Ergodicity and central limit theorem for random interval homeomorphisms

i
μ_∗({a}) = μ_∗({f⁻¹(a)}) for any i ∈ Σ_∗, which is impossible. Indeed, the set

i
{f⁻¹(a) : i ∈ Σ_∗} for admissible iterated function systems is infinite. This, in turn, would imply that μ_∗((0, 1)) = ∞. The proof is complete.
Theorem 2: Let (f₁,... f_N ; p₁,..., p_N ) be an admissible iterated function sys- tem and let P be the corresponding Markov operator. Let μ_∗ ∈ M₁((0, 1)) be its unique invariant measure. Then for any measure μ ∈ M₁((0, 1)) we have

n→∞
lim Pⁿμ, ϕ = μ_∗, ϕ for ϕ ∈ C([0, 1]).

n
Proof. We follow [14] in defining some martingale. Namely, for ϕ ∈ C([0, 1]) we consider the sequence of random variables (ξ^ϕ) defined on the probability space
(Σ, P) by the formula

n

n

1
ξ^ϕ(i) = μ_∗,ϕ ◦ f_{(i ,...,i )} for i = (i₁, i₂,.. .).

n

n
Since μ_∗ is an invariant measure for P , we easily check that (ξ^ϕ) is a bounded martingale and from the Martingale Convergence Theorem it follows that (ξ^ϕ)

n
is convergent P-a.s. Since the space C([0, 1]) is separable, there exists a subset Σ₀ ⊂ Σ with P(Σ₀) = 1 such that (ξ^ϕ(i)) is convergent for any ϕ ∈ C([0, 1])
and i ∈ Σ₀. By the Riesz Representation Theorem for any i ∈ Σ₀ there exists

n
a measure μ_i ∈ M₁([0, 1]) such that

(3.8) lim
n→∞
ξ^ϕ(i) = μ_i, ϕ for every ϕ ∈ C([0, 1].

Now we are going to show that μ_i is supported at some point υ(i) ∈ [0, 1] for P-a.e. i ∈ Σ. To do this it is enough to show that for any ε > 0 there exists Σ_ε ⊂ Σ₀ with P(Σ_ε) = 1 satisfying the following property: for every i ∈ Σ_ε there exists an interval I of length |I|≤ ε such that μ_i(I) ≥ 1 − ε. Hence we obtain that μ_i = δ_υ(i) for all i from the set

∞
^{Σ˜0 = Σ}1/n^.
n=1
Obviously P(Σ^˜₀) = 1.

Fix ε > 0 and let a, b ∈ (0, 1) be such that μ_∗([a, b]) > 1 − ε. Let l ∈ N be such that 1/l < ε/2. Since for any x ∈ (0, 1) there exists i ∈ {1,..., k} such that f_i(x) < x, we may find a sequence (j_n), j_n ∈ Σ_∗, such that f_jn (b) → 0 as n → ∞. Therefore, there exist i₁,... , i_l such that f_im ([a, b]) ∩ f_in ([a, b]) = ∅ for m, n ∈ {1,..., l}, m = n. Put n^∗ = max_m≤l |i_m| and set J_m = f_im ([a, b]) for m ∈ {1,..., l}. Now observe that for any sequence u = (u₁,..., u_n) ∈ Σ_∗ there exists m ∈ {1,..., l} such that |f_u(J_m)| < 1/l < ε/2. This shows that for any cylinder in Σ, defined by fixing the first initial n entries (u₁,... , u_n), the
conditional probability that (u₁,... , u_n,..., u_n+k) are such that
|f_{(un+k ,...,un,...,u1 )}([a, b])|≥ ε/2 for all k = 1,... , n^∗
is less than 1 − q for some q > 0. Hence there exists Σ_ε ⊂ Σ with P(Σ_ε) = 1 such that for all (u₁, u₂,.. .) ∈ Σ_ε we have |f_{(un,...,u1 )}([a, b])| < ε/2 for infinitely many n. Since [0, 1] is compact, we may additionally assume that for infinitely
many n’s the set f_{(un ,...,u1)}([a, b]) is contained in some set I with |I| ≤ ε. Since μ_∗ is an invariant probability measure and a, b ∈ (0, 1) are chosen in such a way that μ_∗([a, b]) > 1 − ε, we obtain that μ_i(I) ≥ 1 − ε. The proof of our assertion that μ_i is supported at some point υ(i) ∈ [0, 1] for P-a.e. i ∈ Σ is finished.

n n
To show that the sequence (Pⁿμ) for μ ∈ M₁((0, 1)) converges weakly to μ_∗ it is enough to prove, since the Lipschitz functions are dense in C([0, 1]), that for any Lipschitz function ϕ and arbitrary two points x, y ∈ (0, 1) we have

lim
n→∞
| P δ_x, ϕ − P δ_y, ϕ | = 0.

In fact, we would obtain then that
n

n n
| P μ,ϕ − | μ_∗, ϕ |

^≤ (0,1) (0,1)
| P δ_x, ϕ − P δ_y, ϕ |μ(dx)μ_∗(dy) → 0 as n → ∞.

Fix x, y ∈ (0, 1) and let x < y. Fix ε > 0. Since μ_∗ is invariant, by the proof of uniqueness in Theorem 1, 0 and 1 belong to its support and consequently
μ_∗((0, x)) > 0 and μ_∗((y, 1)) > 0.

μ ◦ ⁻f_∗
We know by (3.8) that for P almost every i = (i₁, i₂,.. .) ∈ Σ the measures

1 ⁽ⁱn ^,...,i1⁾
, n ∈ N, converge weakly to δ_υ(i). Consequently, for every ε > 0,

μ ◦ ⁻f_∗
1 ⁽ⁱn ^,...,i1⁾
((υ(i) − ε/2, υ(i)+ ε/2) ∩ (0, 1)) → 1 as n → ∞.

Since μ_∗((0, x)) > 0 and μ_∗((y, 1)) > 0, there exist u_n ∈ (0, x) and v_n ∈ (y, 1)

⁽ⁱn ^{,...,i )}1
such that u_n, v_n ∈ f⁻¹ ((υ(i) − ε/2, υ(i)+ ε/2) ∩ (0, 1)) for all n sufficiently
large. Hence f_{(in ,...,i1 )}(u_n), f_{(in ,...,i1 )}(v_n) ∈ (υ(i) − ε/2, υ(i) + ε/2) and conse-
quently f_{(in,...,i1 )}(x), f_{(in ,...,i1)}(y) ∈ (υ(i) − ε/2, υ(i)+ ε/2) for all n sufficiently large. Since ε > 0 was arbitrary, we obtain that for P-a.e. i = (i₁, i₂,.. .) ∈ Σ the following convergence holds:

lim
n→∞
^|f(in ^,...,i1^{)(x) − f(i}n ^,...,i1^{)(y)| = 0.}

By (2.2) and the fact that Pⁿδ_z, ϕ = Uⁿϕ(z) for z ∈ [0, 1] we have

n

n
| P
δ_x,ϕ − P δ_y, ϕ |

(3.9)
=|Uⁿϕ(x)−Uⁿϕ(y)|

≤L |f_{(i1 ,...,in )}(x)−f_{(i1 ,...,in )}(y)|p_i1 ··· p_in for x, y∈ (0, 1),
⁽ⁱ1 ^,...,in ^)∈Σn
where L is the Lipschitz constant of ϕ. We are going to show that the right- hand side of the above inequality converges to 0 as n → ∞. To do this for