1. Introduction

First we will specify some basic notations. By E we denote the Lebesgue measure of ER. For any f:IR, where I is an interval, by fE we denote the restriction of f to EI and the symbol fap(x) stands for approximate derivative of f at x.

Definition 1 (see [<xref ref-type="bibr" rid="B5">1</xref>]).

Let DR be measurable. We say that f:DR has Lusin’s property (N), if the image f(E) of every set ED of Lebesgue measure 0 has Lebesgue measure 0.

This condition was studied exhaustively; some of results can be found in . For the present paper the most important is the following.

Theorem 2 (Third Banach Theorem, [<xref ref-type="bibr" rid="B5">1</xref>] Theorem <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M25"><mml:mrow><mml:mn>7.3</mml:mn></mml:mrow></mml:math></inline-formula>).

If f:[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 [<xref ref-type="bibr" rid="B3">2</xref>, <xref ref-type="bibr" rid="B4">3</xref>]).

We say that f:DR, defined on a measurable set DR, has N-1-property, if the inverse image f-1(E) of every set ER of Lebesgue measure 0 has Lebesgue measure 0.

Some of results concerning N-1-property are presented in [2, 3]. In  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 .

2. Main Results

Our goal is to construct a continuous function f:[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.

Let B1={(2k-1)/2n:k{1,2,,2n-1},nN}, B2={2k-1/2n+1/(3·2n):k{1,2,,2n-1},nN}, and A=(0,1)(B1B2). There exists a homeomorphism f:AA such that

f=f-1,

f has Lusin’s property (N) and N-1-property,

f has no approximate derivative (finite or not) at any xA.

Proof.

Let x=0.i1i2in¯ denote a binary decomposition of x(0,1). It is easily seen that (2k-1)/2n+1/(3·2n)=0.i1i2in-1101010¯. Therefore, x(0,1) and x=0.i1i2in¯ belongs to A if and only if it has a binary decomposition x=0.i1i2in¯ such that(1)nN:i2n=0¯¯=0=nN:i2n=1¯¯  ornN:i2n-1=0¯¯=0=nN: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 nN. Obviously, Ak=02n-1Δnk for every nN. Moreover,(2)AΔnk=xA:x=0.i1i2in¯wherej=1n2n-jij=k.

Define f:AA by(3)fx=0.i1i2i3i4i2n-1i2n¯,where x=0.i1i2in¯ and ij=1-ij. In other words, f(x)=0.m1m2mn¯, where mj=ij for odd j and mj=1-ij for even j. By (1), f(x)A for xA and f is well-defined. Moreover, directly from the definition of f, it follows that f is a bijection and the composition ff is the identity function, whence f-1=f. Moreover, by (2), for each nN and k{0,1,,2n-1}, k=j=1n2n-jij, ij{0,1}, we have(4)fAΔnk=AΔnk,where k=j=1n2n-jmj.

We claim that f is continuous. Fix x0A and ɛ>0. Choose n0N such that 1/2n0<ɛ. There exists k02n0-1 for which x0Δn0k0. By (4), f(AΔn0k0)=AΔn0k0. Since AΔn0k0 is a neighborhood of x0 and |y1-y2|1/2n0<ɛ for all y1,y2Δn0k0, 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). Let HA be any set of Lebesgue measure zero. Fix any ɛ>0. There exists an open in A set UA such that HU and |U|<ɛ. Let (5)B=ΔnkA:k0,1,,2n-1,  nN.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=jJ(ΔnjkjA), where J is at most countable and Δnj1kj1Δnj2kj1= for j1,j2J, j1j2. Then, by (4), jJf(ΔnjkjA)=jJ(ΔnjkjA) is an open in A set containing f(H) and (6)jJΔnjkjA=jJΔnjkjA=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 that f has no approximate derivative at any xA. Fix xA and an even nN. Then xΔnk for some k2n-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)AΔnk=AΔn+24kAΔn+24k+1AΔn+24k+2AΔn+24k+3,fAΔn+24k=AΔn+24k+1,fAΔn+24k+1=AΔn+24k,fAΔn+24k+2=AΔn+24k+3,fAΔn+24k+3=AΔn+24k+2(remember that n is even).

Note that (8)fy-fxy-x<0if xΔn+24k and yAΔn+24k+1 or xΔn+24k+2 and yAΔn+24k+3. Moreover, (9)fy-fxy-x>2n3·2n=13if xΔn+24k and yAΔn+24k+2 or xΔn+24k+1 and yAΔn+24k+3.

Thus, if nN is even and xAΔnk, we can find B,CΔnkA such that(10)B=C=14Δnk,(11)fx-fyx-y<0yB,(12)fx-fyx-y>13yC.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 Theorem 2, we easily get the following.

Corollary 5.

Any function f, 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 Theorem 4, we have the following.

Theorem 6.

There exists a bijection g:[0,1][0,1] such that

g has Lusin’s property (N) and N-1-property,

the set of discontinuity points of g is countable,

g has no approximate derivative at any point.

Proof.

Let B1, 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.i1in001010¯ or 0.i1in1101010¯ for some nN, except 1/6, 1/3, 2/3, 5/6. Define φ:{0,1}B1B2{0,1}B1B2 by (13) φ x = x for   x 0,1 , 1 6 , 1 3 , 2 3 , 5 6 , 0 . j 1 j 2 j n 001010 ¯ for   x = 0 . i 1 i n 001010 ¯ , 0 . j 1 j 2 j n 110101 ¯ for   x = 0 . i 1 i n 110101 ¯ , 0 . j 1 j 2 j n - 1 1 ¯ for   x = 0 . i 1 i n 1 ¯ , 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=fxfor  xA,φxfor  x0,1B1B2.Fix x(0,1)(B1B2), x=0.i1i2in¯, and ɛ>0. Let m be a positive integer such that 1/2m<ɛ. The set (15)C=0,1,16,13,23,562k-12n:k1,2,,2n-1,  nm0.l1ln001010¯:nm0.l1ln1101010¯:nmis finite and C{0,1}B1B2. Hence, we can find δ(0,1/2m+3) for which (x-δ,x+δ)C=. Take any y(x-δ,x+δ)(B1B2). Since |x-y|<1/2m+3 and yB1B2, we conclude y=0.i1i2im001010¯ or y=0.i1i2im110101¯ or y=0.i1i2im1¯. Hence, g(x)=f(x)=0.j1j2jm¯ and g(y)=φ(y)=0.j1j2jm¯ or g(y)=φ(y)=0.j1j2jm1¯. 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}B1B2. Therefore, g satisfies (b1), (b2), and (b3).

Theorem 7.

For each ɛ(0,1) there exist a closed nowhere dense set F(0,1) and a homeomorphism h:FF such that

|F|>1-ɛ,

h=h-1,

h has Lusin’s property (N) and N-1-property,

h has no approximate derivative (finite or not) at any xF (more precisely, if h~:[0,1][0,1] is any extension of h then h~ has no approximate derivative (finite or not) at any xF).

Proof.

Let B1, B2, A, and f be the same as in Theorem 4. Let {xn}n=1=B1B2. Fix ɛ>0 and choose a sequence (mn)n0 of even natural numbers satisfying(16)4n=012mn<ɛ,(17)4j=n+112mj<18·12mnn0.For each n1 there exists kn{1,,2mn-1} such that xn(kn-1)/2mn,(kn+1)/2mn. Let (18)B=Δm000Δm02m0-11n=1Δmnkn-1kn2mnΔmnkn.Since(19)Δmnkn-1kn2mnΔmnkn=kn-12mn,kn+12mn,B is an open subset of [0,1]. Moreover, B1B2{0,1}B and, by (16),(20)B22m0+n=122mn<ɛ2.

By (4), in the proof of Theorem 4, for each n1 there exist un,vn{1,,2mn} such that (21)fΔmnkn-1A=ΔmnunA,fΔmnknA=ΔmnvnA.Moreover,(22)fΔm00A=Δm0i0A,fΔm02m0-1A=Δm0i1Afor some i0,i1{0,1,,2m0}. Hence, (23)C=fBA=fΔm00AΔm02m0-1An=1Δmnkn-1AΔmnknA=Δm0i0AΔm0i1An=1ΔmnunAΔmnvnA=Δm0i0Δm0i1n=1ΔmnunΔmnvnA.Again, applying (16), we have |C|=|B|<ɛ/2. Moreover, since 0,1AIntB, the set (24)BC=0,1ABΔm0i0Δm0i1n=1ΔmnunΔmnvnis open in [0,1].

Finally, put H=[0,1]  (BC). It is clear that HA, 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)fBCA=fBAfCA=CBA=BCA.It follows that f(H)=H and h=fH is a homeomorphism.

Fix x0H and nN. There exists k{0,1,,2n-1} such that x0Δmnk. Certainly, ΔmnkBC. Therefore, by (17), |(BC)Δmnk|<(1/8)|Δmnk|. By (10), (11), and (12), (26)xΔmnk:fx-fx0x-x0<0>14·18Δmnk,xΔmnk:fx-fx0x-x0>13>14·18Δmnk.Therefore, any extension h~:[0,1][0,1] of h has no approximate derivative at x0.

Lemma 8.

Let a,b,c,dR, 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

|H|>(1/2)(b-a),

g-1:g(H)H is continuous,

g has Lusin’s property (N) and N-1-property,

if g~:[a,b][c,d] is any extension of g, then g~ has no approximate derivative (finite or not) at any xH,

|g(minH)-c|<ɛ, |d-g(maxH)|<ɛ, and |g(bn)-g(an)|<ɛ for all nN, where {(an,bn):nN} is the set of all connected components of (a,b)H.

Proof.

Fix ɛ>0 and choose nN 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:FF 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 kn. 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=ϕkhψk. 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 xFk. Finally, let H=k=1nFk and define g:H[c,d] by g(x)=hk(x) for xFk, k{1,,n}.

It is clear that H 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-gαψyk-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-gαψyk-ψxk-1=2d-cn+1<ɛ.Similarly, (31)gminF-cψy1-c=d-cn+1+d-cn+12=n+2n+12d-c<2d-cn+1<ɛ.Analogously, |d-g(maxF)|<ɛ. This completes the proof.

Now, we can prove the main theorem of the present paper.

Theorem 9.

There exists a continuous function f:[0,1][0,1] such that f has N-1-property, but fap exists almost nowhere.

Proof.

We will construct inductively a sequence (Fn)nN of closed subsets of [0,1] and a sequence (fn)nN of continuous functions fn:[0,1][0,1] such that

FnFn+1 and |Fn|>1-1/2n   for all n1,

fnFk=fkFk for all n>k,

|fn(x)-fn+1(x)|<1/2n for n{1,2,} and x[0,1],

every fn restricted to Fn has N-1-property,

every extension of fnFn has no approximate derivative at any xFn.

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 ψE=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 Theorem 7, there exist a closed set F(0,1), |F|>1/2, and a bijection g1:FF 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 f1F1=g1 has no approximate derivative at any xF1.

Let ((ak1,bk1))k1 be the family of all connected components of [0,1](F1{0,1}). Moreover, for every kN, let Jk1 be an open interval with endpoints f1(ak1) and f1(bk1). By Lemma 8, for each kN there exist closed Fk1(ak1,bk1) and gk1:Fk1Jk1 satisfying conditions (d1)–(d5) with ɛ=1/2. Let F2=F1k=1Fk1 and let g2:F2F1k=1Jk1 be defined by g2(x)=f1(x) for xF1 and g2(x)=gk1(x) for xFk1, k{1,2,}. We claim that g2 is continuous. The continuity of g2 at each point of k=1Fk1 is obvious. Fix x0F1 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 xF1(x0,x0+δ] and F2(x0,x0+δ)=(F1(x0,x0+δ))kKFk1 for some KN. Since (32)g2x-g2x0<maxg2ak1-g2x0,g2bk1-g2x0for xJk1, we have |g2(x)-g2(x0)|<ɛ for xF2(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.

Let f2 be the linear extension of g2. It is clear that F1F2, |F2|>1-1/4, f2F1=f1F1, f2 restricted to F2 has N-1-property, and every extension of f2F2=g2 has no approximate derivative at any xF2. Moreover, |f2(x)-f1(x)|<1/2 for x[0,1].

Assume that closed sets F1,,Fn(0,1), F1Fn, 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 frFr has no approximate derivative at any xFr, |fr(x)-fr-1(x)|<1/2r-1 for each x[0,1], and frFs=fsFs for every s{1,,r-1}.

Let ((akn,bkn))k1 be the family of all connected components of [0,1](Fn{0,1}). Moreover, for every kN let Jkn be an open interval with endpoints fn(akn) and fn(bkn). By Lemma 8, for each kN there exist closed Fkn(akn,bkn) and gkn:FknJkn satisfying conditions (d1)–(d5) with ɛ=1/2n. Let Fn+1=Fnk=1Fkn and let gn+1:Fn+1F1k=1Jkn be defined by gn+1(x)=fn(x) for xFn and gn+1(x)=gkn(x) for xFkn, k{1,2,}. Similarly, as in the case of g2, we can check that gn+1 is continuous.

Let fn+1 be the linear extension of gn+1. It is clear that FnFn+1, |Fn+1|>1-1/2n+1, fn+1Fn=fnFn, fn+1 restricted to Fn+1 has N-1-property, and every extension of fn+1Fn+1 has no approximate derivative at any xFn+1. Moreover, |fn+1(x)-fn(x)|<1/2n for x[0,1]. Thus, we have proved inductively that there exist a sequence (Fn)nN of closed subsets of [0,1] and a sequence (fn)nN 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 nN, the sequence (fn)nN is uniformly convergent to some continuous function f:[0,1][0,1]. Moreover, fFn=fnFn for all nN. Therefore, by (5), f has no approximate derivative at any point from n=1Fn. Since, by (1), |n=1Fn|=1, fap exists almost nowhere.

It remains to prove that f has N-1-property. Take any E[0,1] of the Lebesgue measure zero. Then, by (2), (33)f-1En=1Fnf-1E0,1n=1Fn=n=1Fnfn-1E0,1n=1Fn.Applying (1) and (4), we conclude that |f-1(E)|=0. Thus, f has N-1-property.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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