[Solved] Dynamical Midterm

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Problem 1.

Let

x <

T(x) =

1 4

2 x x

Note then that,
x21 T (x) = x < 14

x

In addition, note that for x , we have T and for x, we have

  1. T. Thus, for any y in the image of T, we will need to consider two pre-images under the inverse function T 1(x).

Now let C be the set of intervals in . This is a sufficient semi-ring for the domain. Let us fix some interval I with endpoints a,b C with b a (whether it is open or closed will not change its Lebesgue measure). We will assume the interval is open for now. Then,

(I) = b a

Now let us consider T 1(I). We have two inverse images to consider,

T11(I) = (a2,b2)

and,

T21(I) = (a2 + a + ,b2 + b + )

Note that the inverse of T preserves the interval structure, so T 1(I) is a measurable set.

Now let us check (T 1(I)). We have,

(T 1(I)) = (T11(I)) + (T21(I))

= b2 a2 + (b2 + b (a2 + a + ))

= b2 a2 b2 + b + a2 a

= b a

= (I)

Hence, by Theorem 3.4.1, T is measure preserving.

Problem 2.

We have that T is continuous and measure preserving and that f is continuous and f(T(x)) f(x).

Now suppose T is recurrent. Then for every measurable set A of positive measure, there is a null set N A such that for all x AN there is an integer n = n(x) > 0 with Tn(X) A.

Fix x0 R2 and suppose f(T(x0)) > f(x0). Since f,T are continuous, for points x1 near x0, we would expect to see f(T(x1)) > f(x1) as well.

Problem 3.

  • Let Cm,n be the set of points x X for which m and n are consecutive visit times to A. Suppose Cm,n is not measurable. Then Cm,n 6 S. But since T is measurable and T 1(Tm+1(X)) = Tm(X) S. Furthermore, since A measurable, we must have ATm(X) is measurable as well. The same applies to T1(Tn+1(X)) = Tn(x). Hence, A Tn(X) is measurable as well. Then we have that

A Tn(X) A Tm(X)

must be measurable. Finally, we have,

nm !

A Tm(X) A Tn(X) [ Tm+i(X) A

i=1

is measurable since we are taking a finite union of measurable sets and then set minusing this finite union from another measurable set. However, note that this is exactly Cm,n, and so Cm,n is measurable.

  • Now let us take Cm,n and define the following set:

nm1 !

Cm,n [ Tm+i(X) B

i=1

That is, we have taken the intersection of Cm,n with the union of the set of points in X such that Ti(X) B for some i with m < i < n. Note that, by the same arguments as above Tm+i(X) B is measurable for every i. Furthermore, we are taking a finite union of measurable sets, so Sni=1m1 Tm+i(X) B is measurable as well. And lastly,

Cm,n is measurable by (a), so

nm1 !

Dm,n = Cm,n [ Tm+i(X) B

i=1

is measurable.

(c) We have that x E if and only if x Cm,n implies x Dm,n for all integers m and n with 0 m < n. That is,

Problem 4.

Let us show that T is continuous. That is, we want to show that for each x , we have that for all > 0, there exists > 0 such that d(T(x),T(y)) < whenever y and d(x,y) < . First let us fix x and > 0. Let m be the length of the initial constant sequence in x. Let = 21m. Then for any y with d(x,y) < , we have that the first position where x 6= y is some k such that

k 1

1/2 < =

2m

= 2m2k <

= 2mk <

= 1/2km <

That is, after removing the first m elements of x and y (call these new elements x0, y0), we still have that d(x0,y0) < . Now note that T(x) will remove exactly the first m elements. Since k m and xi = yi for all 0 i k 1, we thus also have that T(y) removes exactly the first m elements of y. Hence, by what we have just shown, for any fixed x and

> 0, we have that for all y. Thus, T is continuous.

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[Solved] Dynamical Midterm
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