New York Journal of Mathematics

New York J. Math. 6 (2000) 87–93.

An Example of a Conservative Exact Endomorphism which is Not Lim Sup Full

Julia A. Barnes and Stanley J. Eigen

Abstract. We show how to modify a construction of Hamachi to obtain a conservative exact endomorphism which is not lim sup full.

Contents

1. Introduction 87 2. Preliminaries 88 3. The Shape of the Measure P 88 4. The Induction Process 89 5. Nonsingularity, Conservativity, and Not Lim Sup Full 90 References 92

1. Introduction

In showing certain endomorphisms are exact, Rohlin [7] developed the notion of full and showed that in the finite measure preserving case these two notions are equivalent. However, there are open questions of conservativity and exactness when measure preserving is not known. To this end, Barnes [1] developed the notion of lim sup full, and showed that lim sup full, nonsingular, n-to-1 endomorphisms are conservative and exact whether or not the map is finite measure preserving. She applied this to certain classes of rational maps, obtaining conservativity and exactness when measure preserving is not known. The purpose of this note is to show by example that a conservative exact map need not be lim sup full.

The example is based on a construction (see Section 3) studied by Hamachi [3] and Krengel [6] and depends strongly on a theorem of Kakutani [5]. As a class, the general construction gives a family of maps which are conservative, exact shift maps on the one-sided symbol space with different product measures. By a construction of Bruin and Hawkins [2] conservative examples can be moved to the Riemann sphere where they are invariant for certain rational maps whose Julia set is the entire sphere.

87^{ISSN 1076-9803/00}

Received April 30, 1999.

Mathematics Subject Classification. 37.

Key words and phrases. exact, full, lim sup full.

2. Preliminaries

The transformation constructed by Hamachi was a Bernoulli shift on a two-sided product space. The example in this paper is a one-sided factor of such a map.

Set Ω = ∏_{i=−∞}^{∞}{0, 1}_{i}. The transformation T is the shift (T ω)_{n} = ω_{n+1}. Most of the work presented here deals with the construction of the product measure P = ∏_{k=−∞}^{∞} P_{k}, where P_{k}(0) = \frac{1}{1+λ_{k}} and P_{k}(1) = \frac{λ_{k}}{1+λ_{k}} for some λ_{k} ≥ 1. Since Ω is fixed throughout the paper, we refer to the measure space (Ω, P, T) as P. We say T is nonsingular with respect to the measure P if the measures P and P ∘ T^{−1} are equivalent; conservative if for every set A of positive measure, there is a k > 0 such that P (A∩T^{−k}(A)) > 0; measure preserving if the measures P and P ∘T^{−1} are identical; ergodic if for any measurable set A, P (A △ T^{−1}A) = 0 ⇒ P (A) is zero or one.

We are interested in a one-sided shift map defined on a factor of Ω. Let X = ∏_{i=−∞}^{0}{0, 1}_{i} with the product measure Q = ∏_{k=−∞}^{0} P_{k}. Define S = T^{−1}, i.e., (⋯ x_{−2}x_{−1}x_{0}) ↦ (⋯ x_{−3}x_{−2}x_{−1}). As above, we refer to the measure space (X, Q, S) as S. We use B to denote the sigma algebra of measurable sets in X. The same definitions as above are used to define S as nonsingular, conservative, measure preserving, and ergodic. In addition, we call S exact if ∩_{0}^{∞}S^{−1}B contains only sets of Q-measure one or zero; full if lim_{j→∞} Q(S^{j}B) = 1 for all B ∈ B of positive measure; lim sup full if lim sup_{j→∞} Q(S^{j}B) = 1 for all B ∈ B of positive measure [1].

3. The Shape of the Measure P

The measure P is determined by a fixed probability distribution μ on {0, 1}, a sequence of probability distributions ν_{i} = \{\frac{1}{1+λ_{i}} , \frac{λ_{i}}{1+λ_{i}} \}, and two sequences of nonnegative integers n_{i} and m_{i}. Hamachi assigns values to λ_{i}, n_{i} and m_{i} by induction. He does this in a way which guarentees that the resulting measure he obtains is nonsingular, conservative, ergodic and preserves no equivalent finite or infinite invariant measure for the two-sided shift. We present a simpler induction process in the next section. Our goal is for a nonsingular, conservative exact measure which is not lim sup full for the one-sided shift. Since it is not lim sup full, it follows that it has no equivalent finite invariant measure. It remains open whether it has an equivalent infinite invariant measure; it also remains open whether the one-sided version of Hamachi’s example is lim sup full.

In general the distribution μ occurs on all positive coordinates and on each of the negative coordinates 0 through −m_{1} +1 of the sequence space. The distribution ν_{1} occurs on each of the next n_{1} negative coordinates. Then μ occurs on each of the next m_{2} negative coordinates. Then ν_{2} occurs on the next n_{2} — and so on.

The following figures illustrate the measure P displayed on the coordinates of Ω and the measure Q displayed on the coordinates of X.

P = ⋯ \overbrace{\underbrace{⋯}_{n_{3}}}^{ν_{3}} \overbrace{\underbrace{⋯}_{m_{3}}}^{μ} \overbrace{\underbrace{⋯}_{n_{2}}}^{ν_{2}} \overbrace{\underbrace{⋯}_{m_{2}}}^{μ} \overbrace{\underbrace{⋯ − m_{1}}_{n_{1}}}^{ν_{1}} \overbrace{\underbrace{⋯ 0}_{m_{1}}}^{μ} \overbrace{\underbrace{1 · · · · · · · · ·}_{∞}}^{μ}

Q = ⋯ \overbrace{\underbrace{⋯}_{n_{3}}}^{ν_{3}} \overbrace{\underbrace{⋯}_{m_{3}}}^{μ} \overbrace{\underbrace{⋯}_{n_{2}}}^{ν_{2}} \overbrace{\underbrace{⋯}_{m_{2}}}^{μ} \overbrace{\underbrace{⋯ − m_{1}}_{n_{1}}}^{ν_{1}} \overbrace{\underbrace{⋯ 0}_{m_{1}}}^{μ} .

Here μ = \{\frac{1}{2}, \frac{1}{2}\} is fixed, and ν_{i} = \{\frac{1}{1+λ_{i}} , \frac{λ_{i}}{1+λ_{i}} \}, where 2 > λ_{i} > 1 for all i with lim_{i→∞} λ_{i} = 1.

As in Hamachi [3], the values of λ_{i}, n_{i} and m_{i} are chosen by an inductive process - though not quite in the obvious order. While doing this induction we want to control three things, the first two of which already appear in [3]: (i) T should be nonsingular with respect to the measure P; (ii) T should be conservative with respect to the measure P; (iii) S should be not lim sup full with respect to the measure Q. The nonsingularity and conservativity of S follow from the corresponding properties for T.

In order to guarantee the nonsingularity of T with respect to P, we use Hamachi’s version [3] of Kakutani’s theorem [5]. In particular, we choose the λ_{i} such that ∑_{i=1}^{∞} ( log(λ_{i}))^{2} < ∞.

In order to guarantee the conservativity of T with respect to P, (again as in [3]) we force the Radon-Nikodym derivatives to sum to infinity, i.e., ∑_{i=0}^{∞} \frac{dP T^{i}}{dP} (ω) = ∞, almost everywhere mod P. The Radon-Nikodym deriatives for the measure P are analyzed by \frac{dP T^{i}}{dP} (ω) = ∏_{k=−∞}^{∞} \frac{dP T^{i}(ω_{k})}{dP (ω_{k})} = ∏_{k=−∞}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})}.

In order to guarantee that S is not lim sup full, we construct a distinguished set of positive measure E in X with the property that lim_{j→∞} Q(S^{j}E) = 0. The set E will be the intersection of a collection of sets A_{i} based on the disjoint coordinates associated to the measures ν_{i}.

4. The Induction Process

We construct the measure P by finding appropriate values for the sequences n_{i}, m_{i}, and λ_{i}, as well as for sets A_{i} used in constructing the set E described above. Along the way, we construct some auxilary increasing sequences N_{i}, K_{i}, L_{i} of integers and set m_{i} = N_{i} + N_{i+1}, n_{i} = ∑_{j=1}^{i} N_{j}, and L_{i} = ∑_{j=1}^{i} m_{j} + n_{j}. We also inductively construct reals I_{i,j} > 0 which are used in controlling the Radon-Nikodym derivatives.

Fix the sequence 1 > ϵ_{i} > 0 such that ∏_{i=1}^{∞}(1 − ϵ_{i}) > 0. The sequence ϵ_{i} is used in choosing the sets A_{i} and eventually in the construction of E.

We use the following lemma to select variables A_{i} and N_{i} below.

Lemma 4.1. Given measures ν ≠ μ on {0, 1}, and given 1 > ϵ > 0, there is an N > 0, such that for any natural number n > N there is a set

A ⊂ ∏_{j=1}^{n}{0, 1}_{j}

with the property that for the finite product measures,

(∏_{1}^{n} μ)(A) < ϵ (∏_{1}^{n} ν)(A) > 1 − ϵ.

Proof. Without loss of generality, we assume that μ(1) < ν(1). Put α = \frac{μ(1)+ν(1)}{2}, and define A_{n} = \{ω ∈ ∏_{i=1}^{n}\{0, 1\}_{i} : \frac{1}{n} ∑_{i=1}^{n} ω_{i} > α\}. Then (∏_{1}^{n} μ)(A_{n}) → 0 and (∏_{1}^{n} ν)A_{n} → 1. Thus, for any natural number n > N there is a set A ⊂ ∏_{j=1}^{n}{0, 1}_{j} with the property that for the finite product measures, (∏_{1}^{n} μ)(A) < ϵ and (∏_{1}^{n} ν)(A) > 1 − ϵ. □

Step 1. Choose 1 < λ_{1} < 2 to obtain ν_{1}. From Lemma 4.1, we obtain N_{1} > 0 and a set A_{1} satisfying (∏_{1}^{N_{1}} μ)(A_{1}) < ϵ_{1} and (∏_{1}^{N_{1}} ν_{1})(A_{1}) > 1 − ϵ_{1}. Set n_{1} = N_{1}, and initialize L_{0} = N_{1}. Choose K_{1} such that K_{1}λ_{1}^{−n_{1}} > 1. Define I_{1,1} = K_{1}λ_{1}^{−n_{1}}.

Note that we have specified n_{1}, λ_{1}, A_{1} and K_{1}, but we have not yet specified values for m_{1} or L_{1}. This is done in Step 2.

Step 2. Choose 1 < λ_{2} < 1 + 1/2 such that λ_{2}^{−(L_{0}+K_{1})}I_{1,1} > 1, and define I_{2,1} = λ_{2}^{−(L_{0}+K_{1})}I_{1,1}.

From Lemma 4.1, we obtain N_{2} > L_{0}+K_{1} and a set A_{2} satisfying (∏_{1}^{N_{2}} μ)(A_{2}) < ϵ_{2} and (∏_{1}^{N_{2}} ν_{2})(A_{2}) > 1 − ϵ_{2}. Now set m_{1} = N_{2} + N_{1}, and n_{2} = N_{2} + N_{1}. Let L_{1} = n_{1} + m_{1} and observe that L_{1} > n_{2}. Choose K_{2} such that K_{2}λ_{2}^{−n_{2}}λ_{1}^{−n_{1}} > 1. Define I_{2,2} = K_{2}λ_{2}^{−n_{2}}λ_{1}^{−n_{1}}.

The values for m_{2} and L_{2} will be specified in the next step.

Step 3. Choose 1 < λ_{3} < 1+1/3 such that λ_{3}^{−(L_{0}+K_{1})}I_{2,1} > 1, and λ_{3}^{−(L_{1}+K_{2})}I_{2,2} > 1. This is possible since I_{2,1} and I_{2,2} are both greater than 1. Define I_{3,1} = λ_{3}^{−(L_{0}+K_{1})}I_{2,1}, and I_{3,2} = λ_{3}^{−(L_{1}+K_{2})}I_{2,2}. From Lemma 4.1, we obtain N_{3} > L_{1}+K_{2} and a set A_{3} satisfying (∏_{1}^{N_{3}} μ)(A_{3}) < ϵ_{3} and (∏_{1}^{N_{3}} ν_{3})(A_{3}) > 1 − ϵ_{3}. Set m_{2} = N_{3} + N_{2}, n_{3} = N_{3} + N_{2} + N_{1} and L_{2} = n_{2} + m_{2} + L_{1}. Again L_{2} > n_{3}. Choose K_{3} such that K_{3}λ_{3}^{−n_{3}}λ_{2}^{−n_{2}}λ_{1}^{−n_{1}} > 1. Define I_{3,3} = K_{3}λ_{3}^{−n_{3}}λ_{2}^{−n_{2}}λ_{1}^{−n_{1}}.

Step t + 1. Suppose that m_{t−1}, L_{t−1}, n_{t}, λ_{t}, A_{t}, and K_{t} have been fixed. Also, I_{t,1}, I_{t,2}, ..., I_{t,t} are all defined and are greater than 1. Choose λ_{t+1} such that 1 < λ_{t+1} < 1 + 1/(t + 1) and λ_{t+1}^{−(L_{j−1}+K_{j})}I_{t,j} > 1 for all j ∈ {1, 2, ..., t}, and define I_{t+1,j} = λ_{t+1}^{−(L_{j−1}+K_{j})}I_{t,j}. Use Lemma 4.1 to obtain an integer N_{t+1} > L_{t−1}+K_{t} and a set A_{t+1} satisfying (∏_{1}^{N_{t+1}} μ)(A_{t+1}) < ϵ_{t+1} and (∏_{1}^{N_{t+1}} ν_{t+1})(A_{t+1}) > 1 − ϵ_{t+1}. Set m_{t} = N_{t+1} + N_{t}, L_{t} = n_{t} + m_{t} + L_{t−1}, and n_{t+1} = N_{t+1} + n_{t}. Choose K_{t+1} such that K_{t+1} ∏_{j=1}^{t+1} λ_{j}^{−n_{j}} > 1. Let I_{t+1,t+1} = K_{t+1} ∏_{j=1}^{t+1} λ_{j}^{−n_{j}}.

Remark 4.2. From the definitions, for fixed i ≥ 1 the sequence I_{t,i} is a decreasing sequence and lim_{t→∞} I_{t,i} ≥ 1.

5. Nonsingularity, Conservativity, and Not Lim Sup Full

In this section, we show that the one-sided shift map with the previously constructed measure P is a conservative, exact 2-to-1 endormorphism which is not lim sup full.

Lemma 5.1. T is nonsingular with respect to P.

Proof. From the construction of the measure P we have a sequence λ_{t} with the property that λ_{t} < 1 + 1/t for all t > 0. Therefore,

∑_{i=1}^{∞}(log(λ_{i}))^{2} ≤ ∑_{i=1}^{∞}(log(1 + 1/i))^{2} < ∑_{i=1}^{∞}(1/i)^{2} < ∞.

Thus T is nonsingular with respect to P [5]. □

It follows that S is also nonsingular with respect to Q.

Lemma 5.2. S not lim sup full with respect to Q.

Proof. We now construct a set E from the collection of sets A_{i} obtained in the previous section.

Let A′_{i} denote the set of all sequences in X with A_{i} placed on the leftmost N_{i} coordinates of n_{i}. This can be observed by the following illustration:

P = ⋯ \overbrace{\underbrace{⋯}_{N_{4}+N_{5}}}^{μ} \overbrace{\underbrace{A_{3} ⋯}_{N_{3}+N_{2}+N_{1}}}^{ν_{3}} \overbrace{\underbrace{⋯}_{N_{3}+N_{4}}}^{μ} \overbrace{\underbrace{A_{2} ⋯}_{N_{2}+N_{1}}}^{ν_{2}} \overbrace{\underbrace{⋯}_{N_{2}+N_{3}}}^{μ} \overbrace{\underbrace{A_{1}}_{N_{1}}}^{ν_{1}} \overbrace{\underbrace{⋯}_{N_{1}+N_{2}}}^{μ} \overbrace{\underbrace{· · · · · · · · ·}_{∞}}^{μ} .

Hence, for N_{1} ≤ j ≤ N_{1} + N_{2}, Q(S^{j}A′_{1}) < ϵ_{1}; for N_{1} + N_{2} ≤ j ≤ N_{1} + N_{2} + N_{3}, Q(S^{j}A′_{2}) < ϵ_{2}; and in general, Q(S^{j}A′_{k}) < ϵ_{k} for ∑_{i=1}^{k} N_{i} ≤ j ≤ ∑_{i=1}^{k+1} N_{i}.

Let E = ∩_{i=1}^{∞}A′_{i}. Since A′_{i} forms a disjoint collection of sets based in ν_{i} blocks, Q(E) = ∏_{i=1}^{∞} Q(A′_{i}) > ∏_{i=1}^{∞}(1 − ϵ_{i}) > 0.

Let ϵ > 0. Since lim_{i→∞} ϵ_{i} = 0, there is a natural number K such that for all k > K, ϵ_{k} < ϵ. Then for all j > N_{K}, there is a k > K such that ∑_{i=1}^{k} N_{i} ≤ j ≤ ∑_{i=1}^{k} N_{i+1}. So, Q(S^{j}(E)) = ∑_{i=1}^{∞} Q(S^{j}(A′_{i})) ≤ Q(S^{j}(A′_{k})) < ϵ_{k} < ϵ. Thus, lim_{j→∞} Q(S^{j}(E)) = 0, and S is not lim sup full. □

Lemma 5.3. T is conservative with respect to P.

Proof. To show conservativity we demonstrate that the sum of the Radon-Nikodym derivatives sum to infinity.

It is clear that

∑_{i=1}^{∞} \frac{dP T^{i}}{dP} (ω) ≥ ∑_{i=L_{0}}^{L_{0}+K_{1}−1} \frac{dP T^{i}}{dP} (ω) + ⋯ + ∑_{i=L_{t}}^{L_{t}+K_{t+1}−1} \frac{dP T^{i}}{dP} (ω) + · · · .

We will show that for all t ≥ 0, ∑_{i=L_{t}}^{L_{t}+K_{t+1}−1} \frac{dP T^{i}}{dP} (ω) ≥ 1. For each i we have \frac{dP T^{i}}{dP} (ω) = ∏_{k=−∞}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})}. Since the sum ∑_{i=L_{t}}^{L_{t}+K_{t+1}−1} \frac{dP T^{i}}{dP} (ω) has K_{t+1} terms,

we will show that for all t ≥ 0 the product ∏_{k=−∞}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ≥ K_{t+1}^{−1}.

Let a_{j} denote the left-end coordinate of the ν_{j} block and b_{j} denote the right-end coordinate of the ν_{j} block. Then

∏_{k=−∞}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} = ⋯ ∏_{b_{j+1}+i+1}^{a_{j}−1} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ∏_{k=a_{j}}^{b_{j}+i} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ⋯

… ∏_{b_{2}+i+1}^{a_{1}−1} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ∏_{k=a_{1}}^{b_{1}+i} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ∏_{k=b_{1}+i+1}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} .

It is immediate that ∏_{k=b_{1}+i+1}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} = 1 since all the P_{l}(ω_{k}) with k > b_{1} + i + 1 are the same, i.e., \{\frac{1}{2}, \frac{1}{2}\}.

Assume L_{0} ≤ i < L_{0} + K_{1}.

Remark 5.4. This means that there are at most L_{0} + K_{1} coordinates of each ν_{j} block moving to the left into a μ block, and there are at most L_{0} + K_{1} coordinates from the μ block on the right moving into the ν_{j} block. For the ν_{1} block there are only n_{1} < L_{0} +K_{1} coordinates to move and these move completely into the μ block to the left.

It is again immediate that for each j, ∏_{b_{j+1}+i+1}^{a_{j}−1} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} = 1, since all the P_{l}(ω_{k})

with b_{j+1} + i + 1 ≤ k ≤ a_{j−1} are the same, i.e., \{\frac{1}{2}, \frac{1}{2}\}. The term ∏_{k=a_{1}}^{b_{1}+i} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ≥ λ_{1}^{−n_{1}} = I_{1,1}K_{1}^{−1} > K_{1}^{−1} because the entire ν_{1} block, which is of length n_{1}, is moved into a μ block and simultaneously covered by a μ block. For j > 1, the terms ∏_{k=a_{j}}^{b_{j}+i} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ≥ λ_{j}^{−i} ≥ λ_{j}^{−(L_{0}+K_{1})}. This follows from the above remark. Hence,

the product ∏_{k=−∞}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ≥ lim_{t→∞} I_{t,1}K_{1}^{−1} ≥ K_{1}^{−1}.

Using a similar argument, we see that for s > 1, L_{s−1} ≤ i ≤ L_{s−1} + K_{s} − 1, ∏_{k=a_{s}}^{b_{1}+i} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ≥ ∏_{j=1}^{s} λ_{j}^{−n_{j}} = I_{s,s}K_{s}^{−1}. For all t > s, ∏_{k=a_{t}}^{b_{t}+i} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ≥ λ_{t}^{−i} ≥

λ_{t}^{−(L_{s−1}+K_{s})}. Therefore, ∏_{−∞}^{∞} \frac{P_{k−i}(ω_{k})}{P_{k}(ω_{k})} ≥ lim_{t→∞} I_{t,s}K_{s}^{−1} ≥ K_{s}^{−1}. □

Lemma 5.5. S exact with respect to Q.

Proof. Associated to (S, X, Q) is the usual odometer map R except that the coordinates of X are negative: if x is a point of X with x_{−i} = 1 for i = 0, 1, 2, ⋯, k − 1 and x_{−k} = 0 then (Rx)_{−i} = 0, i = 0, ⋯, k − 1, (Rx)_{−k} = 1 and (Rx)_{−j} = x_{−j} for all j > k.

From the construction of Q, it follows that (R, X, Q) is a nonsingular ergodic transformation [4]. It then follows from the Kolmogorov zero-one law that S is exact. □

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Western Carolina University, Cullowhee, NC 28723 jbarnes@wpoff.wcu.edu http://wcuvax1.wcu.edu/˜ jbarnes/

Northeastern University, Boston MA 02115 eigen@neu.edu http://www.math.neu.edu/˜ eigen/

This paper is available via http://nyjm.albany.edu:8000/j/2000/6-5.html.