STABILITY OF THE DISTRIBUTION FUNCTION FOR PIECEWISE MONOTONIC MAPS ON THE INTERVAL

. For piecewise monotonic maps the notion of approximating distribution function is introduced. It is shown that for a mixing basic set it coincides with the usual distribution function. Moreover, it is proved that the approximating distribution function is upper semi-continuous under small perturbations of the map.


1.
Introduction. The notion of distributional chaos has been introduced by Schweizer and Smítal in [12]. To this end distribution functions were introduced. Given t ∈ R and two points x, y in a dynamical system one counts the relative number of times when the distance of the orbits of x and y is smaller than t. Denote the limit superior of this sequence by U x,y (t) and its limit inferior by F x,y (t). In this way one obtains two functions U, F : R → [0, 1], called the upper and lower distribution function. We will call the lower distribution function simply the distribution function. A dynamical system is called distributionally chaotic, if there are x, y and t > 0 with F x,y (t) < 1 and U x,y (s) = 1 for all s > 0. More details on distributional chaos can be found in the nice survey paper [11] and also in [3].
We like to consider the distribution function for piecewise monotonic maps T : [0, 1] → [0, 1]. This means that there exists a finite partition of [0, 1] into intervals such that the restriction of T to such an interval is continuous and strictly monotonic. However, T needs not be continuous at the endpoints of intervals of monotonicity. Because of these discontinuities it is much harder to work with the distribution function. Therefore we introduce the notion of approximating distribution functions. It is defined similar as the usual distribution function but taking into account only points x, y in Markov subsets avoiding the points of discontinuity and turning points.
For a piecewise monotonic map a set B is called basic set, if it is a maximal topologically transitive subset with positive topological entropy. It will be proved in Theorem 1 that the approximating distribution function of a mixing basic set coincides with usual one in all points where the usual distribution function is right continuous. Hence they differ in at most countably many points. In particular defining distributional chaos one can use the approximating distribution function instead of the usual one.
The main question addressed in this paper is the behaviour of the approximating distribution function under small perturbations of the piecewise monotonic map. For continuous interval maps (but not necessarily piecewise monotonic) the upper semi-continuity of distribution functions has been proved by Francisco Balibrea, Bert Schweizer, Abe Sklar and Jaroslav Smítal in [2] (Theorem 4.5 and Corollary 4.6 of [2]). We will prove in Theorem 2 that for mixing basic sets of (not necessarily continuous) piecewise monotonic maps the approximating distribution function is upper semi-continuous. This implies that distributional chaos (defined using the approximating distribution functions) is stable under small perturbations of the map.
Finally we consider again a mixing basic set B (as defined above) of a piecewise monotonic map and its approximating distribution function G. Consider finitely many points x 1 , x 2 , . . . , x u ∈ [0, 1] and a decreasing sequence (J n ) of unions of u intervals forming a neighbourhood of {x 1 , . . . , x u } such that ∞ n=1 J n = {x 1 , x 2 , . . . , x u }. For each n let G n be the approximating distribution function of the set of all points of B whose orbits omit J n . In Theorem 3 it is proved that lim n→∞ G n (t) = G(t) for all t where G is right continuous.
The authors like to thank Franz Hofbauer for useful discussions on the topic of this paper, in particular for a lot of useful suggestions concerning the proofs.
for t ∈ R, t > 0. Obviously both F x,y and U x,y are increasing functions with values in [0, 1], and F x,y ≤ U x,y . Moreover, note that F x,x (t) = U x,x (t) = 1 for all t > 0. We define the lower distribution function F of the dynamical system (M, T ) by then we call the lower distribution function F of A, T A the lower distribution function of A. Throughout this paper we will call the lower distribution function simply the distribution function.
The distribution function has been introduced by Bert Schweizer and Jaroslav Smítal in [12] for continuous maps. A dynamical system is said to be distributionally chaotic, if the distribution function F satisfies F (t) < 1 for some t > 0. For continuous interval maps it is proved in [12] that distributional chaos is equivalent to positive topological entropy and to the existence of a basic set. In the case of maps on a graph this has been proved by Roman Hric and Michal Málek in [6]. This result is no longer true in higher dimensions as shown by MartaŠtefánková in [1] for two dimensional skew products.
A map T : [0, 1] → [0, 1] is called piecewise monotonic with respect to c 0 = 0 < c 1 < · · · < c N = 1, if T is continuous and strictly monotonic on (c j−1 , c j ) for j ∈ {1, 2, . . . , N }. Note that it may be discontinuous in E := {c 0 , c 1 , . . . , c N }. If T is not continuous, then it is much more difficult to deal with the distribution function as defined in (2). Therefore we use another approach in order to overcome these problems.
Set Z := {(c j−1 , c j ) : 1 ≤ j ≤ N } and for n ∈ N set Furthermore, for n ∈ N let C n be the union of {c 0 , c 1 , . . . , c N } and those intervals in Z n having an endpoint in {c 0 , c 1 , . . . , c N }. Observe that Note that A n is closed for any n, and If there exists an n with A n = ∅, then we define the approximating distribution function G of A by x, y ∈ A n for some n and U x,y (s) = 1 for all s > 0} (4) We call a closed T -invariant set A ⊆ [0, 1] topologically transitive, if the map T A is topologically transitive, this means there exists an element in A whose orbit is dense in A. If there does not exist a topologically transitive set A 0 with A A 0 , then A is called maximal topologically transitive.
where A n is defined as in (3). Now set If there exists an n with A n = ∅, then we obtain by (4), (5) and (6) that for all t ∈ R with t > 0, where G is the approximating distribution function of A. We give two more similar definitions which will be useful later. For these definitions we assume that there exists an n with A n = ∅. Set Q(A) := ∞ n=1 A n × A n and define for all t > 0. Obviously L is increasing and L(t) ≤ G(t) for all t > 0. For n ∈ N let P n (A) be the set of all periodic points in A having period n. If for a q ∈ N there is an n ≥ q with P n (A) = ∅, then set P q (A) := ∞ n=q P n (A) × P n (A), and define for all t > 0. Also in this case it is obvious that H q is increasing. As introduced above let Z be the collection of intervals of monotonicity of T . We call Y a refinement of Z, if Y consists of finitely many pairwise disjoint open intervals such that for every Y ∈ Y there is a Z ∈ Z with Y ⊆ Z, and [0, 1] ⊆ Y ∈Y Y . In order to find suitable approximations of basic sets, we define Markov subsets.
T -invariant, disjoint from C n for some n ≥ 1, and has a finite Markov partition refining Z.
In the proofs of our next results we need as our main tool the Markov diagram of T . It has been introduced by Franz Hofbauer and describes the orbit structure of T . A description of the Markov diagram can be found e.g. in [5]. Now we recall its definition and its most important properties.
Assume that Y is a refinement of Z. Let D be a nonempty open subinterval of an element of Y.
If C is irreducible and C ∈ C, then define N (C) as the set of all n such that there is a path D 0 → D 1 → · · · → D n in C with D 0 = C and D n = C. Then the greatest common divisor of N (C) equals the greatest common divisor of N (D) for any D ∈ C. If the greatest common divisor of N (C) equals 1, then C is called aperiodic. We call an irreducible For a maximal irreducible C ⊆ D let K(C) be the set of all x such that x is represented by an infinite path in C and x is not contained in the interior of an interval I satisfying that T n I is monotonic for all n. Suppose that B is a basic set of T . By Theorem 11 of [5] there exists a maximal irreducible C ⊆ D such that B = K(C). For a C ⊆ C define K(C ) as the set of all x ∈ K(C) which can be represented by an infinite path in C . If C is irreducible, then K(C ) is topologically transitive. Moreover, for an irreducible C the set K(C ) is mixing if and only if C is aperiodic.
As we will also need the notion of a variant (A, →) of the Markov diagram as introduced in [7] (cf. also [10]), we briefly describe this concept. It is an orientated graph together with a function A : Furthermore the construction in [7] shows that for every irreducible C ⊆ A having an element with more than one successor in C the property that A(C) is aperiodic implies that C is aperiodic. The versions of the Markov diagram introduced in [8] (and also used in [9]) are essentially the same as variants, but contain also the orbits of some single points. Since these additional orbits of single points are not interesting for our purpose we can work with variants. Then in the conclusions of Lemma 2 in [8] and Lemma 1 in [9] we can work with variants instead of versions, and the conclusions concerning A(c) having only one element can be omitted.
Remark. Consider a finite union X of closed subintervals of [0, 1] and a map T : X → [0, 1] such that there exists a finite family Z of pairwise disjoint open intervals with Z∈Z Z = X and T Z is continuous and strictly monotonic for all Z ∈ Z. Then one can define the Markov diagram (and also variants and versions of the Markov diagram) of T with respect to a refinement Y of Z in the same way as above (see also for example [7]). This can also be seen by extending T to a map on [0, 1] defining it so on each maximal open subinterval I of [0, 1] \ X that T I ⊆ I and each point in I is attracted by a fixed point. Hence all results apply also in this slightly more general case.
is a piecewise monotonic map, assume that B is a mixing basic set of T , and let m ∈ N. Moreover, let x 1 , x 2 , . . . , x r ∈ B m . Then there exists an integer n ≥ m and there exists a mixing Markov subset X of B n with x 1 , x 2 , . . . , x r ∈ X.
Proof. Without loss of generality we may assume that m ≥ 2. Since our assumptions imply that Then C ∞ is a union of at most N + 1 intervals, and If I ⊆ C ∞ is a nonempty open interval which does not contain an element of {c 0 , c 1 , . . . , c N }, then T n is strictly monotonic on I for each n, and therefore it cannot contain a point in a basic set of T . Hence B ⊆ X ∞ . For sufficiently large n the set C n is a union of v pairwise disjoint open intervals. Therefore there are 0 = a Moreover, for any l ∈ {1, 2, . . . , 2v} we have lim n→∞ a (n) l = a l . This means that "C n converges to C ∞ " in the sense described in [9]. Set X n : Now let D n be the smallest set containing Y and with the property that D ∈ D n and D → C imply C ∈ D n . Let ( D ∞ , →) be the Markov diagram of T X∞ with respect to Y m . Since B ⊆ X ∞ is mixing there exists a maximal irreducible and aperiodic C ⊆ D ∞ such that As Y ∩ C m = ∅ we get that Y ∈ D ∞ . Moreover, using again Y ∩ C m = ∅, one gets that T (Y ) ∈ Y m−1 and its closure is the union of closures of elements of Y m . This means that all successors of Y having nonempty intersection with B m are again in E. Therefore x j can be represented by an infinite path C 2 → · · · in C. By Theorem 1 in [5] there is an n j such that D Choose a C ∈ r j=1 E j . As C is aperiodic there exists an s, and there exist paths E 0 → E 1 → · · · → E s and D 0 → D 1 → · · · → D s+1 in C with E 0 = E s = D 0 = D s+1 = C. Let C ⊆ C be a finite and irreducible set containing r j=1 E j ∪ {E 0 , E 1 , . . . , E s } ∪ {D 0 , D 1 , . . . , D s+1 }. Since both E 0 → E 1 → · · · → E s and D 0 → D 1 → · · · → D s+1 are paths in C beginning and ending in C the set C is aperiodic.
By Lemma 2 in [10] and Lemma 1 in [9] there exists an n ≥ m such that the following properties hold. There exist variants (A, →) and ( A, →) of D ∞ , resp. of D n , there exist C ⊆ A and an irreducible and aperiodic C ⊆ A with A C = C , and there exists a bijective function ϕ : . The set C is irreducible and aperiodic, because C is irreducible and aperiodic. There exists a maximal irreducible B ⊆ D n with A C ⊆ B. Then the corresponding set K( B) is a subset of B n . Define X := K( C) (the set of all points in K( B) which can be represented by a path in C). Then X ⊆ B n is mixing, as C is aperiodic. Assume that j ∈ {1, 2, . . . , r}. We get that d → · · · is a path in C. As this path represents an x ∈ X, and A(c) ⊆ A ϕ(c) we get that x j ∈ X.
Because of the finiteness of C the set E of all endpoints of elements of A C is finite, and also the set (a) For every ε > 0 there is an M ∈ N such that for any k ∈ N, any k points x 1 , x 2 , . . . , x k ∈ X, any nonnegative integers there exists a point x ∈ X of period p satisfying |T n (x) − T n−aj (x j )| ≤ ε for all n ∈ {a j , a j+1 , . . . , b j } and all j ∈ {1, 2, . . . , k}. (b) Suppose that ε > 0. Then there exists an M ∈ N such that for any points x 1 , x 2 , . . . ∈ X and any integers 0 Proof. Fix ε > 0 and define B ε (x) := (x − ε, x + ε). Moreover, let Y be a Markov partition of X refining Z. Now we show that there is an M ∈ N such that T M X ∩ B ε (x) = X for all x ∈ X. Set Y n = n−1 j=0 T −j Y for n ≥ 1. By the compactness of X there exists a k ∈ N such that for each x ∈ X the set B ε (x) contains an element of Y k (notice that the interior of a periodic or wandering interval does not belong to X). Since X is a mixing Markov subset, for every Y ∈ Y k there is an n(Y ) such that T n (Y ∩ X) = X for all n ≥ n(Y ). Set M := max{n(Y ) : Y ∈ Y k }. Hence X = T n (Y ∩ X) ⊆ T M X ∩ B ε (x) ⊆ X for any x ∈ X. Now one can prove (a) in the same way as in [4] (see the proof of Theorem 8.7 in [4]).
In order to prove (b) we use (a). If k ∈ N, then there is a (periodic) point y k ∈ X satisfying |T n (y k ) − T n−aj (x j )| ≤ ε for a j ≤ n ≤ b j and 1 ≤ j ≤ k. Let x be a limit point of the sequence (y k ) k≥1 . Then x satisfies (b), since T is continuous on X.
For w ∈ [0, 1] \ ∞ j=0 T −j {c 0 , c 1 , . . . , c N } let Z n (w) be the unique interval in Z n containing w. We have Z n+1 (w) ⊆ Z n (w) for n ≥ 1 and Lemma 3. Assume that B is a mixing basic set or a mixing Markov subset of the piecewise monotonic map T and fix a q ∈ N. Let t > 0 and ε ∈ (0, t). Then there is an integer p ≥ q and there are u and v in P p (B) such that 1 p N u,v (0, p, t − ε) < L(t) + ε. In particular, F u,v (t − ε) < L(t) + ε.
Proof. By the definition of L there is a (x, y) ∈ Q(B) with F x,y (t) < L(t) + ε 2 . Then there is an m ≥ 1 with x, y ∈ B m because of the definition of Q(B). If B is a mixing basic set then by Lemma 1 there is a mixing Markov subset X of B containing x and y. Otherwise set X = B, which is a mixing Markov subset. In particular, X satisfies the requirements of Lemma 2. Let M be the constant for ε 2 in (a) of Lemma 2. Because of F x,y (t) = lim inf n→∞ This shows the first assertion, since p > q.
As u and v have period p, it follows that Note that for maps T : [0, 1] → [0, 1] one always has F x,y (t) = 1 for t ≥ 1. Therefore F (t) = U (t) = L(t) = H q (t) = 1 for any t ≥ 1. This means that one may consider these functions as functions from [0, 1] → [0, 1], if one is only interested in the non-trivial part of them.

Lemma 4.
Suppose that A is a mixing basic set or a mixing Markov subset of the piecewise monotonic map T . Then for every ε > 0 there is (u, v) ∈ R(A) such that F u,v is in the upper ε-neighbourhood of L.
Proof. Choose t 0 = 0 < t 1 < · · · < t r = 1 with t j − t j−1 < ε 2 for 1 ≤ j ≤ r. By the definition of L there are (x j , y j ) ∈ Q(A) with F xj ,yj (t j ) < L(t j ) + ε 4 for 1 ≤ j ≤ r. Using the definition of Q(A) there is an m ≥ 1 such that x j and y j are in A m for 1 ≤ j ≤ r. If A is a mixing basic set we apply Lemma 1 and find a Markov subset X of A with x j and y j in X for 1 ≤ j ≤ r. Otherwise A is a Markov subset, and we set X = A. Let α be the minimal distance in the Hausdorff metric between U ∩ X and V ∩ X for two different U, V ∈ Z. Since X is a Markov subset there is an n ≥ 1 with X ⊆ A n and hence α > 0. As X is uncountable, there is a w ∈ X with lim n→∞ |Z n (w)| = 0, where |I| denotes the length of the interval I. Set u l (n) := max k ≤ n : T j (Z n (w)) < 1 l for 0 ≤ j ≤ k . Since T is continuous on the intervals in Z, we have that lim n→∞ u l (n) = ∞ for all l ≥ 1.
Let M be the constant in (b) of Lemma 2 for the set X and for min ε 4 , α 2 instead of ε. Set x j+lr := x j , y j+lr = y j and t j+lr = t j for l ≥ 1 and 1 ≤ j ≤ r. We choose integers a 1 ≤ b 1 < c 1 ≤ d 1 < a 2 ≤ b 2 < c 2 ≤ d 2 < · · · in the following way. Define holds. This is possible, since we have F x k ,y k (t k ) < L(t k ) + ε 4 . Then set for c k ≤ n ≤ d k , k ≥ 1. Now (11) and the definition of b k give Using the definition of t k for k > r this implies Hence for every t ∈ (0, 1] there is a t j with |t−t j | < ε and F u,v (t) < L(t j ) + ε. This means that F u,v is in the upper ε-neighbourhood of L.
It follows from (12) and from the definition of α that T n (u) and T n (v) are in the same element of Z as T n−c k (w) for c k ≤ n ≤ d k . This implies that T c k (u) and Since this holds for all k ≥ 1, we get U u,v (t) = 1 for all t ∈ (0, 1]. Since X is a Markov subset there is an n ≥ 1 with X ⊆ A n . Hence we have also proved that (u, v) ∈ R(A).
If the piecewise monotonic map T has points in E which are eventually periodic or are attracted by periodic orbits, then let γ be the maximum of the periods of these periodic points. Otherwise set γ := 0. Proposition 1. Fix q > γ and let B be a mixing basic set or a mixing Markov subset of the piecewise monotonic map T . Then H q (t) = L(t) = G(t) for all points t ∈ (0, 1), where L is right continuous.
Proof. Since R(B) ⊆ Q(B) we have L(t) ≤ G(t) for all t ∈ (0, 1]. It follows from Lemma 4 that G is contained in every upper ε-neighbourhood of L. The intersection of all these upper ε-neighbourhoods is {(a, b) : 0 < a < 1, b ≤ lim z→a + L(z)}, which must then contain the points (t, G(t)) for 0 < t < 1. Therefore, we have L(t) = G(t) for all points t ∈ (0, 1), where L is right continuous.
By the definition of γ, for a periodic point x whose period is larger than γ there is an n such that x and an element of E cannot lie in the closure of the same element of Z n . This implies P q (B) ⊆ Q(B), and hence L(t) ≤ H q (t) for all t ∈ (0, 1]. Choose t ∈ (0, 1) such that L is right continuous in t. For every ε ∈ (0, 1 − t) we have that H q (t) < L(t + ε) + ε by Lemma 3, which implies H q (t) ≤ L(t).

4.
The approximating distribution function of continuous piecewise monotonic maps. For piecewise monotonic maps we work with the approximating distribution function G instead of the distribution function F . Therefore we will prove in our next result that for continuous piecewise monotonic transformations these two functions essentially coincide. More exactly, they coincide in all points where F is right continuous. Proof. Choose an arbitrary ε > 0 and fix some q > γ. Then there is a δ > 0 such that F (t) ≤ F (s) < F (t) + ε for all s ∈ [t, t + δ). Now we choose an s ∈ [t, t + δ) such that F is continuous at s and L is right continuous at s. Then by Theorem 4.3 in [2] there is a pair of periodic points u, v ∈ B with F u,v (s) < F (t) + ε. Using the specification property we may assume that the periods of u and v are larger than q. Hence H q (s) ≤ F u,v (s). As L is right continuous at s Proposition 1 implies that G(s) = H q (s). Therefore we get Since ε > 0 was arbitrary this implies F (t) = G(t).

5.
Stability results for the approximating distribution function of piecewise monotonic maps. Next we show that for (not necessarily continuous) piecewise monotonic maps the approximating distribution function is upper semi-continuous. If ε > 0 then a piecewise monotonic map T with respect to c 0 = 0 < c 1 < · · · < c N = 1 is called to be ε-close to a piecewise monotonic map T with respect to c 0 = 0 < c 1 < · · · < c N = 1, if N = N (this means they have the same number of intervals of monotonicity) and T ( cj−1, cj ) is in an ε-neighbourhood of T (cj−1,cj ) for all j ∈ {1, 2, . . . , N }. Proof. Fix an ε > 0 and a q > γ. Since G is increasing we can choose a 0 = 0 < t 1 < a 1 < t 2 < a 2 < · · · < a k−1 < t k < a k = 1 with a j − a j−1 < ε 5 , G(t) > G(t j ) − ε 3 for all t ∈ (a j−1 , a j ] and L is right continuous at t j for 1 ≤ j ≤ k. Then G(t j ) = H q (t j ) for 1 ≤ j ≤ k by Proposition 1. Hence there exist periodic points x j and y j of period at least q with F xj ,yj (t j ) < G(t j ) + ε 3 . We choose a refinement Y of Z with |Y | < ε 5 , and [5] we get that the points x j and y j can be represented by periodic paths in the Markov diagram (D, →) of T with respect to Y. Furthermore there exists an aperiodic maximal irreducible C ⊆ D with B = K(C). Therefore using also Lemma 5 of [7] there is an r ∈ N such that any x ∈ M can be represented by a periodic path in D r and for every variant (A, →) of D there is an aperiodic irreducible C 1 ⊆ A r with A(C 1 ) ⊆ C and each x ∈ M is represented by a periodic path in C 1 .
By Lemma 6 in [7] there exists a δ > 0 such that every piecewise monotonic map T which is δ-close to T satisfies the following properties. There exists a variant (A, →) of the Markov diagram of T , a finite partition Y refining the partition of intervals of monotonicity of T , a variant ( A, →) of the Markov diagram of T with respect to Y, and an injective function ϕ : A r → A r with c → d in A is equivalent to ϕ(c) → ϕ(d) in A and A ϕ(c) is ε 5 -close to A(c) in the Hausdorff-metric for all c ∈ A r .
Then ϕ(C 1 ) is an aperiodic irreducible subset of A r . Hence it is contained in an aperiodic maximal irreducible C ⊆ A and B := K( C) is a mixing basic set of T . Let x ∈ M . Then there is an n ≥ 1 such that x is represented by a path c 0 → c 1 → c 2 → · · · in C 1 ⊆ A r with c n+k = c k for all k. Therefore ϕ(c 0 ) → ϕ(c 1 ) → ϕ(c 2 ) → · · · is a periodic path in ϕ(C 1 ) ⊆ A r . Hence it represents an x ∈ B, and we obtain that T n x = x. Moreover we have | x − x| < 2ε 5 . Since T s x ∈ ϕ Y (T s x) we get also | T s x − T s x| < 2ε 5 for all s ≥ 0. This implies that F xj , yj (t j − 4ε 5 ) ≤ F xj ,yj (t j ) for all j ∈ {1, 2, . . . , k}. Using Lemma 1 and Lemma 2 we obtain that there are 3 for all j ∈ {1, 2, . . . , k}. Now let t ∈ (0, 1). Choose j ∈ {1, 2, . . . , k} so that a j−1 < t ≤ a j . Note that a j < t j + ε 5 . Then there exists an s > t with s − ε 5 < t j such that L is right continuous at s−ε. By Proposition 1 we get G(s−ε) = L(s−ε). Observe that t − ε < s − ε < t j − 4ε 5 and G(t j ) < G(t) + ε 3 . Therefore we obtain This shows that G is in an upper ε-neighbourhood of G.
In our next result we show that the approximating distribution function does not change essentially if one adds artificially intervals of monotonicity. Observe that there are at most countably many points where G is not right continuous, since G is increasing. Hence G n (t) → G(t) for all t where G is right continuous implies that lim n→∞ G n (t) = G(t) for all t ∈ R \ E where E is an at most countable set (note that the distribution functions are equal to 0 for t ≤ 0 and equal to 1 for t ≥ 1). In particular, G n → G almost everywhere (with respect to the Lebesgue measure).
Theorem 3. Suppose that T : [0, 1] → [0, 1] is a piecewise monotonic map with h top (T ) > 0, and assume that B is a mixing basic set of T . Let G be the approximating distribution function of T on B. Moreover, suppose that u ∈ N and x 1 , x 2 , . . . , x u ∈ [0, 1], and for n ≥ 1 let J n ⊆ [0, 1] be a union of at most u pairwise disjoint open intervals such that J n+1 ⊆ J n holds for all n ≥ 1 and ∞ n=1 J n = {x 1 , x 2 , . . . , x u }. Assume that G n is the approximating distribution function of T on ∞ k=0 B \ (T −k (J n )). Then lim n→∞ G n (t) = G(t) for all t ∈ (0, 1) where G is right continuous.
Proof. It is obvious that G n (t) ≥ G(t) for all n and all t. Moreover, for any t the sequence (G n (t)) is decreasing. Hence it remains to show that every t ∈ (0, 1) where G is right continuous, and for every ε > 0 there is an n with G n (t) < G(t) + ε. For n ∈ N set B n := ∞ k=0 B \ (T −k (J n )). Denote the endpoints of intervals of monotonicity of T by c 0 , c 1 , . . . , c N . Furthermore let L n be the function defined in (8) for B n . Again it is obvious that the sequence (L n (t)) is decreasing for any t and L n (t) ≥ L(t) for any t and any n. x 2 , . . . , x u } is eventually periodic or attracted by a periodic orbit, then let γ 0 be the maximum of the periods of these periodic points. Otherwise set γ 0 := 0. Now fix q > γ 0 .
Let t ∈ (0, 1) be so that G is right continuous at t and let ε > 0. As G is right continuous at t there is a δ > 0 with G(s) < G(t) + ε 2 for all s ∈ (t, t + δ). Choose t 1 ∈ (t, t + δ) so that L is right continuous at t 1 . From Proposition 1 we obtain that G(t 1 ) = H q (t 1 ). Therefore there exist periodic points x, y ∈ B whose periods are at least q, such that F x,y (t 1 ) < H q (t 1 ) + ε 2 = G(t 1 ) + ε 2 < G(t) + ε .
By the choice of q there is an m with x, y / ∈ ∞ k=0 T −k (C m ). Moreover, again by the choice of q there is an l ∈ N with x ∈ B l (otherwise T k x ∈ J l for some k) and y ∈ B l . Hence (x, y) ∈ Q( B l ).
Using Theorem 8 of [5] we obtain that the x and y can be represented by periodic paths in the Markov diagram (D, →) of T with respect to Y. Moreover, there is an aperiodic maximal irreducible C ⊆ D with B = K(C). Therefore using also Lemma 5 of [7] there exists an r such that x and y can be represented by a periodic path in D r and for every variant (A, →) of D there is an aperiodic irreducible C 1 ⊆ A r with A(C 1 ) ⊆ C and x and y can be represented by a periodic path in C 1 . For n ∈ N define X n := ∞ k=0 [0, 1] \ (T −k (J n )). Because of our assumptions "J n converges to {x 1 , x 2 , . . . , x u }" in the sense described in [9]. Hence Lemma 2 in [10] and Lemma 1 in [9] yield the existence of an n ≥ l and variants (A, →) of D and ( A, →) of the Markov diagram of T Xn and an injective function ϕ : A r → A r with c → d in A is equivalent to ϕ(c) → ϕ(d) in A. Moreover, in this case we get A ϕ(c) ⊆ A(c) for all c ∈ A r . Then there is an aperiodic irreducible C 1 ⊆ A r with A(C 1 ) ⊆ C and x and y can be represented by a periodic path in C 1 . Therefore ϕ(C 1 ) is aperiodic and irreducible, and x and y can be represented by a path in ϕ(C 1 ). This yields that ϕ(C 1 ) is contained in an aperiodic maximal irreducible C ⊆ A and therefore K( C) is a mixing basic set of T Xn . By Lemma 1 it contains an aperiodic Markov subset X ⊆ B n with x, y ∈ X. Then (x, y) ∈ Q(X). Denote by G, respectively L, the functions defined in (4), respectively (8), for the set X. We get L(t 1 ) ≤ F x,y (t 1 ) < G(t) + ε. Now choose an s with t < s < t 1 such that L is right continuous at s. As L is increasing we get L(s) ≤ L(t 1 ) < G(t) + ε. Furthermore Proposition 1 implies that G(s) = L(s), which leads to G(s) < G(t) + ε. Since X ⊆ B n we get G n (s) ≤ G(s). Now G n (t) ≤ G n (s) ≤ G(s) < G(t) + ε completes the proof.