Full text of DOI:10.1017/S0143385700000651

Ergod. Th. & Dynam. Sys. (2000), 20, 1215–1229
c
Printed in the United Kingdom
ꢀ 2000 Cambridge University Press
A non-singular inverse Vitali lemma with
applications
GENEVIEVE MORTISS
School of Mathematics, The University of New South Wales, NSW 2052, Australia
(Received 12 May 1998 and accepted in revised form 24 November 1999)
Abstract. A proof for a non-singular version of the inverse Vitali lemma is given. The
result is used to describe non-singular orbit equivalence within the framework of Rudolph’s
restricted orbit equivalence and in the construction of an alternative proof of the Hurewicz
ergodic theorem.
1. Introduction
In extending ergodic theory from a measure-preserving to a non-singular setting many of
the problems we encounter are technical rather than fundamental. Frequently, once these
technical barriers have been overcome, it is straightforward to establish a theorem for non-
singular actions by generalizing the corresponding measure-preserving result.
It is in this light that we examine the inverse Vitali lemma. As we show in this paper, a
non-singularized version of this result opens the door to a number of possibilities.
While the basic form of the non-singularizedproof is taken from the measure-preserving
proof given in [5], the ideas which allow us to extend the result are due to Halmos [1].
The first use of our main result is demonstrated in §3, in which we establish the
foundations of a non-singular version of Rudolph’s restricted orbit equivalence. This part
of our work is still in its early stages and does not yet deal with any form of entropy.
However, we are able to define non-singular orderings and sizes, and we can show that
much of the basic machinery still works in a non-singular context. In particular, we present
a size for non-singular orbit equivalence, which we denote as m0. The inverse Vitali lemma
is used to show that m0 is in fact a size and that all orbit equivalent dynamical systems give
rise to orderings which are m0 equivalent.
It will be assumed that the reader has some familiarity with [4], from which most of
our technical work in §3 is adapted. It should be noted that although the restricted orbit
equivalence as presented in its original form in [4] deals exclusively with Z actions, the
theory has since been extended to the actions of more general groups. (See [2] and [3].) At
this preliminary stage of development we have chosen to stay close to the original format
as we believe that in doing so we can better highlight the problems involved in extending
the theory to include non-singular actions.

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G. Mortiss
Our second application is an alternate non-Halmos type proof of the Hurewicz ergodic
theorem. Now that we have a non-singular inverse Vitali lemma at our disposal,
the Ornstein–Weiss style proof of the Birkhoff ergodic theorem given in [5] is easily
generalized.
2. The inverse Vitali lemma
Our proof of the non-singular version of the inverse Vitali lemma is based on that given for
the standard measure-preserving case in [5]. We first state the result.
LEMMA 2.1. (Inverse Vitali lemma) Let T be an ergodic, non-singular, non-periodic
invertible automorphism on (X, , µ). Further suppose that for some measurable A ⊂ X
with µ(A) > 0 there are measurable integer valued functions ik(x), jk(x) such that
limk→∞ jk(x) − ik(x) = ∞ for all x ∈ A. Then there is some subset B ⊂ A such
that for all x ∈ B there are functions i(x) ≤ 0 ≤ j(x) where i(x) = ik(x) , j(x) = jk(x)
j(x)
for some k depending on x, so that the orbit intervals I(x) = ∪i(x) T i(x) are all pairwise
disjoint and
µ(A \ ∪x∈BI(x)) < ꢀ.
There is only one real problem in trying to generalize the proof given in [5] to the
non-singular case. In covering a set with orbit intervals we obviously need to know how
much measure we are including with each step. For a measure-preserving transformation
T we know that if the length of an orbit interval is extended by a certain factor then its
associated mass will also be increased by the same proportion. In our non-singular proof
we need some similar control over the Radon derivatives. For this reason we first establish
an intuitively obvious fact.
P∞
LEMMA 2.2. Let ωi(x) = (dT −iµ/dµ)(x), and suppose i=0 ωi(x) = ∞ for µ almost
ωn(x)
all x ∈ X. Then
lim
= 0
n→∞ P
n−1 ωi(x)
i=0
for µ almost all x ∈ X.
Proof. For a real valued function q(x) and n ∈ Z+ let
n−1
X
qn(x) =
q(T i(x))ωi(x).
i=0
Then if qc(x) = ω1(x) − (1 + c), we have
qcn(x) = ωn(x) − c
n−1
X
ωi(x) − (1 + c).
i=1
For A ⊆ X a T invariant set and c ∈ R, let
∞
[
EA+(c) =
{x ∈ A : qcn(x) ≥ 0}
n=1

A non-singular inverse Vitali lemma with applications
1217
and
∞
[
EA−(c) =
{x ∈ A : qcn(x) ≤ 0}.
n=1
Then
Z
Z
qc(x) dµ ≥ 0 and
−qc(x) dµ ≥ 0.
EA−(c)
EA+(c)
For proof see [1]. So
Z
Z
ω1(x) dµ ≥ (1 + c)µ(EA+(c)) and (1 + c)µ(EA−(c)) ≥
ω1(x) dµ.
EA+(c)
EA−(c)
Now the sequence
ωn(x) − 1
i=0
an(x) =
P
n−1 ωi(x)
is bounded µ-almost everywhere. For if, for some set B ⊆ X, x ∈ B implies that an(x) is
unbounded then
\
[ \
B =
EX+(c)
EX−(−c)
c>0
c>0
and so by the above inequalities, µ(B) = 0.
Note that
ω
n+1(x) − ω1(x)
i=0
an(T (x)) =
P
n−1 ωi+1(x)
and so lim sup an(x) and lim inf an(x) are µ-almost everywhere T invariant. Hence if we
let
Aα,β = {x : lim inf an(x) < α < β < lim sup an(x)}
for α, β ∈ Q, then Aα,β is a T invariant set. So
Z
Z
ω1(x) dµ ≥ (1 + c)µ(E+ (c)) and (1 + c)µ(EA− (c)) ≥
ω1(x) dµ
Aα,β
α,β
EA+α,β (c)
EA−α,β (c)
for c ∈ R.
Now EX−(α) ∩ Aα,β = EA− (α), but Aα,β ⊆ EX−(α), so Aα,β = EA− (α). Similarly,
α,β
α,β
α,β
Aα,β = EA+ (β). So
Z
ω1(x) dµ ≥ (1 + β)µ(Aα,β
)
Aα,β
and
Z
(1 + α)µ(Aα,β) ≥
ω1(x) dµ.
Aα,β
But as α < β this is not possible unless µ(Aα,β) = 0. Hence limn→∞(ωn(x) − 1)/
P
n−1 ωi(x) exists µ-almost everywhere so it is easy to show that
i=0
ωn(x)
n−1 ωi(x)
lim
n→∞ P
= 0
i=0
on a set of full measure.

1218
G. Mortiss
P
Note that this proof could easily be extended to show that limn→∞ n+p ωi(x)/
ωi(x) = i0=.n
P
P−n
n
i=0 ωi(x) = 0 for any finite p ∈ N, or that limn→∞ −n(x)/
ω
i=0
LEMMA 2.3. Suppose that for x ∈ A ⊆ X, µ(A) > 0 there are bounded integer valued
functions i(x) ≤ 0 ≤ j(x) such that N1 < j(x)−i(x)+1 < N2 where N1 is so large that
j+1
j
dT −rµ
dµ
dT −rµ
dµ
X
X
(x) < (1 + 2ꢀ)
(x)
i−1
i
for all i, j ∈ N with j − i + 1 > N1 and ꢀ < µ(A)/16. Then there is subset A0 ⊂ A so
S
jk(x)
that if I(x) = i=i (x) T i(x) then I(x) is disjoint from I(x0) for x, x0 ∈ A0 where x = x0
k
and
ꢀ
ꢁ
[
µ(A) ≤ 4µ
I(x) .
x∈A0
Proof. Find a subset X1 ⊂ X of measure at least 1 − ꢀ, and N3 ∈ N so that for x ∈ X1,
ꢂ
N3
N3
dT −rµ
dT −rµ
X
X
(x)
(x) < ꢀ.
dµ
dµ
r=N3−N2
r=0
By using the induced transformation on the set X1 and building a Rokhlin tower of height
N3, construct a Kakutani skyscraper for T with base B ⊆ X1 such that each column has
height h(x) ≥ N3 for all x ∈ B.
For each x ∈ A there is a y ∈ B such that x = T p(x)(y) where 0 ≤ p(x) ≤ h(y) and
an interval [i(x), j(x)] to which we ascribe the ‘length’
j(x)
dT −(p(x)+r)
X
l(x) =
(y).
dµ
r=i(x)
h(y)−N2
We cover the points in A contained in P(y) = ∪
T r(y), (y ∈ B) by the following
r=0
method: list all x ∈ A ∩ P(y) in descending order of length l(x) i.e. label the points
x1, x2, . . . , etc, where l(xr) ≥ l(xr+1). Start with x1 and cover an interval of points
j(x1)
I(x1) = ∪
T r(x1)
i(x1)
in P(y) with length l(x1). Next choose xs where l(xs) is the largest length remaining with
I(xs) disjoint from I(x1).
Continue with this method of disjoint covering until no further intervals can be chosen
without intersecting with those already selected. For y ∈ B, define R(y) ⊂ [0, h(y) − N2]
as the set of integers such that k ∈ R(y) if and only if T k(y) ∈ A and I(T k(y)) is chosen
as a covering interval.
Let L(x) = 3l(x) and define
j0(x)
X
I0(x) =
T r(x)
i0(x)

A non-singular inverse Vitali lemma with applications
where i0(x) is the greatest integer such that
1219
i(x)−1
dT −(p+r)µ
dµ
X
(y) ≥ l(x)
r=i0(x)
and j0(x) is the least integer such that
j0(x)
dT −(p+r)µ
X
(y) ≥ l(x)
dµ
r=j(x)+1
where x = T p(y) for y ∈ B.
Now
A ∩ P(y) ⊆ ∪k∈R(y)I0(T k(y)).
For if we suppose otherwise then for some x0 ∈ P(y) we would have I(x0) disjoint from
all the selected orbit intervals I(T k(y)) (k ∈ R(y)). But this would mean that I(x0) would
already have been included in our covering, which would be a contradiction.
Defining A0 as the set
∪
y∈B
∪
k∈R(y) T k(y),
we can see that as
h(y)−N2
h(y)
dT −rµ
dµ
dT −rµ
dµ
X
X
(y) ≥ (1 − ꢀ)
(y)
r=0
r=0
for y ∈ B, then
Z
y∈B
X
µ(A) − ꢀ ≤
L(T k(y)) dµ(y)
k∈R(y)
0
≤ 3(1 + 2ꢀ)µ(∪x∈A I(x))
and as ꢀ < µ(A)/16,
0
µ(A) ≤ 4µ(∪x∈A I(x)).
Now we come to the real heart of the inverse Vitali argument. The following lemma
tells us that we can cover all but ꢀ in measure of a set with disjoint orbit intervals providing
we start with a suitable finite set of integer valued function pairs ik(x) ≤ 0 ≤ jk(x).
The covering is achieved by an inductive process, using Lemma 2.3 to cover a quarter
in measure of what remains uncovered at each stage. The final proof of the inverse
Vitali lemma is the construction of an appropriate set of functions from the infinite series
ik(x), jk(x).
LEMMA 2.4. Choose 1/2 > ꢀ > 0 and a large M ∈ N such that ꢀM/16 > 1. Suppose
we have a set B ⊆ X, µ(B) > ꢀ/2 such that for x ∈ B we have bounded integer valued
functions ik(x) ≤ 0 ≤ jk(x), (k = 1, 2, . . . , M). Let
niknf = inf (jk(x) − ik(x) + 1) and nksup = sup(jk(x) − ik(x) + 1)
x∈B
x∈B

1220
G. Mortiss
and let N1 < mink∈(1,2,...,M){niknf} be so large that
j+1
j
ꢃ
ꢄ
dT −rµ
ꢀ
dT −rµ
dµ
X
X
(x) < 1 +
(x)
dµ
16
i−1
i
for all x ∈ B and all i, j with i ≤ 0 ≤ j and j − i + 1 > N1. Further suppose that
jk(x)+nsup
k+1
ꢃ
ꢄ
dT −rµ
dµ
ꢀ
jk(x) dT −rµ
X
X
(x) < 1 +
(x)
2
dµ
ik(x)−nsup
ik(x)
k+1
for k = 1, 2, . . . , M − 1. Then there is a subset B0 ⊆ B, µ(B0) > 0 and measurable
functions i(x) ≤ 0 ≤ j(x) for x ∈ B0 with (i(x), j(x)) = (ik(x), jk(x)) for some k,
(k = 1, 2, . . . , M) depending on x so that the sets
j(x)
I(x) = ∪r=i(x)T r(x)
are pairwise disjoint for x ∈ B0 and
µ(B \ ∪x∈B0 I(x)) < ꢀ.
Proof. First of all we define
ik0 (x) = ik(x) − sup(jk+1(x) − ik+1(x) + 1)
jk0 (x) = jk(x) +xs∈uBp(jk+1(x) − ik+1(x) + 1)
x∈B
for k = 1, 2, . . . , M − 1 and iM0 (x) = i(x), jM0 (x) = jM(x). Using Lemma 2.3 we find a
subset B10 ⊂ B with
j10 (x)
I10(x) = ∪i=i0 (x)T i(x)
1
pairwise disjoint and
µ(B)
µ(∪x∈B0 I0(x)) ≥
.
1
4
Now take
B2 = B \ ∪ 0 I10(x).
x∈B1
If µ(B2) > ꢀ/2 we can reapply Lemma 2.3 to construct B20 ⊆ B2 so that
µ(B2)
µ(∪x∈B0 I20 (x)) ≥
.
2
4
Continue the induction, setting
k−1
Ii0(x).
0
Bk = B \ ∪
∪
x∈Bi
i=1
As long as µ(Bk) > ꢀ/2 we can find Bk0 ⊆ Bk so that
µ(Bk)
µ(∪x∈B0 Ik0 (x)) ≥
.
k
4

A non-singular inverse Vitali lemma with applications
1221
We do this up to M times using M function pairs ik0 (x), jk0 (x).
M
Now B0 = ∪k=1Bk0 is a disjoint union hence it is not hard to show that
j(x)
j(x0)
I(x) = ∪i=i(x)T i(x) and I(x0) = ∪i=i(x0)T i(x0)
are disjoint for x = x0. (Here i(x), j(x) are defined to be ik(x), jk(x) where k is the unique
integer 1 ≤ k ≤ M such that x ∈ Bk0 .)
Let BM+1 be the remaining piece of B which we have not yet covered with disjoint
orbit intervals I0(x), i.e.
M
I0(x).
0
BM+1 = B \ ∪k=1
∪
x∈Bk
We wish to prove that µ(BM+1) < ꢀ/2. Suppose µ(Bk) ≥ ꢀ/2 for all k = 1, 2, . . . , M+1.
Now
µ(∪x∈B0 I0(x)) < (1 + ꢀ/2)µ(∪x∈B0 I(x))
k
k
< µ(∪x∈B0 I(x)) + ꢀ/2µ(∪x∈B0 I0(x)).
k
k
So
µ(∪x∈B0 I(x)) ≥ (1 − ꢀ/2)µ(∪x∈B0 I0(x))
µ(Bk)
k
k
≥ (1 − ꢀ/2)
4
for all k = 1, 2, . . . , M and because of the disjointness of the Bk0 and the fact that
µ(BM+1) ≤ µ(Br) we can say
ꢀ
Mꢀ
µ(∪x∈B0 I(x)) ≥ M(1 − ꢀ/2)
>
> 1
8
16
which is a contradiction.
Hence
µ(B \ ∪x∈B0 I(x)) ≤ µ(BM+1) + µ(∪x∈B0 I0(x) \ I(x))
< ꢀ/2 + ꢀ/2µ(∪x∈B0 I(x))
< ꢀ
as required.
Proof of the inverse Vitali lemma. Choose ꢀ/2 > 0 and M ∈ N such that ꢀM/32 > 1.
We need to construct M function pairs on a set AM ⊂ A which satisfy the hypotheses of
Lemma 2.4.
First find an N ∈ N and a set A0 ⊂ A with µ(A0) > µ(A) − ꢀ/22, such that for any
j − i + 1 > N, i ≤ 0 ≤ j and all x ∈ A0,
j+1
j
ꢃ
ꢄ
dT −rµ
dµ
ꢀ
dT −rµ
dµ
X
X
(x) < 1 +
(x).
32
i−1
i
Now on a subset A1 ⊂ A0 with µ(A1) > µ(A0) − ꢀ/23, define bounded functions
ıˆ1(x) = ik(x)(x) and ˆ1(x) = jk(x)(x) where the k(x) are chosen measurably so that
jk(x)(x) − ik(x)(x) + 1 > N.

1222
G. Mortiss
Next construct ıˆ2(x) = ik(x), ˆ2(x) = jk(x)(x) bounded on A2 ⊂ A1 with µ(A2) >
µ(A1) − ꢀ/24 and the k(x) are chosen to ensure ˆ2(x) − ıˆ2(x) + 1 > N and that for all
x ∈ A2,
ˆ2(x)+nsup
ꢃ
ꢄ
dT −kµ
dµ
ꢀ
ˆ2(x) dT −kµ
1
X
X
(x) < 1 +
(x)
4
dµ
ıˆ2(x)−nsup
ıˆ2(x)
1
where ns1up = supx∈A ˆ1(x) − ıˆ1(x) + 1.
1
Continuing in this fashion we obtain M bounded function pairs ıˆk(x), ˆk(x) on AM.
Now for x ∈ AM the ıˆk(x), ˆk(x) listed in reverse order satisfy the required conditions for
Lemma 2.4. Hence we can cover all but ꢀ/2 in measure of AM with disjoint orbit intervals
S
I(x) = ˆ(x) T k(x) for x ∈ A0 ⊂ AM. But
ıˆ(x)
M+2
X
1
2k
µ(AM) > µ(A) − ꢀ
k=2
> µ(A) − ꢀ/2
and so clearly we are done.
3. Non-singular orderings and sizes
To begin our modification of restricted orbit equivalence we start with the definition of an
ordering.
Definition 3.1. For a non-singular ergodic transformation system (X, , µ, T ) an ordering
will be a map
α : X × X → Z
0
where x0 = T α(x,x )(x).
We will only be comparing orderings within orbit equivalence classes; that is, if we
write m(α1, α2) for some size m or any other expression involving the two orderings then
we are assuming that there is some invertible orbit map φ : X1 → X2 with µ1 ◦ φ ∼ µ2
and
φ(T1i(x)) = T j(i,x)(φ(x)).
2
Also, any orbit map between two orderings is assumed to induce an orbit equivalence.
Definition 3.2. We say that α1 and α2 differ by a co-boundary if there is an orbit map
φ : X1 → X2 where
φ(T1i(x)) = T j(i,x)(φ(x))
2
and a measurable function f : X1 → Z such that
α1(x, x0) − α2(φ(x), φ(x0)) = f (x) − f (x0).
LEMMA 3.3. The orderings α1, α2 differ by a co-boundary if and only if T1 and T2 are
conjugate.
The proof of this is an easy extension of the calculation of [4, pp. 4–6].

A non-singular inverse Vitali lemma with applications
1223
Notation. Let φ : X1 → X2 be an orbit map for orderings α1, α2 on X1, X2 respectively,
x,φ
and x ∈ X1. Define f α ,α : Z → Z by
1
2
fxα,2φ,α2 (i) = j
whenever α2(φ(x), φ(T1i(x))) = j. Denote
α1,α2
x,φ,(i,j)
5
as the permutation of (i, j) which re-orders the interval in the same order as fxα,1φ,α2 . We
will now discuss non-singular sizes. For a size m, the value
α1,α2
x,φ,(i,j)
m(5
)
will no longer depend just on the permutation. It may also depend on the interval (i, j),
on x, and on the action of T1 on (X1, 1, µ1). So the notation is in fact something of a
α1,α2
x,φ,(i,j)
α1,α3
x,ψ,(i,j)
α1,α2
x,φ,(i,j)
α1,α3
x,ψ,(i,j)
shorthand. Note that if 5
= 5
then m(5
) = m(5
).
For a size m (axioms described below) let
m(fxα,1φ,α2 ) = i→l−im∞,ijn→f ∞ m(5α1,α2
x,φ,(i,j)).
By axiom ii(a) below this is constant for almost all x by ergodicity; we shall denote it as
α1,α2
φ
m(fφα ,α ). When we write m(α1, α2) we mean m(f
) where φ is the identity map.
1
2
We now present our definition of a size in a non-singular context, generalizing axioms
i–vi given in [4, pp. 7–8]. Note that all our axioms are stated in terms of m(fφα ,α
)
1
2
α1,α2
x,φ,(i,j)
or m(5
) rather than in terms of m(α1, α2). Of course, these axioms may be in
need of refinement for future developments, however they are sufficient for our immediate
purposes.
Definition 3.4. We call m a size if it satisfies the following axioms:
i(a) m(id) = 0.
α1,α2
T1(x),φ,(i,j)
α1,α2
x,φ,(i+1,j+1)
ii(a) m(5
) = m(5
).
iii(a) If for orderings α1, α2 for all ꢀ > 0 there exists an orbit map φ : X1 → X2 such that
m(fφα ,α ) < ꢀ then for each ꢀ > 0 there exists an orbit map ψ : X2 → X1 such that
1
2
m(fψα ,α ) < ꢀ.
2
1
α1,α2
x,φ,(i,j)
iv(a) For all ꢀ > 0 there is a δ so that if m(5
I ⊂ (i, j) with
(x) < ꢀ
) < δ then for all but a subset
j
dT1−kµ1
dT1−kµ1
dµ1
X
X
(x)
dµ1
I
i
we have
5αx1,φ,α2 (k + 1) = 5αx,1φ,α2 (k) + 1.
vi(a) For all ꢀ > 0 there is a δ1 so that for each α2 with m(f α ,α ) < δ1 there is a δ2 such
1
2
φ
that for all α3 with m(f α ,α ) < δ2 we have m(f
) < ꢀ.
α1,α3
ψφ
2
3
ψ
We have included in this list of axioms only those which are either technically necessary
(i(a), ii(a), iii(a) and vi(a)) for our work here or intuitively desirable (iv(a)). In [4], axiom v
is needed mainly for the construction of m-entropy and its use in the m-equivalence
theorem. As we have not yet developed a non-singular version of m entropy we have
not given an analogue of axiom v.

1224
G. Mortiss
Definition 3.5. We say orderings α1 and α2 are m-equivalent (written α1m˜ α2) if for each
ꢀ > 0 there is an α2ꢀ differing from α2 by a co-boundary such that m(α1, α2ꢀ) < ꢀ.
By generalizing the proof of Theorem 2.1 in [4] we have the following.
LEMMA 3.6. Non-singular m-equivalence is an equivalence relation.
We now present our size for non-singular orbit equivalence.
Definition 3.7. (Size for non-singular orbit equivalence) Let
ꢂ j−1
X
dT1−kµ1
dµ1
dT1−kµ1
dµ1
X
m0(5αx,φ,α,(i,j)) =
(x)
(x)
1
2
k∈I
k=i
where I ⊂ (i, j − 1) is such that k ∈ I implies that
5αx1,φ,α2 (k + 1) = 5αx,1φ,α2 (k) + 1.
Our new m0 is still consistent with the measure-preserving size for orbit equivalence
given in [4]. Clearly m0 satisfies axioms i(a) and iv(a). For axiom ii(a) simply note that
ꢂ
dT −nµ
dµ
dT −(n+1)µ
dµ
dT −1µ
dµ
(T x) =
(x)
(x).
The following three results are modified versions of Lemmas 2.6, 2.7 and 2.8 from [4].
LEMMA 3.8. For orderings α1, α2, m0(f α ,α ) < a if and only if there is a set A ⊂ X1
1
2
φ
with µ1(A) > 1 − a such that x ∈ A implies that
fxα,1φ,α2 (1) = 1.
Proof. Let m0(f α ,α ) < a¯ < a < 1.
1
2
φ
So for almost all x ∈ X1 we have a sequence of intervals ik(x) → −∞, jk(x) → ∞
such that
m0(5αx,φ,α,(i (x),j (x))) < a¯.
1
2
k
k
Hence for each k ∈ N, and almost all x ∈ X1, there is a subset Sk,x ⊂ (ik(x), jk(x)) with
jk(x)
dT1−iµ1
dµ1
dT1−iµ1
dµ1
X
X
(x) > (1 − a¯)
(x)
i∈Sk,x
i=ik(x)
such that if l ∈ Sk,x then
α1,α2
x,φ,(ik(x),jk(x))
α1,α2
x,φ,(ik(x),jk(x))
5
(l + 1) = 5
(l) + 1.
Set ꢀ and N. Find a set A0 ⊂ X1 with µ1(A0) > 1 − ꢀ and an N1(ꢁN) so that if
j − i + 1 ≥ N1 and I ⊂ (i, j) has #I ≤ N then for x ∈ A0
j−N
j
dT1−lµ1
dµ1
dT1−lµ1
dµ1
X
X
(x) > (1 − ꢀ)
(x).
i+N
i

A non-singular inverse Vitali lemma with applications
1225
Restricting ourselves to function pairs ik(x), jk(x) with jk(x) − ik(x) + 1 > N1, we apply
the inverse Vitali lemma to cover all but ꢀ in measure of A0 with disjoint T1 orbit intervals;
i.e.
CN,ꢀ = ∪x∈B
∪
j(x) T1i(x), BN,ꢀ ⊂ A0
N,ꢀ
i(x)
is a disjoint union and µ1(A0 \ CN,ꢀ) < ꢀ. Now let
AN,ꢀ = ∪x∈B
As BN,ꢀ ⊂ A0 we have
∪
∩(i(x)+N,j(x)−N) T i(x).
Sk,x
N,ꢀ
1
j(x)
i(x)
µ1(∪S ∩(i(x)+N,j(x)−N)T1i(x)) > (1 − a¯ − ꢀ)µ1(∪ T1i(x)).
k,x
So
µ1(AN,ꢀ) > (1 − a¯ − ꢀ)µ1(CN,ꢀ
> 1 − a¯ − 3ꢀ.
)
Put ꢀ = 2−N . If A = lim supN AN,2−N then µ1(A) ≥ 1 − a¯ > 1 − a.
Now x ∈ A if and only if x ∈ AN ,2−N for an infinite sequence {Nr}∞r=1. If x = T1l(x¯)
r
r
for x¯ ∈ BN ,2−Nr , l ∈ Sk,x¯ ∩(i(x¯)+Nr, j(x¯)−Nr), then putting ur = i(x¯)−l, vr = j(x¯)−l
r
we have
α1,α2
x,φ,(ur ,vr )
α1,α2
x¯,φ,(i(x¯),j(x¯))
α1,α2
x,φ,(i(x¯),j(x¯))
f
(i) = f
(l + i) − f
(l).
α1,α2
x,φ,(ur ,vr )
Hence as f
(0) = 0 and l ∈ Sk,x¯ we have
α1,α2
x,φ,(ur ,vr )
5
(1) = 1,
and as r → ∞
α1,α2
x,φ,(ur,vr )
5
(1) → fxα,1φ,α2 (1) = 1
for x ∈ A with µ1(A) > 1 − a.
If such an A exists then if T1k(x) ∈ A then
α1,α2
x,φ
f α ,α (1) = f
(k + 1) − fxα,1φ,α2 (k)
1
2
T1k(x),φ
= 1.
So
5αx1,φ,α2 (k + 1) = 5αx,1φ,α2 (k) + 1.
But by the Hurewicz ergodic theorem [1],
ꢂ
dT1−kµ1
dµ1
dT1−kµ1
dµ1
n−1
n−1
X
X
lim
χA(T1k(x))
(x)
(x) = µ1(A) > 1 − a.
n→∞
k=0
k=0
Hence m0(f α ,α ) < a.
1
2
φ
LEMMA 3.9. The non-singular size m0 satisfies axiom vi(a).

1226
G. Mortiss
α2,α3
ψ
Proof. Choose ꢀ > 0. Suppose m0(f α ,α ) < ꢀ/2, and m0(f
) < δ where δ is chosen
1
2
φ
so that µ1(φ−1(E)) < ꢀ/2 for µ2(E) < δ, for all measurable E ⊂ X2.
By the previous lemma we have sets S1 ⊂ X1 and S2 ⊂ X2 with µ1(S1) > 1 − ꢀ/2 and
µ2(S2) > 1 − δ such that
fxα,1ψ,α◦3φ(1) = f α2,α3 (fxα,1φ,α2 (1)) = 1
φ(x),ψ
for x ∈ S1 ∩ φ−1(S2) ⊂ X1. Hence m0(fψα◦1,φα3 ) < ꢀ.
¯
LEMMA 3.10. For all α1, α2 and for all ꢀ > 0 we can find an orbit map φ such that
m0(fφα ,α ) < ꢀ. Hence m0 satisfies axiom iii(a) and further
2
1
−1
¯
α1m˜ 0α2.
Proof. Fix ꢀ > 0. Choose a δ > 0 such that for any measurable set E ⊂ X1 if
µ1(E) > 1 − δ then µ2(φ(E)) > 1 − ꢀ/3.
Choose N0 so large that for a set φ(A) with µ2(φ(A)) > 1 − ꢀ/3 if x ∈ φ(A) then
|α1(φ−1(x), φ−1(T2(x)))| < N0.
Now choose N(ꢁN0) so that on a set B ⊂ X1 with µ1(B) > 1 − δ/4, for all x ∈ B,
N1 > N and for any index set I ⊂ [0, N1 − 1] with #I ≤ N0 we have
dT1−iµ1
dµ1
N −1 dT1−iµ1
1
X
X
δ
(x) <
(x).
8
dµ1
I
i=0
Take the induced transformation TB where
TB(x) = T r (x)(x) for x ∈ B.
B
1
Build a Rokhlin Tower of height N1 covering all but δ/4 of B; i.e.
∪
N1−1TBi (C) ⊂ B, C ⊂ B
i=0
is a disjoint union and
µ1(∪N1−1TBi (C))
i=0
> 1 − δ/4.
µ1(B)
We now use this TB tower as the basis for a series of T1 towers of differing heights.
Partition C into sets Cp where
Cp = {x ∈ C : T N −1(x) = T1p−1(x)}.
T i(Cp) are disjoint unions by the disjointness of ∪N −1TB(C). Further
1
B
p−1
1
The towers ∪
i=0
1
i=0
q−1
∪
p−1T1i(Cp) ∩ ∪ T1i(Cq) = ∅ a.e.
i=0
i=0
if p = q.
¯
Now we construct a new orbit map φ between T1 and T2 as follows.

A non-singular inverse Vitali lemma with applications
1227
(i) If x ∈/ ∪p≥N
∪
p−1 T1i(Cp) then let
1
i=0
¯
φ(x) = φ(x).
p−1
i=0
(ii) Let x ∈ ∪
T i(Cp) for some p ≥ N1.
1
p−1
For each orbit segment ∪
T i(z), z ∈ Cp, take
i=0
1
p−1
i=0
p−1 j(i,z)
T (φ(x))
2
φ(∪
T1i(z)) = ∪
i=0
and rearrange the right-hand side of this expression so that the j(i, z) are in ascending
i
¯
order. Then let φ(T1 (z)) be the ith element in this rearranged sequence.
We define our new ordering:
α1ꢀ : X2 × X2 → Z
by
ꢀ
0
−1
α1(y, y ) = α1(φ (y), φ−1(y0)).
¯
¯
Clearly α1 and α1ꢀ differ by a coboundary. Let
p−N0−1
¯
T1 (Cp)) ∩ φ(∪p≥N ∪
i=N0
p−2 T1i(Cp)).
i
S = φ(A) ∩ φ(∪p≥N
∪
1
1
i=0
Now
α2,αꢀ
y,id
f
1 (1) = α1ꢀ(y, T2(y))
−1
= α1(φ (y), φ−1(T2(y))).
¯
¯
We shall prove that this expression is equal to 1 for y ∈ S.
If y ∈ S then y ∈ φ(A) so
|α1(φ−1(y), φ−1(T2(y)))| < N0.
Thus
φ
−1(T2(y)) = T1k(φ−1(y))
where |k| < N0 so as φ−1(y) = T1j (z) for some j, N0 ≤ j ≤ p − N0, z ∈ Cp, some
p ≥ N1. It follows that
p−1
T2(y) ∈ φ(∪
T1i(z)).
i=0
i
Let x = φ−1(y). Then x = T1 (z) for some i, 0 ≤ i ≤ p − 2.
¯
p−1
Hence T1(x) ∈ ∪i=0 T1i(z), from which we deduce
α1(φ (y), φ−1(T2(y))) = 1,
−1
¯
¯
for all y ∈ S.
All that remains to prove is that the set S is sufficiently large.
We know that
µ1(∪p−N0−1T1i(Cp)) > (1 − δ/4)µ1(∪p−1T1i(Cp))
i=N0
i=0
as Cp ⊂ B, and so
µ1(∪p≥N
∪
p−N0−1 T1i(Cp)) > 1 − δ.
1
i=N0

1228
G. Mortiss
Hence
µ2(φ(∪p≥N
∪
p−N0−1 T1i(Cp))) > 1 − ꢀ/3.
1
i=N0
Also for z ∈ Cp,
φ(∪p−2T1i(z)) = φ(∪p−1T1i(z) \ T k(z)(z))
¯
i=0
i=0
1
for some k(z) ∈ [0, 1, . . . , p − 1]. Hence, as
µ1(∪z∈C
∪
p−1 T1i(z) \ T k(z)(z)) > (1 − δ/4)µ1(∪p−1T1i(Cp)),
p
i=0
1
i=0
we have
µ1(∪p≥N
∪
z∈Cp
∪
p−1T1i(z) \ T k(z)(z)) > (1 − δ/4)(1 − δ/4)µ1(B) > 1 − δ.
1
i=0
1
So
p−2 T i(Cp))) = µ2(φ(∪p≥N
∪
∪
p−1T1i(z) \ T k(z)(z)))
z∈Cp
i=0
¯
µ2(φ(∪p≥N
∪
1
1
1
i=0
1
> 1 − ꢀ/3.
Lastly, of course, we have chosen φ(A) so that µ2(φ(A)) > 1 − ꢀ/3. Hence µ2(S) >
1 − ꢀ.
4. The Hurewicz ergodic theorem
THEOREM 4.1. Suppose T is non-singular on the Lebesgue probability space (X, , µ)
and let f ∈ L1(µ). Then if
ꢂ
n−1
An(f, x) = n−1 f (T i(x))ωi(x)
ωi(x)
X
X
i=0
i=0
then An(f, x) converges to a finite limit for almost all x ∈ X. Further if An(f, x) → f ?(x)
R
R
then E f ? dµ = E f dµ for all T -invariant sets E.
Proof. First we deal with periodic points. If Xn is the set of points in X with period n then
it is easy to show that
ꢂ
n−1
n−1
X
X
f ?(x) = lim Ak(f, x) =
f (T i(x))ωi(x)
ωi(x)
k→∞
i=0
i=0
for all x ∈ Xn.
As each Xn is an invariant set we can deal with the points in X by assuming that
∞
ˆ
ˆ
ˆ
T is aperiodic. Let f (x) be defined as limn→∞ sup An(f, x). As f (T (x)) = f (x) we
ˆ
ˆ
will assume that f ≥ 0. Let E be a T -invariant set. We will examine f on the sets
ˆ
ˆ
EM = {x ∈ E : f (x) ≤ M} and E = {x ∈ E : f (x) = ∞}. For any M > 0 and ꢀ > 0
∞
we can define functions ik(x) = 0 and jk(x) so that
(1)
A
j (x)+1(f, x) > 1/ꢀ, for x ∈ E and
∞
k
ˆ
(2) |Aj (x)+1(f, x) − f (x)| < ꢀ for x ∈ EM.
k

A non-singular inverse Vitali lemma with applications
1229
By the inverse Vitali lemma there are functions i(x) = ik(x)(x), j(x) = jk(x)(x) and a set
j(x)
F ⊂ X with µ(F) > 0 such that the intervals I(x) = ∪i(x) T i(x) are disjoint for x ∈ F
and µ(B) = µ(∪x∈F I(x)) > 1 − ꢀ. Now
Z
j(x)+1
X
µ(E∞ ∩ B) =
ωi(x) dµ
E
∞∩F
i=0
Z
j(x)+1
X
≤ ꢀ
A
j(x)+1(f, x)
ωi(x) dµ
E
∞∩F
i=0
Z
= ꢀ
f dµ.
E
∞∩B
So letting ꢀ → 0 we see that µ(E ) = 0. For each EM and ꢀ > 0 we have
ꢅ∞
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
ꢅ
Z
Z
j(x)
ꢅ
ꢅ
ꢅ
ꢅ
X
ˆ
ˆ
f − f dµ ≤
(f (x) − Aj(x)+1(f, x))
ω (x) dµ
i
ꢅ
EM ∩B
EM ∩F
i=0
Z
≤
|f (x) − Aj(x)+1(f, x)| j(x) ωi(x) dµ
X
ˆ
EM ∩F
Z
i=0
j(x)
X
< ꢀ
ωi(x) dµ
EM ∩F
i=0
≤ ꢀµ(EM ∩ B) ≤ ꢀ.
R
R
ˆ
Again, let ꢀ → 0 and M → ∞ we get E f dµ = E f dµ for all T -invariant sets E.
Repeating the same procedure for lim infAn(f, x) we see that limn→∞ An(f, x) = f ?
R
R
exists as a finite limit and further that E f ? dµ = E f dµ for all T -invariant sets E.
Acknowledgements. Much of this work forms part of the author’s PhD thesis. The author
wishes to thank Tony Dooley and Dan Rudolph for their comments and suggestions and
the Australian Research Council for their financial support.
REFERENCES
[1]
[2]
P. R. Halmos. An ergodic theorem. Proc. Natl. Acad. Sci. USA 32 (1946), 156–161.
d
J. W. Kammeyer and D. J. Rudolph. Restricted orbit equivalence for ergodic
actions I. Ergod. Th. &
Dynam. Sys. 17 (1997), 1083–1129.
[3]
J. W. Kammeyer and D. J. Rudolph. Restricted orbit equivalence for actions of discrete amenable groups.
Preprint.
[4]
[5]
D. J. Rudolph. Restricted orbit equivalence. Mem. Amer. Math. Soc. 323 (1985).
D. J. Rudolph. Fundamentals of Measurable Dynamics. Oxford University Press, New York, 1990.