In mathematics, the Rokhlin lemma, or Kakutani–Rokhlin lemma is an important result in ergodic theory. It states that an aperiodic measure preserving dynamical system can be decomposed to an arbitrary high tower of measurable sets and a remainder of arbitrarily small measure. It was proven by Vladimir Abramovich Rokhlin and independently by Shizuo Kakutani. The lemma is used extensively in ergodic theory, for example in Ornstein theory and has many generalizations.
Rokhlin lemma belongs to the group mathematical statements such as Zorn's lemma in set theory and Schwarz lemma in complex analysis which are traditionally called lemmas despite the fact that their roles in their respective fields are fundamental.
A Lebesgue space is a measure space
(X,lB,\mu)
We consider only measure-preserving maps. As typical in measure theory, we can freely discard countably many sets of measure zero.
An ergodic map is a map
T
T-1(A)=A
A
X-A
An aperiodic map is a map such that the set of periodic points is measure zero:A Rokhlin tower is a family of sets that are disjoint. is called the base of the tower, and each is a rung or level of the tower. is the height of the tower. The tower itself is
R:=(S\cupTS\cup...\cupTN-1S)
There are several Rokhlin lemmas. Each states that, under some assumptions, we can construct Rokhlin towers that are arbitrarily high with arbitrarily small error sets.
The Rokhlin lemma can be used to prove some theorems. For example, (Section 2.5)
(Section 4.6)
Ornstein isomorphism theorem (Chapter 6).
Let
style(X,T)
styleX
styleT:X → X
style(X,T)
styleT
Tkx=x
x\inX
k\in{Z
k=0
style(Y,S)
style(X,T)
style\varphi:X → Y
style\varphi(Tx)=S\varphi(x)
stylex\inX
Elon Lindenstrauss proved the following theorem:[3]
Theorem: Let
style(X,T)
stylen\in\N
stylef\colonX → \R
styleE=\{x\inX\midf(Tx) ≠ f(x)+1\}
styleE,TE,\ldots,Tn-1E
Gutman proved the following theorem:[4]
Theorem: Let
(X,T)
\varepsilon>0
f\colonX → \R
styleE=\{x\inX\midf(Tx) ≠ f(x)+1\}
\operatorname{ocap}(styleE)<\varepsilon
\operatorname{ocap}
Proofs taken from.
Proposition. An ergodic map on an atomless Lebesgue space is aperiodic.
Proof. If the map is not aperiodic, then there exists a number , such that the set of periodic points of period has positive measure. Call the set . Since measure is preserved, points outside of do not map into it, nor the other way. Since the space is atomless, we can divide into two halves, and maps each into itself, so is not ergodic.
Proposition. If there is an aperiodic map on a Lebesgue space of measure 1, then the space is atomless.
Proof. If there are atoms, then by measure-preservation, each atom can only map into another atom of greater or equal measure. If it maps into an atom of greater measure, it would drain out measure from the lighter atoms, so each atom maps to another atom of equal measure. Since the space has finite total measure, there are only finitely many atoms of a certain measure, and they must cycle back to the start eventually.
Proposition. If is ergodic, then any set satisfies (up to a null set)Proof. is a subset of , so by measure-preservation they are equal. Thus is a factor of , and since it contains , it is all of .
Similarly, is a subset of , so by measure-preservation they are equal, etc.
Let be a set of measure . Since is ergodic, , almost any point sooner or later falls into . So we define a “time till arrival” function: with if never falls into . The set of is null.
Now let .
By a previous proposition,
X
If we can pick a near-zero set with near-full coverage, namely some such that , then there exists some , such that , and since
T-i(TnA)\supsetTn-iA
i=0,1,2,...
TnA
Thus, our task reduces to picking a near-zero set with near-full coverage.
Pick . Let be the family of sets such that are disjoint. Since preserves measure, any has size .
The set nonempty, because . It is preordered by iff . Any totally ordered chain contains an upper bound. So by a simple Zorn-lemma–like argument, there exists a maximal element in it. This is the desired set.
We prove by contradiction that . Assume not, then we will construct a set , disjoint from , such that , which makes no longer a maximal element, a contradiction.
Since we assumed , with positive probability,
x\not\in\cupk\inTkA
Since is aperiodic, with probability 1,And so, for a small enough , with probability ,And so, for a small enough , with probability , these two events occur simultaneously. Let the event be .
It suffices to prove the case where only the base of the tower is probabilistically independent of the partition. Once that case is proved, we can apply the base case to the partition .
Since events with zero probability can be ignored, we only consider partitions where each event has positive probability.
The goal is to construct a Rokhlin tower with base , such that
\mu(S'\capPi)=
1-\epsilon | |
N |
\mu(Pi)
Given a partition and a map , we can trace out the orbit of every point as a string of symbols , such that each . That is, we follow to , then check which partition it has ended up in, and write that partition’s name as .
Given any Rokhlin tower of height , we can take its base , and divide it into equivalence classes. The equivalence is defined thus: two elements are equivalent iff their names have the same first- symbols.
Let be one such equivalence class, then we call a column of the Rokhlin tower.
For each word , let the corresponding equivalence class be .
Since is invertible, the columns partition the tower. One can imagine the tower made of string cheese, cut up the base of the tower into the equivalence classes, then pull it apart into columns.
Let be very small, and let be very large. Construct a Rokhlin tower with levels and error set of size . Let its base be . The tower has mass .
Divide its base into equivalence classes, as previously described. This divides it into columns where ranges over the possible words .
Because of how we defined the equivalence classes, each level in each column falls entirely within one of the partitions . Therefore, the column levels almost make up a refinement of the partition , except for an error set of size .
That is,The critical idea: If we partition each equally into parts, and put one into a new Rokhlin tower base , we will have
Now we construct a new base as follows: For each column based on , add to , in a staircase pattern, the setsthen wrap back to the start:and so on, until the column is exhausted.The new Rokhlin tower base is almost correct, but needs to be trimmed slightly into another set
S''
\mu(S''\capPi)=
1-\epsilon | |
N |
\mu(Pi)
The new Rokhlin tower , contains almost as much mass as the original Rokhlin tower. The only lost mass is due to a small corner on the top right and bottom left of each column, which takes up proportion of the whole column’s mass. If we set , this lost mass is still . Thus, the new Rokhlin tower still has a very small error set.
Even after accounting for the mass lost from cutting off the column corners, we still have
\end
Since there are only finitely many partitions, we can set , we then haveIn other words, we have real numbers such that
\mu(S'\capPi)=
1-ci\epsilon | |
N |
\mu(Pi)
Now for each column , trim away a part of into , so that
\mu(S''\capPi)=
1-\epsilon | |
N |
\mu(Pi)