Dynamical -Final - Solved

Recall that a rotation of the plane is a linear map of the form

R : R2 → R2;
 ! x

R    = y
 

cosθ sinθ
−sinθ! x! cosθ y
This map preserves Lebesgue measure, but we want to show that it is never ergodic. We have that,

 ! ! ! x cosθ −sinθ x

                                                                      R          =

                                                                              y           sinθ      cosθ        y

    ! xcosθ − y sinθ

=

xsinθ + y cosθ

                                                                 !                                                                                  ! !

                                                    R−1     xy = cos 2 θ +1 sin2 θ −cossinθθ  cossinθθ          xy

      ! ! cosθ sinθ x

=

                                                                                  −sinθ     cosθ      y

     ! xcosθ + y sinθ

=

−xsinθ + y cosθ

Let us now consider the unit disk in R2, denoted by A = {(x,y) ∈ R2 : −1 ≤ x2+y2 ≤ 1}. We have that Lebesgue measure is a generalization of area in R2 and, since the unit disk has a well-defined area, we know its Lebesgue measure must be equal to 1.

Let us fix (x,y) ∈ A. Then −1 ≤ x2 + y2 ≤ 1. Applying R to this point, we get,

x0 = xcosθ − y sinθ y0 = xsinθ + y cosθ And note that,

(x0)2 + (y0)2 = (xcosθ − y sinθ)2 + (xsinθ + y cosθ)2

= x2 cos2 θ − 2xy cosθ sinθ + y2 sin2 θ + x2 sin2 θ + 2xy sinθ cosθ + y2 cos2 θ

= x2 cos2 θ + y2 sin2 θ + x2 sin2 θ + y2 cos2 θ

= x2(cos2 θ + sin2 θ) + y2(cos2 θ + sin2 θ)

= x2 + y2

 !

And so we have −1 ≤ (x0)2 + (y0)2 ≤ 1 as well. Thus, R                  x ∈ A and, since (x,y) ∈ A y

 ! x

was arbitrary, we have that (x,y) ∈ A =⇒ R                    ∈ A.

y

Now applying R−1 to an arbitrary (x,y), we get,

x0 = xcosθ + y sinθ y0 = −xsinθ + y cosθ

And note that,

(x0)2 + (y0)2 = (xcosθ + y sinθ)2 + (−xsinθ + y cosθ)2

= x2 cos2 θ + 2xy cosθ sinθ + y2 sin2 θ + x2 sin2 θ − 2xy sinθ cosθ + y2 cos2 θ

= x2 cos2 θ + y2 sin2 θ + x2 sin2 θ + y2 cos2 θ

= x2(cos2 θ + sin2 θ) + y2(cos2 θ + sin2 θ)

= x2 + y2

 !

And so we have −1 ≤ (x0)2 +(y0)2 ≤ 1 as well. Thus, R−1                      x           ∈ A and, since (x,y) ∈ A y

 !

was arbitrary, we have that (x,y) ∈ A =⇒ R−1                    x     ∈ A.

y

Thus, from the above derivations, we must have that A is strictly R-invariant. However, as we said above, note that A has measure 1. Moreover, Ac = {(x,y) ∈ R2 : −1 ≤ x2 +y2 ≤ 1}, which is a set of infinite measure (the area of the plane minus the unit disk). Hence, although A is strictly R-invariant, we do not have that µ(A) = 0, nor that µ(Ac) = 0. As a result, R is not ergodic for any value of θ.

Problem 2.

Let (X,S,µ) be a measure space and suppose that S : X → X and T : X → X are measure-preserving transformations. We want to show that their composition, S ◦ T, preserves measure as well.

Since S and T are measure-preserving transformations, for any A ∈ S(X) we have that µ(S−1(A)) = µ(A) and µ(T −1(A)) = µ(A). Let us consider (S ◦T)−1 = T −1 ◦S−1. We have, (T −1 ◦ S−1)(A) = T −1(S−1(A))

We have that, since S is measure-preserving, then µ(S−1(A)) = µ(A). Moreover, since S−1(A) is measurable, we have that S−1(A) ∈ S(X) and hence. Let us call this set B. Then we have,

T −1(S−1(A)) = T −1(B)

and, since T is measure-preserving, we have

µ(T −1)(B) = µ(B)

= µ(S−1(A))

= µ(A)

Sine A ∈ S(X), this holds for every set in S(X) and so S ◦ T is measure-preserving.

Problem 3.

a) Let us begin by showing that D is measurable using the Dyadic squares. Fix a set A in the Dyadic squares and consider D−1(A). Observe that for x,y ∈ [0,1/2), we have that 2x ∈ [0,1) and y ∈ [0,1). Hence, we can consider D(x,y) on [0,1/2) × [0,1/2) to be just D(x,y) = (2x,2y). Now, for x,y ∈ (1/2,1), we have that D(x,y) ∈ (0,1). Note that for x,y in this interval, we have 2x mod 1 = 2x−1 and 2y mod 1 = 2y−1. Thus, for x,y ∈ (1/2,1) we have D(x,y) = (2x − 1,2y − 1). Hence, rephrasing D, we

have,



(2x,2y)

(2x,2y − 1)

D(x,y) =

(2x − 1,2y)
x,y ∈ [0,1/2) x ∈ [0,1/2),y ∈ (1/2,1) x ∈ (1/2,1),y ∈ [0,1/2)


                                                                          (2x − 1,2y − 1)       x,y ∈ (1/2,1)

Hence, we can think of D(x,y) as a cross product of the tent map on [0,1) with µ = 2.

We can now formulate the inverse of D(x,y) as,



(x/2,y/2) (x/2,y/2 + 1/2)

−1

D       (x,y) =

(x/2 + 1/2,y/2)
x,y ∈ [0,1/2) x ∈ [0,1/2),y ∈ (1/2,1) x ∈ (1/2,1),y ∈ [0,1/2)
                                                                       (x/2 + 1/2,y/2 + 1/2)        x,y ∈ (1/2,1)

Denote each of the four cases of D−1(x,y) as Di−1(x,y) for i ∈ {1,...,4} and observe that λ(A) = 1/2k · 1/2k = 1/4k. Now let us check D−1(A). We have, λ(D−1(A)) = λ(D1−1(A)) + λ(D2−1(A)) + λ(D3−1(A)) + λ(D4−1(A))

Let us now split up each Di into Dix and Diy for the x and y components. In addition, write Ax for the x component of A and Ay for the y component. Then,

λ(Di−1(A)) = λ(Di−x1(Ax)) · λ(Di−y1(Ay))

Further, observe that λ(Di−x1(Ax)) =  λ(Ax) and λ(Di−y1(Ay)) =  λ(Ay) by the proof of Theorem 3.3.1 in the text. Note that λ(Ax) = 1/2k and λ(Ay) = 1/2k, so

                                                                                 λ(Di−1(A)) =                     λ(Ay)

=

  k

This is true for every i and thus,

λ(D−1(A)) = λ(D1−1(A)) + λ(D2−1(A)) + λ(D3−1(A)) + λ(D4−1(A))

                                                                    1     1       1     1      1     1       1     1

                                                                =      ·       +     ·       +     ·       +     ·

                                                                                k        4    4k        4    4k        4    4k

=

4k

= λ(A)

Since A was an arbitrary Dyadic square, we thus have that D is measure-preserving.

b) We want to show that D is ergodic. Let A1 and B1 be any sets of positive measure in L(X). Then, since the Dyadic squares form a sufficient semiring for Lebesgue measure, there are squares I and J such that,

  and  

Furthermore, we may assume that I and J are of the same measure. Write A = A1 ∩I and B = B1 ∩ J.

Now define I = [2pk1, p12+1k )×[2q1k, q12+1k ) and J = [2pm2 , p22m+1)×[2qm2 , q22+1m ). Since the doubling map is chaotic in 1 dimension, we have that its orbit is dense in both the x and y axes.

Thus, there is an integer n such that  . Thus, we have,

λ(Dn(A) ∩ B) ≥ λ(Dn(I) ∩ J) − λ(I \ A) − λ(J \ B)

 

and, by Lemma 3.7.2(5), we have that D is ergodic.

Problem 4.

Recall that B(R) is the smallest σ-algebra containing the open sets. Since X ∈ B(R), we have that B(X) = {A ∩ X : A ∈ B(R)}.

We are given that T : X → X is a Borel-measurable transformation for a set X ∈ B(R) (that is, X is a Borel set). Since T is Borel measurable, we have T −1(B) ∈ B(X) for all B ∈ B(X).

Define p : X → [−∞,+∞] by,



                                                          k             if x is periodic and has least period k

x 7→

                                                           +∞       if x is not periodic

where least period k means that k ≥ 1 is the minimal integer such that Tk(x) = x.

We want to prove that p is Borel measurable. Recall that p : X → R∗ is B(X) measurable iff p−1([−∞,a)) ∈ B(X) for every a ∈ R. Equivalently, p : X → R∗ is B(X)-measurable iff p−1(C) ∈ B(X) for every C ∈ B(R∗) = B(R) t P({±∞}).

Note that x ∈ X is period k if and only if Tk(x)−x = 0. So consider the set [−∞,a) for some a ∈ R. We have p−1([−∞,a)) is the set of x ∈ X with Tk(x) − x = 0 for some k ∈ N with k < a. Since we know k ≥ 1 and p(x) →7 +∞ when x is not periodic, we have that

p−1([−∞,a)) = p−1([1,a))

for a ≥ 1 and p−1([−∞,a)) = ∅

for a < 1. Since the empty set is trivially measurable, let us only consider a ≥ 1.

For a ≥ 1, we have that,

p−1([−∞,a)) = p−1([1,a))

a

= [{x ∈ X : Ti(x) − x = 0}

i=1

a

= [{x ∈ X : Ti(x) = 0}

i=1

Now, note that by Lemma 4.2.6, since T : X → X is a measurable transformation and X ⊂ R, we then have that T ◦ T = T2 is measurable. Proceeding inductively, we have that Tj is measurable for any j ≥ 1. Moreover, by Exercise 4.4.9, we have that since Tj is Borel measurable (and hence Lebesgue measruable), then {x ∈ X : Tj(x) = a} is measurable for any a. In particular, {x ∈ X : Tj(x) = x} is measurable. Hence, for a ≥ 1, we have that p−1([−∞,a)) = p−1([1,a)) is just a countable union of elements in B(X) and hence, since

B(X) is a σ-algebra, we have,

p−1([−∞,a)) = p−1([1,a))

  a

= [{x ∈ X :

i=1

Thus, p−1([−∞,a)) is measurable for any a ≥ 1. We have now covered every case of a ∈ R and so, we have that p−1([−∞,a)) ∈ B(X) for every a ∈ R. Thus, p is measurable.

Problem 5.

a)     Recall that a point x is periodic if Tk(x) for some k ≥ 1. Since almost every point of x is periodic under T, there is a set N with λ(N) = 0 which contains all of the point which are not periodic under T. Now fix a measurable set A ⊂ X. Consider the set A \ N. By the definitions of T and N, we have that A \ N must consist only of points which are recurrent. Thus, for every x ∈ A \ N, there exists k = k(x) > 0 such that Tk(x) = x ∈ A\N. Since λ(N) = 0 and A was arbitrary, we have that T is recurrent.

b)     Fix A ⊂ X and a point x ∈ A\N. Hence, n is periodic with period k ≥ 1. Clearly we have that {x,T(x),T2(x),...,Tk−1(x)} is a T-invariant set. Applying T −1 gives us,

{T −1(x),x,T(x),...,Tk−2(x)} = {Tk−1(x),x,T(x),...,Tk−2(x)}

= {x,T(x),T2(x),...,Tk−1(x)}

Observe that, since there are a finite number of elements in this orbit, we have that λ(A) = 0 and λ(X \ (A ∪ N)) = λ(X). Since λ(X) > 0, there must be “more” of these orbits outside of A in some sense. Moreover, observe that the orbits of each point are either disjoint or equal to each other. To show this, suppose that the orbits of two points x and y have a non-empty intersection. Then Tj(x) = Ti(y) for some i,j ≥ 1. But note that, if x has period k, then Tk−j(Tj(x)) = Tk(x) = x and thus Tk−j(Ti(y)) = Tk−j+i(y) = x. Thus we must have,

Tk−j+1(Tj(x)) = Tk+1(x) = T2(x)

= Tk−j+1(Ti(y))

= Tk−j+1+i(y)

Continuing by induction, we see that the orbit of y equals the orbit of x for every point after Ti(y). But note that y is also periodic with period `. Then there is some point in the orbit of x such that Tm(x) = y. Hence, for all n such that 1 ≤ n < i, we must have that Tn(y) is equal to an element of the orbit of x as well. Thus, these are the same orbit.

In the case where the intersection of the orbits is empty, it is clear that the two orbits are disjoint.

Let us now divide X into two sets, A containing some of the orbits and B containing orbits which are disjoint from all of the orbits in A. Observe that it is possible to construct two such sets of positive measure. If it were not, it would mean that every element, up to a set of 0 measure, would be contained within a single orbit. But note that each orbit has finite elements and thus has measure 0. Hence, this is a contradiction and there must be more than 1 orbit.

Thus, we have λ(A) > 0 and λ(B) > 0, with A ∩ B = ∅. Observe further that Ac = B ∪ N because B contains all the orbits which are not in A and N contains all the points which are not part of any orbit (we know the points of N are not part of any orbit because they are not periodic and every point of an orbit is periodic).

Thus, we have that λ(A) > 0 and λ(Ac) = λ(B t N) = λ(B) > 0 as well. In addition, by our discussion above, we have that every orbit in A is T-invariant and thus T −1(A) = A. Moreover, all of the orbits in B are T-invariant. Lastly, as previously stated, no element of N can be an element of an orbit. Since the orbits are exactly the set Nc, we have that N is T-invariant and so T −1(N) = N as well. Thus,

T −1(A) = A

and

T −1(Ac) = T −1(B t N) = B t N

but λ(A) > 0 and λ(Ac) > 0. As a result, we have violated the definition of ergodicity and T is not ergodic.

Problem 6.

a) We have that the set of intervals with rational endpoints forms a sufficent semi-ring for R. Hence, in order to show that S preserves Lebesgue measure, we must show that S−1(I) is measurable and that λ(S−1(I)) = λ(I) for any interval I with rational endpoints.

Observe that,

S−1(y1) = {y ∈ R : y = x + n for some x ∈ [0,1), n ∈ Z with T(x) + n + φ(x) = x1 + n1 = y1}

Now let us fix an interval with rational endpoints I = (n1 + p1/q1,n2 + p2/q2) with n1,n2,p1,q1,p2,q2 ∈ Z and 0 ≤ p1 < q1 and 0 ≤ p2 < q2. Assume n2 ≥ n1 and p2/q2 ≥ p1/q1 without loss of generality. We have that,

λ(A) = n2 + p2/q2 − (n1 + p1/q1)

and we must show that λ(S−1(I)) = λ(I) in order to show that S preserves Lebesgue measure. Observe that T is measure-preserving and is also the only part of S which maps onto the fractional part of any x ∈ I. Hence if we consider T −1(I), the range of values upon which T is acting is only the fractional components, (p1/q1,p2/q2) and so, we must have that,

λ(T −1(I)) = λ(T −1((p1/q1,p2/q2))

= λ((p1/q1,p2/q2)) = p2/q2 − p1/q1

b) Suppose T is ergodic and φ is integrable with R[0,1) φdλ 6= 0. We want to show that S is not recurrent. That is, we want to show that there is a measurable set A of positive measure such that there is no null set N ⊂ A with the property that for all x ∈ A \ N there is n = n(x) > 0 with Tn(x) ∈ A.

Let A be some set of positive measure and fix y ∈ A with y = x + n where x is the fractional part and n is the integral part. We have that Sk(y) = Tk(x) + n +

Pk−1 φ(Tix). Then we have, i=0

                                                                       1 k                1     k                       kX−1         i

 S (y) =  (T (x) + n + φ(T x)) k k i=0

and so, by the Birkhoff Ergodic Theorem, we have,

                                                          1 k                          1     k                       kX−1         i

lim         S (y) = lim      (T (x) + n +    φ(T x)) k→∞ k  k→∞ k    i=0

                                                                                           1                      k1 kX−1                                  i

= lim  T n + lim   φ(T x) k→∞ k k→∞ k k→∞ k i=0

Z

                                                                            = 0 + 0 +       φdλ

> 0

But note then that limk→∞ k1Sk(y) > 0 implies that Sk(y) → ∞. Thus, if we fix y in any interval with rational endpoints (a,b), then Sk(y) will leave that interval as k → ∞. But by Lemma 3.5.1, if S were recurrent, then there would be an increasing sequence mi > 0 with Smi(y) ∈ A \ N for some null set N for all i ≥ 0. However, if this were the case, then limk→∞ k1Sk(y) would equal 0. Thus, we have a contradiction and so S cannot be recurrent.