Dynamical -Homework #8 - Solved

1              Questions from Silva
1.1            Section 3.11
Problem 3. Revised problem: Let (X,S,µ,T) be an invertible, recurrent, finite measurepreserving dynamical system. If A is a set of positive measure such that the transformation TA is ergodic on A and µ(X \ ∪n≥0T−n(A)) = 0, then T is ergodic.

Note: X \ ∪n≥0T−n(A) is the set of points that never hit A.

1.2            Section 4.2
Problem 3.

Suppose A is measurable and consider IA. Observe that IA(x) = 1 if x ∈ A and IA(x) = 0 if x 6∈ A. Then we must have that,

{x ∈ X : IA(x) > 0} = A

Hence, this set must be measurable and thus, by Proposition 4.2.1, we have that IA is measurable.

Now suppose IA is measurable. Then again by Proposition 4.2.1, we can say that {x ∈ X : IA(x) > 0} is measurable. Since this set is equal to A by the definition of IA, we must have that A is measurable as well.

Problem 5.

Note that f−1(A ∪ B) = {x ∈ X : f(x) ∈ A ∪ B}. But note that f(x) ∈ A ∪ B is the same as f(x) ∈ A or f(x) ∈ B (inclusive or). Hence we have that,

f−1(A ∪ B) = {x ∈ X : f(x) ∈ A ∪ B}

= {x ∈ X : f(x) ∈ A or f(x) ∈ B}

= {x ∈ X : f(x) ∈ A} ∪ {x ∈ X : f(x) ∈ B}

= f−1(A) ∪ f−1(B)

Now consider f−1(A ∩ B). We have that,

f−1(A ∩ B) = {x ∈ X : f(x) ∈ A ∩ B}

= {x ∈ X : f(x) ∈ A and f(x) ∈ B}

= {x ∈ X : f(x) ∈ A} ∩ {x ∈ X : f(x) ∈ B}

= f−1(A) ∩ f−1(B)

Lastly, consider f−1(R \ A). We have that,

f−1(R \ A) = {x ∈ X : f(x) ∈ R \ A}

= {x ∈ X : f(x) ∈ Ac}

Note that the final line in the aboe is the set of x in X that map to Ac in R. In other words, it is the complement of the set of points that map to A in R. Hence, we have,

{x ∈ X : f(x) ∈ Ac} = {x ∈ X : f(x) ∈ A}c

= f−1(A)c

= X \ f−1(A)

as required.

Now consider f(A ∪ B). Let x ∈ f(A ∪ B). Then x ∈ {y ∈ R : f−1(y) ∈ A ∪ B). Thus, x ∈ A or x ∈ B, or both. That is, x ∈ {y ∈ R : f−1(y) ∈ A) or x ∈ {y ∈ R : f−1(y) ∈ A) or both. Hence, we have that f(A ∪ B) ⊂ f(A) ∪ f(B). Now fix x ∈ f(A) ∪ f(B). Then f−1(x) ∈ A or f−1(x) ∈ B. But this is exactly the definition of f(A∪B) and so x ∈ f(A∪B). Thus, f(A) ∪ B ⊂ f(A ∪ B) and hence, f(A ∪ B) = f(A) ∪ f(B).

For f(A ∩ B), let us use a counterexample. Consider f(x) = x2 and A = [−1,0], B = [0,1]. Then f(A ∩ B) = f({0}) = 0, but f(A) ∩ f(B) = [0,1] ∩ [0,1] = [0,1]. Hence, these are not equal and this does not hold.

Lastly, consider f(X \ A). Again let us use f(x) = x2 as a counterexample with X = R and A = (0,1]. Then f(X \ A) = R. However, f(X) \ A = R \ [0,1]. Hence, these are not equal and so this does not hold.

Problem 6.

Suppose that f is Lebesgue measurable. Then, by Proposition 4.2.1 and Lemma 4.2.2, the inverse image under f of any interval is a measurable set. Now fix G ∈ R such that G is an open set. Since every open subset of R is a countable union of disjoint open intervals, we have that G  Ik for some disjoint open intervals Ik. Now note that disjoint set in a function’s image must have disjoint preimages. Otherwise, there would be an x such that f(x) has two outputs, which is not possible for a validly defined function. Hence, we must have that,

f−1(G) = f−1 (t∞k=1Ik)

= t∞k=1f−1(Ik)

Since each Ik is an interval, we have that f−1(Ik) is measurable. And since the countable union of measurable sets is measurable, we have that f−1(Ik) = f−1(G) is measurable, as required.

Now suppose f−1(G) is measurable for every open set G ⊂ R. Then, in particular, the preimage of every open interval in R is measurable. Hence,

{x ∈ X : f(x) < a}

is measurable for all a ∈ R and so f is measurable.

Problem 7.

Suppose g : R → R is Lebesgue measurable and f : R → R is continuous. We want to show that f ◦ g = f(g(x)) is Lebesgue measurable.

Note that R is a metric space and f is continuous on R. Hence, by Lemma 4.2.3, f is Lebesgue measurable as well. Now fix B ∈ B(R). We have that,

(f ◦ g)−1(B) = g−1(f−1(B))

We have that f−1(B) ∈ L(R) since f is Lebesgue measurable. We need to use the continuity of f to show that f−1(B) is Borel.

A Borel set is any set that can be formed from open sets through the operations of countable union, countable intersection, and complement. The inverse images of open sets under a continuous function are open sets and inverse images of a countable union is the countable union of the inverse images. The same notions hold true for complements and countable intersections. Hence, we can write f−1(B) as an expression of open sets through countable unions, countable intersections, and complements.

Hence, f−1(B) is a Borel set and so g−1(f−1(B)) is Lebesgue measurable, and so we have that f ◦ g is measurable, as required.

Problem 8.

Suppose f,g : R → R are Lebesgue measurable and g is such that for all null sets N, g−1(N) is measurable. We want to show that f ◦ g = f(g(x)) is Lebesgue measurable.

Fix B ∈ B(R). We have that,

(f ◦ g)−1(B) = g−1(f−1(B))

We have that f−1(B) ∈ L(R) since f is Lebesgue measurable. We need to use the property of g to show that g−1(f−1(B)) is measurable.

Since f−1(B) is measurable, then there exists a Gδ set G∗ and a null set N such that f−1(B) = G∗ \ N = G∗ ∩ NC. Note that G∗ is a countable intersection of open sets, and hence is Borel. Hence, we have from Problem 5 that,

g−1(f−1(B)) = g−1(G∗ ∩ NC)

= g−1(G∗) ∩ g−1(N)C

Since G∗ is a Borel set, we have that g−1(G∗) is measurable and since N is a null set, we have that g−1(N)C is measurable. Thus, we have that g−1(f−1(B)) is a finite intersection of measurable sets and hence is measurable. As a result, f ◦ g is Lebesgue measurable.

Problem 9.

Suppose that f is a Lebesgue measurable function. Then by Proposition 4.2.1, we have that

{x ∈ X : f(x) ≥ a}

and

{x ∈ X : f(x) ≤ a}

are both measurable sets for any a ∈ R. Since the Lebesgue measurable sets form a sigma algebra, we can take the intersection of these sets and still have a measurable set. This gives us that,

{x ∈ X : f(x) ≥ a} ∩ {x ∈ X : f(x) ≤ a} = {x ∈ X : f(x) = a}

is measurable for every a ∈ R, as required.

Problem 10.

Let x ∈ {x ∈ X : limn→∞ fn(x) > α} Then limn→∞ fn(x) converges to some number f(x) such that f(x) > α. Hence, if we fix  > 0, there exists N ∈ N such that for all n > N, we have,

|fn(x) − f(x)| < |fn(x) − α| <