A counter–example in ergodic theory

Péter Major Mathematical Institute of the Hungarian Academy of Sciences and Bolyai College of the Eötvös Loránd University, Budapest

address: H–1364, Budapest, P.O.B. 127 (Mathematical Institute of the Hungarian Academy of Sciences) e-mail address: major@math-inst.hu AMS Subject Classification (1991) 47A35

Abstract: We construct a (non-integrable) function f and two measure preserving, ergodic transformations S and T on a measure space (X , A, μ), μ(X) = 1,

in such a way that the ergodic means lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(S^{k}x) and lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} f(T^{k}x)

exist for almost all x, they are finite constants not depending on x, but these constants differ when we are averaging with respect to the operators S and T. This means that in the case of a non-integrable function f and an ergodic transformation T the ergodic mean depends not only on the function f, but also on the transformation T. The construction applies some probabilistic arguments.

The aim of this paper is to construct a probability space (X , A, μ), a function f and two ergodic transformations T and S on it such that

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} T^{k}f = 0 a. s.  (1)

lim_{n→∞} \frac{1}{n} ∑_{k=1}^{n} S^{k}f = a a. s.  (2)

with some constant 0 < a < ∞. Here and in the sequel the same symbol is used for a measure preserving transformation and the linear operator it induces on measurable functions by composition. (Because of the ergodic theorem such an example is possible only with a function f such that ∫ f^{+} dμ = ∞ and ∫ f^{−} dμ = −∞.)

The question about the possibility of such an example was raised by Zoltán Buczolich. He studied the problem that for how large class of functions the classical notion of integral can be extended. He was interested in the relation of this problem to the ergodic theorem (see [1]), and this was the reason for asking such a question. But we hope that this question can also be interesting for its own sake.

We define the probability space (X , A, μ) as X = [0, 1]^{Z}, where Z is the set of integers, A is the Borel σ-algebra on X and μ = λ^{Z}, the direct product of different copies of the Lebesgue measure on [0, 1]. Let T be the shift operator to the left on X, i.e. for x ∈ X, x = (…, x_{−1}, x_{0}, x_{1}, …)

T(…, x_{−1}, x_{0}, x_{1}, …) = (…, x_{0}, x_{1}, x_{2}, …) .

To define the function f we introduce the sequences

A_{n} = [\frac{2^{n+10}}{n^{2}} ] , n = 1, 2, …,

B_{0} = 0, B_{n} = 2(A_{1} + ⋯ + A_{n}), n = 1, 2, …,

where [·] denotes integer part. Put f_{k}(u) = 0 for k ≤ 0, 0 ≤ u ≤ 1, define the functions f_{k}(u) for B_{n−1} < k ≤ B_{n}, n = 1, 2, …, on [0, 1] as

f_{k}(u) = \{ \begin{matrix}n^{3} & for 0 ≤ u < 2^{−n} \\ 0 & otherwise \end{matrix}  if k is odd

f_{k}(u) = \{ \begin{matrix}−n^{3} & for 0 ≤ u < 2^{−n} \\ 0 & otherwise \end{matrix}  if k is even ,

and set

f(x) = ∑_{k=−∞}^{∞} f_{k}(x_{k}) , x = (…, x_{−1}, x_{0}, x_{1}, …) .

The relation

∑_{k=−∞}^{∞} μ(f_{k}(x_{k}) ≠ 0) = ∑_{n=1}^{∞} 2A_{n}2^{−n} ≤ 2^{11} ∑_{n=1}^{∞} \frac{1}{n^{2}} < ∞

holds, hence the Borel–Cantelli lemma implies that for almost all x with respect to the measure μ the sum defining f(x) contains only finitely many non-zero terms. Hence this sum is meaningful. Moreover, since the sum defining f(x) contains finitely many non-zero terms with probability one, the rearrangements of summation appearing in this article are legitimate.

First we show that the above defined T and f satisfy (1), then define an operator S which satisfies (2).

Write

\frac{1}{n} ∑_{k=1}^{n} T^{k}f(x) = \frac{1}{n} ∑_{m=−∞}^{∞} H_{m,n}(x_{m}) = \frac{1}{n} ∑_{m=−n+1}^{B_{L}} H_{m,n}(x_{m})

+ \frac{1}{n} ∑_{m=B_{L}+1}^{∞} H_{m,n}(x_{m}) = Z_{n}^{(1)}(L) + Z_{n}^{(2)}(L)

with arbitrary L > 0 and

H_{m,n}(u) = ∑_{k=m+1}^{m+n} f_{k}(u) , 0 ≤ u ≤ 1 .

The relation

μ (Z_{n}^{(2)}(L) ≠ 0 for some n ≥ 1) ≤ ∑_{k=L}^{∞} ∑_{m=B_{k}+1}^{B_{k+1}} μ(x_{m} ∈ [0, 2^{−k−1}])  (3)

= 2 ∑_{k=L}^{∞} A_{k+1}2^{−k} < \frac{const.}{L} , where x = (x_{m}, m ∈ Z)

holds.

Observe that the functions H_{m,n}(x_{m}) are independent for fixed n, and H_{m,n}(u) = j(n + m)f_{n+m}(u) − j(m)f_{m+1}(u), where j(m) = 1 if m is odd, and j(m) = 0 if m is even. Define the moments E_{n}^{(p)}(m) = ∫ H_{m,n}^{p}(x_{m}) dμ(x) of the random variables H_{m,n}(x_{m}), p = 1, 2 …. The identity E_{n}^{(1)}(m) = j(n+m)ℓ^{3}(n+m)2^{−ℓ(n+m)} −j(m)ℓ^{3}(m+1)2^{−ℓ(m+1)} holds, where ℓ(m) denotes the number l for which B_{l−1} < m ≤ B_{l} if m ≥ 1, and ℓ(m) = 0 if m ≤ 0. Also the inequalities |E_{n}^{(p)}(m)| ≤ C(p), p = 1, 2, …, hold with some constant C(p). These relations together with the independence of the variables H_{m,n}(x_{m}) imply that

lim_{n→∞} EZ_{n}^{(1)}(L) = lim_{n→∞} \frac{1}{n} ∑_{m=−n+1}^{B_{L}} E_{n}^{(1)}(m) = 0 ,  (4)

and

E (Z_{n}^{(1)}(L) − EZ_{n}^{(1)}(L))^{4} ≤ \frac{const.}{n^{2}} .

Hence

μ (|Z_{n}^{(1)}(L) − EZ_{n}^{(1)}(L)| > ε) = μ (|Z_{n}^{(1)}(L) − EZ_{n}^{(1)}(L)|^{4} > ε^{4}) ≤ \frac{const.}{n^{2}ε^{4}} ,

and since the right-hand side of the last inequality is summable in n, the Borel–Cantelli lemma implies that

lim_{n→∞} Z_{n}^{(1)}(L) − EZ_{n}^{(1)}(L) = 0 a. s. .  (5)

Since L can be chosen arbitrary large, relations (3), (4) and (5) imply relation (1).

The operator S we shall construct is an appropriate conjugate of T. First we conjugate T with an operator U_{π} which induces a permutation of the coordinates x_{k}. Then we apply another conjugation with an operator G which places the range of the functions f_{k} to other intervals. We describe this construction in more detail.

Let π be a permutation of the positive integers Z^{+} = {1, 2, … }. We denote its extension to Z defined by the formula π(k) = k for k ≤ 0 again by π and define the transformation U_{π} on X as

U_{π}x = (x_{π(k)}, k ∈ Z) , for x = (x_{k}, k ∈ Z) .

For all k ∈ Z define a transformation g_{k}: [0, 1] → [0, 1] such that either g_{k}(u) = u or it is an interval exchange transformation defined in the following way: We consider two disjoint intervals I_{k}^{(1)} = [a(k), b(k)) and I_{k}^{(2)} = [c(k), d(k)) such that 0 ≤ a(k) < b(k) ≤ 1, 0 ≤ c(k) < d(k) ≤ 1, b(k) − a(k) = d(k) − c(k), and put

g_{k}(u) = \{ \begin{matrix}u & if u ∈ [0, 1] \ (I_{k}^{(1)} ∪ I_{k}^{(2)}) \\ u + c(k) − a(k) & if u ∈ I_{k}^{(1)} \\ u + a(k) − c(k) & if u ∈ I_{k}^{(2)} \end{matrix} .  (6)

Given the above defined set of functions g_{k}(u), k ∈ Z, we introduce the following transformation G of X:

G(x) = (g_{k}(x_{k}), k ∈ Z) , for x = (x_{k}, k ∈ Z) .

We remark that G^{−1} = G and define with the help of the above considered permutation π and G the transformation

S = S(π, g_{k}, k ∈ Z) = GU_{π}TU_{π}^{−1}G .

Clearly,

S^{k} = GU_{π}T^{k}U_{π}^{−1}G .

Observe that

S^{k}f(x) = ∑_{m=−∞}^{∞} f_{π(k+π^{−1}(m))}(g_{π(k+π^{−1}(m))}(g_{m}(x_{m}))) for x = (x_{k}, k ∈ Z) ,

and this formula together with the relations π(k) = k and f_{k}(u) ≡ 0 for k ≤ 0 imply that

\frac{1}{n} ∑_{k=1}^{n} S^{k}f(x) = ∑_{m=−n+1}^{∞} U_{m,n}(x_{m})  (7)

with

U_{m,n}(u) = \frac{1}{n} ∑_{k=1}^{n} f_{π(k+π^{−1}(m))}(g_{π(k+π^{−1}(m))}(g_{m}(u))) .

We want to construct the operator S in such a way that

lim_{n→∞} ∑_{m=1}^{∞} U_{m,n}(x_{m}) = 0 a. s. ,  (8)

and

lim_{n→∞} ∑_{m=−n+1}^{0} U_{m,n}(x_{m}) = a > 0 a. s. .  (9)

Relations (7), (8) and (9) imply (2). We want to guarantee (8) by defining π and G in such a way that for large m the probability of the set where U_{n,m}(x_{m}) ≠ 0 is negligibly small and to deduce (9) from the law of large numbers. To accomplish this goal we make such a construction where the cancellations between different functions f_{k} guarantee that U_{m,n}(x_{m}) is strictly positive and relatively small. Before this construction we show that a transformation S defined in the above way is ergodic.

The transformation S is measure preserving. Given a measurable set A ⊂ X define the set B such that A = GU_{π}B. Then S^{−1}A = GU_{π}T^{−1}B. Hence A = S^{−1}A if and only if B = T^{−1}B. The latter relation can hold only if μ(B), hence μ(A) equals zero or one because of the ergodicity of the operator T.

To define the permutation π and functions g_{k} such that the operator S given with their help satisfies relations (8) and (9) we introduce some sequences.

Put M_{n} = n^{3}2^{−n} and

P_{1} = A_{1}, P_{n} = 2 ⌈\frac{P_{n−1}}{2} \frac{M_{n−1}}{M_{n}} ⌉ , n = 2, 3, …,
R_{0} = 0, R_{n} = B_{n−1} + P_{n} , S_{n} = R_{n} + 2(A_{n} − P_{n}) = B_{n} − P_{n} , n = 1, 2, …,

where ⌈x⌉ denotes the smallest integer larger than or equal to x. Simple induction shows that P_{n} ≤ A_{n} for all positive integers n, and P_{n} < A_{n} for n ≥ 2. Really, this relation holds for n = 1, and by induction

P_{n} ≤ 2 (\frac{A_{n−1}}{2} \frac{M_{n−1}}{M_{n}} + 1) = \frac{2(n − 1)^{3}}{n^{3}} A_{n−1} + 2 ≤ 2^{n+10} \frac{(n − 1)}{n^{3}} + 2 < A_{n}

for n > 1. This relation implies that R_{n} ≤ S_{n} ≤ B_{n} ≤ R_{n+1} for all n = 1, 2, …. Now we define the value of the permutation π for 1 ≤ k ≤ P_{1} = S_{1}, R_{n} < k ≤ S_{n}, n ≥ 2, and S_{n} < k ≤ R_{n+1}, n ≥ 1, together with some functions b(k) and j(k) that we need for the definition of the functions g_{k}(u). Since R_{1} = S_{1}, these sets cover Z^{+}.

For 1 ≤ k ≤ P_{1} let π(k) = 2k − 1, j(k) = 0 and b(k) = 0. For R_{n} < k ≤ S_{n}, n ≥ 2, put j(k) = 0, b(k) = n and define

π(k) = k + P_{n} if k − R_{n} is odd
π(k) = k − P_{n} if k − R_{n} is even.

(The numbers π(k), k and k − R_{n} have the same parity.) We have

{π(k), 1 ≤ k ≤ P_{1}} = {k, 1 ≤ k < B_{1}} ∩ {odd numbers}

and

{π(k), R_{n} < k ≤ S_{n}} = [{k, B_{n−1} + 2P_{n} + 1 ≤ k ≤ B_{n} − 1} ∩ {odd numbers}]
⋃[{k, B_{n−1} + 2 ≤ k ≤ B_{n} − 2P_{n}} ∩ {even numbers}] .

For S_{n} < k ≤ R_{n+1}, n ≥ 1, we define π(k) as a map from K^{(n)} = [S_{n} + 1, R_{n+1}] to J_{1}^{(n)} ∪ J_{2}^{(n)} with

J_{1}^{(n)} = [B_{n} + 1, B_{n} + 2P_{n+1} − 1] ∩ {odd numbers}
J_{2}^{(n)} = [B_{n} − 2P_{n} + 2, B_{n}] ∩ {even numbers} .

The set J_{1}^{(n)} has cardinality P_{n+1} and J_{2}^{(n)} cardinality P_{n}. The cardinality of K^{(n)} equals R_{n+1} − S_{n} = P_{n} + P_{n+1}. We define the functions π(k), b(k) and j(k), S_{n} < k ≤ R_{n+1}, by induction. The function j(k) equals either 0 or 1 and b(k) takes the values n or n + 1. Let us assume that we have already defined these functions for all S_{n} < k ≤ L with some L ≥ S_{n} + 1, but not for L < k ≤ R_{n+1}. Put

K_{1}^{(n)}(L) = {k: S_{n} < k ≤ L, b(k) = n + 1}

and

K_{2}^{(n)}(L) = {k: S_{n} < k ≤ L, b(k) = n} .

Set

Π_{1}^{(n)}(L) = max_{k∈K_{1}^{(n)}(L)} π(k) and Π_{2}^{(n)}(L) = max_{k∈K_{2}^{(n)}(L)} π(k) .

If M_{n+1}|K_{1}^{(n)}(L)| ≥ M_{n}(|K_{2}^{(n)}(L)| + 1), where |K_{i}^{(n)}(L)| denotes the cardinality of the set K_{i}^{(n)}(L), then define π(L + 1) = Π_{2}^{(n)}(L) + 2, j(L + 1) = 0 and b(L + 1) = n. In the other case define π(L + 1) = Π_{1}^{(n)}(L) + 2, π(L + 2) = Π_{1}^{(n)}(L) + 4, j(L + 1) = 0, j(L + 2) = 1 and b(L + 1) = b(L + 2) = n + 1. If the set K_{1}^{(n)}(L) or K_{2}^{(n)}(L) is empty, then we define Π_{1}^{(n)}(L) = B_{n} − 1 and Π_{2}^{(n)}(L) = B_{n} − 2P_{n} respectively.

We have to show that the above definition is correct, i.e. the iteration can be stopped in such a way that Π_{1}^{(n)}(L) = B_{n} + 2P_{n+1} − 1 and Π_{2}^{(n)}(L) = B_{n}, what is equivalent to saying that |K_{1}^{(n)}(L)| = P_{n+1} and |K_{2}^{(n)}(L)| = P_{n}. To prove the correctness of this definition first we make the following observation:

\begin{aligned}P_{n}M_{n} = 2\frac{P_{n}}{2} \frac{M_{n}}{M_{n+1}} M_{n+1} & ≤ P_{n+1}M_{n+1} \\ & < 2 (\frac{P_{n}}{2} \frac{M_{n}}{M_{n+1}} + 1) M_{n+1} = P_{n}M_{n} + 2M_{n+1} . \end{aligned}  (10)

Let L̄ be a value for which one of the relations |K_{1}^{(n)}(L̄)| = P_{n+1} or |K_{2}^{(n)}(L̄)| = P_{n} holds. It is enough to show that if the cardinality of the other set is less than the corresponding number, then in the next step of the iteration this set increases.

If |K_{1}^{(n)}(L̄)| = P_{n+1} and |K_{2}^{(n)}(L̄)| < P_{n}, then |K_{2}^{(n)}(L̄)| ≤ P_{n} −1 and |K_{1}^{(n)}(L̄)|M_{n+1} = M_{n+1}P_{n+1}. Hence by the left side of relation (10),

M_{n}(|K_{2}^{(n)}(L̄)| + 1) ≤ P_{n}M_{n} ≤ P_{n+1}M_{n+1} = M_{n+1}|K_{1}^{(n)}(L̄)| ,

and the value of |K_{2}^{(n)}(·)| is increasing in the next step of the iteration.

If |K_{2}^{(n)}(L̄)| = P_{n} and |K_{1}^{(n)}(L̄)| < P_{n+1}, then |K_{1}^{(n)}(L̄)| ≤ P_{n+1} − 2. Hence we get, using the right-hand side of (10), that

M_{n+1}|K_{1}^{(n)}(L̄)| ≤ M_{n+1}(P_{n+1} − 2) < P_{n}M_{n} < M_{n}(|K_{2}^{(n)}(L̄)| + 1) .

This implies that the value of |K_{1}^{(n)}(·)| is increasing in the next step of the iteration. Hence the above definition of the functions π(·), b(k) and j(k) is meaningful.

It is not difficult to check that the above defined function π is an isomorphism from Z^{+} to Z^{+}. Hence we can use it as the permutation π in the definition of S. We define the functions g_{k} needed in this definition in the following way: Let g_{k}(u) = u if k ≤ 0 or j(k) = 0. If j(k) = 1 and b(k) = n, then let g_{k}(u) be the interval exchange transformation described in formula (6) with the intervals I_{k}^{(1)} = [0, 2^{−n}) and I_{k}^{(2)} = [2^{−n}, 2^{−n+1}). (We defined the function j(k) in order to decide whether we want to make an interval exchange transformation g_{k}(u) in the k-th coordinate. The function b(k) identifies in which interval (B_{n−1}, B_{n}] the number π(k) lies.)

Now we turn to the proof of relation (8). First we show that if k ∈ (B_{n−1}, B_{n}], n ≥ 2, then π(k) ∈ (B_{n−2}, B_{n+1}]. This follows from the following observations: If k ∈ (R_{n}, S_{n}] then k ∈ (B_{n−1}, B_{n}] and π(k) ∈ (B_{n−1}, B_{n}], and if k ∈ (S_{n}, R_{n+1}] then k ∈ (B_{n−1}, B_{n+1}] and π(k) ∈ J_{1}^{(n)} ∪ J_{2}^{(n)} ⊂ (B_{n−1}, B_{n+1}]. This relation implies that π(k) > B_{n−1} and π^{−1}(k) > B_{n−1} for k > B_{n}. Hence we get that if m > B_{L} with some L > 1 then π(k + π^{−1}(m)) > B_{L−2} for all k ≥ 0, and

U_{m,n}(u) = 0 if m > B_{L} and 2^{−L+3} ≤ u ≤ 1 .

For 1 ≤ m ≤ B_{L} write

U_{m,n}(u) = \frac{1}{n} ∑_{k=1}^{B_{L}} f_{π(k+π^{−1}(m))}(g_{π(k+π^{−1}(m))}(g_{m}(u)))

+ \frac{1}{n} ∑_{k=B_{L}+1}^{n} f_{π(k+π^{−1}(m))}(g_{π(k+π^{−1}(m))}(g_{m}(u)))

= Σ_{L}^{1}(m, n, u) + Σ_{L}^{2}(m, n, u) .

Then

Σ_{L}^{1}(m, n, u) < \frac{C(L)}{n} ,

and

Σ_{L}^{2}(m, n, u) = 0 if 2^{−L+3} ≤ g_{m}(u) ≤ 1 .

Hence

∑_{m=B_{L}+1}^{∞} U_{m,n}(x_{m}) = 0 on the set

D_{L} = {x = (x_{j}, j ∈ Z), x_{j} ≥ 2^{−(p+L)+3} for B_{L+p} < j ≤ B_{L+p+1}, p = 0, 1, … } ,

∑_{m=1}^{B_{L}} Σ_{L}^{2}(m, n, x_{m}) = 0 on the set

E_{L} = {x = (x_{j}, j ∈ Z), g_{j}(x_{j}) ≥ 2^{−L+3} for 1 ≤ j ≤ B_{L}}

and

lim_{n→∞} ∑_{m=1}^{B_{L}} Σ_{L}^{1}(m, n, u) = 0 .

Since

μ(X \ D_{L}) ≤ 2 ∑_{p=0}^{∞} A_{p+L}2^{−(p+L)+3} < const. \frac{1}{L} ,
μ(X \ E_{L}) ≤ 2^{−L+3}B_{L} < const. \frac{1}{L^{2}} ,

and L can be chosen arbitrary large, the above relations imply (8).

Since π(m) = m and g_{m}(u) = u for m ≤ 0, the expression U_{m,n}(u) is simpler in this case. We can write

U_{m,n}(u) = \frac{1}{n} ∑_{k=1}^{n} f_{π(k+m)}(g_{π(k+m)}(u)) = \frac{1}{n} ∑_{l=1}^{n+m} h_{l}(u) = \frac{H_{n+m}(u)}{n} for 1 − n ≤ m ≤ 0

with

h_{l}(u) = f_{π(l)}(g_{π(l)}(u)) and H_{L}(u) = ∑_{l=1}^{L} h_{l}(u) , (H_{L}(u) = 0 for L ≤ 0) .

Let us study the expression H_{L}(u). It follows from the definition of the functions π(k) and g_{k}(u) that

H_{L}(u) = \{ \begin{matrix}L & for 0 ≤ u < \frac{1}{2} \\ 0 & for \frac{1}{2} ≤ u ≤ 1 \end{matrix} if 1 ≤ L ≤ P_{1} ,

H_{L}(u) − H_{R_{n}}(u) = \{ \begin{matrix}n^{3} & if 0 ≤ u < 2^{−n} and L − R_{n} is odd \\ 0 & if 2^{−n} ≤ u ≤ 1 or L − R_{n} is even \end{matrix} if R_{n} ≤ L ≤ S_{n}, n ≥ 2 .

In particular,

H_{S_{n}}(u) − H_{R_{n}}(u) = 0, for all 0 ≤ u ≤ 1, n ≥ 2 .

We have

H_{R_{n+1}}(u) − H_{S_{n}}(u) = \{ \begin{matrix}\frac{1}{2}P_{n+1}(n + 1)^{3} − P_{n}n^{3} & if 0 ≤ u < 2^{−n} \\ 0 & if 2^{−n} ≤ u ≤ 1 \end{matrix} ,

and the functions P_{n,L}(u) = H_{L}(u) − H_{S_{n}}(u) , S_{n} < L ≤ R_{n+1} , n ≥ 1 , satisfy the identity

P_{n,L}(u) = \{ \begin{matrix}0 & if 2^{−n} ≤ u ≤ 1 \\ \frac{1}{2}|K_{1}^{(n)}(L)|(n + 1)^{3} − |K_{2}^{(n)}(L)|n^{3} & if 0 ≤ u < 2^{−n} \\ & and |K_{1}^{(n)}(L)| is even \\ \frac{1}{2} (|K_{1}^{(n)}(L)| + 1) (n + 1)^{3} − |K_{2}^{(n)}(L)|n^{3} & if 0 ≤ u < 2^{−(n+1)} \\ & and |K_{1}^{(n)}(L)| is odd \\ \frac{1}{2} (|K_{1}^{(n)}(L)| − 1) (n + 1)^{3} − |K_{2}^{(n)}(L)|n^{3} & if 2^{−(n+1)} ≤ u < 2^{−n} \\ & and |K_{1}^{(n)}(L)| is odd \end{matrix} ,

where the functions K_{i}^{(n)}(L), i = 1, 2, appeared in the definition of the function π(k). To bound the function P_{N,L}(u), first we prove the following relation for S_{n} < L ≤ R_{n+1} by induction:

0 ≤ M_{n+1}|K_{1}^{(n)}(L)| − M_{n}|K_{1}^{(n)}(L)| < \{ \begin{matrix}M_{n} + M_{n+1} if |K_{1}^{(n)}(L)| is odd \\ M_{n} + 2M_{n+1} if |K_{1}^{(n)}(L)| is even \end{matrix} . (11)

Indeed, relation (11) holds for L = S_{n} + 1. Taking into consideration the inequality which decides which one of the sets K_{i}^{(n)}(L), i = 1, 2, increases in the next step of the iteration we get the proof of relation (11) by separating the two cases by induction in the following way. If

M_{n} ≤ M_{n+1}|K_{1}^{(n)}(L)| − M_{n}|K_{2}^{(n)}(L)| < M_{n} + 2M_{n+1} ,

then |K_{1}^{(n)}(L + 1)| = |K_{1}^{(n)}(L)| and |K_{2}^{(n)}(L + 1)| = |K_{2}^{(n)}(L)| + 1, hence relation (11) holds for L + 1. If

0 ≤ M_{n+1}|K_{1}^{(n)}(L)| − M_{n}|K_{2}^{(n)}(L)| < M_{n} and |K_{1}^{(n)}(L)| is even ,

then |K_{1}^{(n)}(L + i)| = |K_{1}^{(n)}(L)| + i and |K_{2}^{(n)}(L + i)| = |K_{2}^{(n)}(L)|, i = 1, 2. Hence, in this case relation (11) holds for L + 1 and L + 2.

Since n^{3} = 2^{n}M_{n}, relation (11) implies that

0 ≤ P_{n,L}(u) ≤ 3(n + 1)^{3} .

The above inequalities imply that for B_{p} < L ≤ B_{p+1}

0 ≤ H_{L}(u) ≤ \{ \begin{matrix}0 & if \frac{1}{2} ≤ u ≤ 1 \\ const. s^{4} & for 2^{−s−1} ≤ u < 2^{−s} for 1 ≤ s ≤ p \\ const. p^{4} & for 0 ≤ u < 2^{−p−1} \end{matrix} .

Define the moments of the random variables H_{L}(x_{m})

E_{L}^{(p)}(m) = E_{L}^{(p)} = ∫ H_{L}^{p}(x_{m}) dμ(x) = ∫_{0}^{1} H_{L}^{p}(u) du , for L ≥ 1 and m ≤ 0 .

Because of the bound given for the functions H_{L}(u)

0 ≤ E_{L}^{(p)} < C(p) < ∞ for all p = 1, 2, …

with some appropriate constant C(p). In particular, the first moment E_{L}^{(1)} satisfies the relation

lim_{L→∞} E_{L}^{(1)} = E = A_{1} + ∑_{k=1}^{∞}(P_{k+1}M_{k+1} − P_{k}M_{k}) > 0 .  (12)

Also the relation E = lim_{L→∞} E_{L}^{(1)} < ∞ holds, since E = A_{1} + lim_{k→∞} P_{k}M_{k}, and

P_{k+1}M_{k+1} = ⌈\frac{P_{k}}{2} + \frac{M_{k}}{M_{k+1}} ⌉ M_{k+1} ≤ P_{k}M_{k} + M_{k} ,

hence sup_{k} P_{k}M_{k} ≤ P_{0}M_{0} + ∑_{k=1}^{∞} M_{k} < ∞.

Introduce the random variables ξ_{L}(m) = H_{L}(x_{−m}) − E_{L}^{(1)}, m = 0, 1, 2, …, on the probability space (X , A, μ). The random variables ξ_{L(m)}(m), m = 1, 2, …, are independent for an arbitrary sequence L(m), Eξ_{L}(m) = 0 and Eξ_{L}^{4}(m) ≤ C < ∞ for all L ≥ 0 and m. Hence

E (\frac{ξ_{n}(0) + ⋯ + ξ_{1}(n − 1)}{n})^{4} < \frac{const.}{n^{2}} ,

and

μ (|\frac{ξ_{n}(0) + ⋯ + ξ_{1}(n − 1)}{n}| > ε) < \frac{const.}{n^{2}ε^{4}} for all ε > 0 .

Since the right-hand side of the last expression is summable in n for all ε > 0,

lim_{n→∞} \frac{ξ_{n}(0) + ⋯ + ξ_{1}(n − 1)}{n} = 0 a. s. .  (13)

Relations (12) and (13) imply that

lim_{n→∞} ∑_{m=−n+1}^{0} U_{m,n}(x_{m}) = lim_{n→∞} \frac{1}{n} ∑_{m=0}^{n−1} ξ_{n−m}(m) + lim_{n→∞} \frac{1}{n} ∑_{m=o}^{n−1} E_{n−m}^{(1)} = E > 0 a. s. .

Relation (9), hence relation (2) is proved.

Reference:

[1] Buczolich, Z: Arithmetic averages of rotations of measurable functions. To appear in Journal of Ergodic Theory and Dynamical Systems