# An introduction to ergodic theory - Walters P.

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Hence M(X,T) can be considered to contain the periodic orbits. Some people hi e to think of an invanant probability for a continuous transformation as a generalisation of a periodic orbit.

Theorems 5.11 and 6.16 tell us that for an automorphism A of an /l-torus K" the set M(Kn, A) con’ .is many atomic measures. This is also true for the shift transformations (by Theorem 5.12).

We have already proved the following results (Theorems 5.15, 5.16) relating invariant measures to topological transitivity. If T.X-+X is a homeomorphism of a compact metric space X and if there exists p e M{ X, T) which is ergodic and gives non-zero measure to each non-empty open set then p({x £ A'|0r(x) = X}) = 1. Similarly, if such a p e M(X, T) exists for a continuous T\X -* X with TX = X then /i({x e X|{T"fx)}o is dense in A'}) = 1. From these results we concluded that ergodic affine transformations are one-sided topologically transitive.

§6.5 Unique Ergodicity

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Ii' this section we study those transformations for which M(X, T) is as small as possible i.e., contains only one member. It turns out that this is equivalent to strong behaviour in the ergodic theorem

Definition 6.2. A continuous transformation T:X -* X is a compact mclris-able space X is called uniquely eryodic if there is only one T invariant Borel probability measure on X, i.e., M(X, T) consists of one point.

If T is uniquely ergodic and M(X, T) = {^} then p is ergodic because it is an extreme point of M(X, T) (Theorem b.lO(iii)).

$6.5 Unique Ergodicity

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Unique ergocity is connected to minimality by:

Theorem 6.17. Suppose 7: X -» X is a homeomorphism of the compact metris-able space X. Suppose 7 is uniquely ergoilic and M(X T) = {/<}. Then T is mlmmul Ijj p{V) > 0 for all non-empty open sets U.

Proof. Suppose T is minimal. If U is open, Ц ф ф, then Л' = (J,r=-a Tn{U), so if n( U) = 0 then m(X) = 0, a contradiction.

Conversely, suppose p(U)> 0 for all open non-empty U. Suppose also that T is not minimal. There exists a closed set К such that TK = K, 0 ф К ф X. The homeomorphism T\K has an invariant Borel probability measure pK on К by Corollary 6.9.1 Denne p on X by p(В) = рк(К n В) for all Borel sets B. Then p e MIX, T) and p ф ,t< because p(X\K) > 0. as X\K is nonempty and open, while p(X\K) = 0. This contradicts the unique ergodicity of T. □

The map T\K~* К given by T(e2n,°) = e2"1"2, 0 e [0,1 ], is an example of a uniquely ergodic homeomorphism which is not minimal The point 1 e К is a fixed point for Tand Q(T) = {1} so that M(K, 7) = {i\'r

We have the following resuit (due to H. Furstenberg [1]) about homeomorphisms of the.unit circle K. We always write intervals on К anticlockwise so [z, u] denotes the anticlockwise closed interval beginning at z and ending at w. We shall use the fact, Droved in §6.6, that a minimal rotation of К is uniquely ergodic.

Theorem 6.18. Let T:K-+ К he a homeomorphism with no periodic points. Then T is uniquely ergodic. Moreover

(a) there is a continuous ^injection ф:К К and a minimal rotation S: К —> К with ф'Г = The map ф has the property that for eeth z e К, ф~ 1{z) is either a point or a closed sub-interval of К. ,

(b) if T is minimal the map ф is a homeomorphism so that every minimal homeomorphism of К is topologically conjugate to a rotauon.

Proof. Since T has no periodic points no member of V(K, T) can g’ve positive measure to a point ofK. Let/*], p2 e ЩК, 7”) and put v = j(p1 + p2) e M{K, T). Define ф:К -» К by ф(г) — exp(2Tt/v([l,r])). Since v has no points of positive measure we know that ф is continuous and is surjective.

For any three points zt, z2, z3 of К we have v([zl5z2]) + v([z2,r3]) = v(0i,z3]) mod 1, so

ф{Т{г)) = exp2Jftv([l, T(z)])

= exp2*i(v([l, T(l)]) + v([7(l), 7(z)]))

= е2л'*ф((1)) where a = exp(2ntv([l, 7(1)])).

In other words, if S(z) = е2яшг then фТ = S$.

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6 invariant Measures for Continuous Transformations

We now show a is irrational. If ap = 1 then ф{Тр:) = ф\:) Vr 8 К so thai v([l, Tp(i)]) = v([J,i])mod 1 Vr e K. Hence v([r, Грг]) = Omod 1 Vre/C. Since Tp has no fixed points there exist' <5 > 0such that d(z, Trz) > 6 Vr e K. Therefore each point of К is the end point of an interval of length <5 of zero v-measure. This is not possible. Therefore y. is irrational,

We now know S is minimal and so Haar measure m is the only element of M(K,S). Therefore v ф~х. Hi ° Ф~1, /ь ■ ф~1 all equal m. Let [«,/(] be an interval in K. Then (/>([а.]) = ancj ф- '(/)([и.]) =

[c,r/] where с = inf{r: v([r,a]) = 0} and d = sup{w:v([b,w]) = 0J. Since г(( / "л({[а, ft]) Д [cj, /?J) = 0 we have ц,{ф~ ‘ф([а, b]) Д [a ft]) = 0, / = 1,2, so/i,([a,fc]) = 1ц{Ф~1Ф(1*-1>])) = т(ф([а h])\ Hence /!,([«,/>]) = /ь|[»,/>]) and so nt — /ь. Therefore T is uniquely ergodic.

Let we К and consider the set ф~1(м). Lei г, e ф~х(\\). Then r2 e </>-,(*') ifTv([r,,z2J) = 0 or 1 so that </>_1(iv) is the largest closed interval with zero v-measure which contains z,.

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