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
- 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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