1 Questions from Silva
1.1 Section 3.7 Problem 2.
Observe that the only T-invariant subset of X is X itself. Suppose we have a subset A of X such that A 6= X. Then there exists aj ∈ X such that a(j+1) mod n ∈6 A and hence, T(aj) 6∈ A. Thus A is not T-invariant. Hence, we must have A = X. But AC = XC = ∅. So µ(AC) = 0 whenever A is T-invariant, and so T is ergodic.
Problem 3.
Suppose T is a totally ergodic measure-preserving transformation and suppose that T is invertible. Since T is ergodic, observe that µ(A) = 0 or µ(AC) = 0 for any T-invariant set A. Now fix a T-invariant set A and suppose µ(A) = 0. Since T is measure-preserving, we have that if µ(A) = 0, then µ(T−1(A)) = 0.
For T−2(A), observe that we have
T−2(A) = T−1(T−1(A)) = T−1(B)
where B = T−1(A), a measure 0 set. Again, since T is measure-preserving and µ(B) = 0, we have that µ(T−1(B)) = µ(T−2(A)) = 0. Proceeding inductively, we get that if µ(A) = 0, then µ(Tn(A)) = 0 for all n < 0. Moreover, if A is T-invariant, then A is also T−1-invariant, since T is invertible. This again applies for Tn with any n < 0, so we have that Problem 6.
Recall the the two dimensional Baker’s transformation is defined as
y
(2x, ), if 0 ≤ x < 1/2
T(x,y) = 2
, if 1/2 ≤ x ≤ 1
For any subset A of [0,1] × [0,1], we have that µ(A) =
1
1.2 Section 3.10 Problem 1.
Problem 3.
Problem 4.
Problem 6.
