Abstract
We construct a continuous function such that possesses -property, but 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 does not hold for -property. Some relevant properties are presented.
1. Introduction
First we will specify some basic notations. By we denote the Lebesgue measure of . For any , where is an interval, by we denote the restriction of to and the symbol stands for approximate derivative of at .
Definition 1 (see [1]). Let be measurable. We say that has Lusin’s property , if the image of every set of Lebesgue measure has Lebesgue measure .
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 ). If is continuous and has Lusin’s property , then 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 that , defined on a measurable set , has -property, if the inverse image of every set of Lebesgue measure has Lebesgue measure .
Some of results concerning -property are presented in [2, 3]. In [2] a systematic study of -property for smooth and almost everywhere differentiable functions can be found. Some applications of -property in functional equation and geometric function theory can be found in [4–6].
2. Main Results
Our goal is to construct a continuous function with -property which is not approximately differentiable on a set of full measure. We start with the basic theorem.
Theorem 4. Let , , and . There exists a homeomorphism such that (a1),(a2) has Lusin’s property and -property,(a3) has no approximate derivative (finite or not) at any .
Proof. Let denote a binary decomposition of . It is easily seen that . Therefore, and belongs to if and only if it has a binary decomposition such that(in other words has infinitely many s and infinitely many s at even places or infinitely many s and infinitely many s at odd places). Let for and . Obviously, for every . Moreover,Define bywhere and . In other words, , where for odd and for even . By (1), for and is well-defined. Moreover, directly from the definition of , it follows that is a bijection and the composition is the identity function, whence . Moreover, by (2), for each and , , , we havewhere .
We claim that is continuous. Fix and . Choose such that . There exists for which . By (4), . Since is a neighborhood of and for all , we conclude that is continuous at . Thus, is continuous, because was arbitrary. By the equality , is a homeomorphism.
Now we will show condition (a2). Let be any set of Lebesgue measure zero. Fix any . There exists an open in set such that and . Let Clearly, is a base of the natural topology in . Since either any two sets from are disjoint or one of them is contained in the other, it is easy to see that any open subset of can be represented as a union of some subfamily of pairwise disjoint sets from . Thus, , where is at most countable and for , . Then, by (4), is an open in set containing and Since was arbitrary, and has Lusin’s property (). Since , has also -property.
Finally, we will show that has no approximate derivative at any . Fix and an even . Then for some . Let be such that . Moreover, let be understood as before. It is clear that (remember that is even).
Note that if and or and . Moreover, if and or and .
Thus, if is even and , we can find such thatSince this is true for every even and , we conclude that has no approximate derivative (finite or not) at . The proof is completed.
From Banach’s Theorem 2, we easily get the following.
Corollary 5. Any function , defined on an interval, which possesses Lusin’s condition such that the set of discontinuity points of 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 such that (b1) has Lusin’s property and -property,(b2)the set of discontinuity points of is countable,(b3) has no approximate derivative at any point.
Proof. Let , , , and be the same as in Theorem 4. It is easily seen that every member of is of the form or for some , except , , , . Define bywhere and . It is easy to see that is a bijection. Let be defined by Fix , , and . Let be a positive integer such that . The set is finite and . Hence, we can find for which . Take any . Since and , we conclude or or . Hence, and or . Therefore, . Since is continuous, is continuous at . Thus, we have proved that the set of all discontinuity points of is contained in . Therefore, satisfies (b1), (b2), and (b3).
Theorem 7. For each there exist a closed nowhere dense set and a homeomorphism such that (c1),(c2),(c3) has Lusin’s property and -property,(c4) has no approximate derivative (finite or not) at any (more precisely, if is any extension of then has no approximate derivative (finite or not) at any ).
Proof. Let , , , and be the same as in Theorem 4. Let . Fix and choose a sequence of even natural numbers satisfyingFor each there exists such that . Let Since is an open subset of . Moreover, and, by (16),By (4), in the proof of Theorem 4, for each there exist such that Moreover,for some . Hence, Again, applying (16), we have . Moreover, since , the set is open in .
Finally, put . It is clear that , is a closed subset of , and . Since is a bijection and , we have It follows that and is a homeomorphism.
Fix and . There exists such that . Certainly, . Therefore, by (17), . By (10), (11), and (12), Therefore, any extension of has no approximate derivative at .
Lemma 8. Let , , and . For every there exist a closed nowhere dense set and a continuous injection such that (d1),(d2) is continuous,(d3) has Lusin’s property and -property,(d4)if is any extension of , then has no approximate derivative (finite or not) at any ,(d5), , and for all , where is the set of all connected components of .
Proof. Fix and choose such that . Let be a partition of such that for and for . Let be a linear homeomorphism, . By Theorem 7, there exist a closed nowhere dense set and a homeomorphism satisfying conditions (c2)–(c4) such that . For each define linear homeomorphisms , and ,Moreover, let for . Obviously, each is a closed nowhere dense subset of . Besides, . For each define by . It is easy to see that each is a continuous injection, has Lusin’s property and -property, and, moreover, any extension of to is not approximately differentiable at any point . Finally, let and define by for , .
It is clear that and satisfy conditions (d1)–(d4). Let be any connected component of . If for some then If for some then Similarly, Analogously, . This completes the proof.
Now, we can prove the main theorem of the present paper.
Theorem 9. There exists a continuous function such that has -property, but exists almost nowhere.
Proof. We will construct inductively a sequence of closed subsets of and a sequence of continuous functions such that (1) and for all ,(2) for all ,(3) for and ,(4)every restricted to has -property,(5)every extension of has no approximate derivative at any .First, we give a useful definition. If is closed and , then by the linear extension of we mean such that , , , and is linear on every closed interval contiguous to . It is clear that is continuous if and only if is continuous.
By Theorem 7, there exist a closed set , , and a bijection satisfying conditions (c1)–(c4). Let and be the linear extension of . Then is continuous, has -property, and every extension of has no approximate derivative at any .
Let be the family of all connected components of . Moreover, for every , let be an open interval with endpoints and . By Lemma 8, for each there exist closed and satisfying conditions (d1)–(d5) with . Let and let be defined by for and for , . We claim that is continuous. The continuity of at each point of is obvious. Fix and . If is not isolated from the right in , then there exist such that for and for some . Since for , we have for . Hence, is continuous from the right at . Similarly, we can show that is continuous from the left at . Since was arbitrary, is continuous.
Let be the linear extension of . It is clear that , , , restricted to has -property, and every extension of has no approximate derivative at any . Moreover, for .
Assume that closed sets , , and continuous functions , , are chosen. Moreover, assume that for every we have , restricted to has -property, every extension of has no approximate derivative at any , for each , and for every .
Let be the family of all connected components of . Moreover, for every let be an open interval with endpoints and . By Lemma 8, for each there exist closed and satisfying conditions (d1)–(d5) with . Let and let be defined by for and for , . Similarly, as in the case of , we can check that is continuous.
Let be the linear extension of . It is clear that , , , restricted to has -property, and every extension of has no approximate derivative at any . Moreover, for . Thus, we have proved inductively that there exist a sequence of closed subsets of and a sequence of continuous functions satisfying conditions .
Since for and , the sequence is uniformly convergent to some continuous function . Moreover, for all . Therefore, by , has no approximate derivative at any point from . Since, by , , exists almost nowhere.
It remains to prove that has -property. Take any of the Lebesgue measure zero. Then, by , Applying and , we conclude that . Thus, has -property.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.