[Solved] Dynamical Homework 8

1.1 Section 3.11

Problem 3. Revised problem: Let (X,S,,T) be an invertible, recurrent, finite measurepreserving dynamical system. If A is a set of positive measure such that the transformation T_A is ergodic on A and (X _n0Tⁿ(A)) = 0, then T is ergodic.

Note: X _n0Tⁿ(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 f¹(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,

f¹(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}

= f¹(A) f¹(B)

Now consider f¹(A B). We have that,

f¹(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}

= f¹(A) f¹(B)

Lastly, consider f¹(R A). We have that,

f¹(R A) = {x X : f(x) R A}

= {x X : f(x) A^c}

Note that the final line in the aboe is the set of x in X that map to A^c 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) A^c} = {x X : f(x) A}^c

= f1(A)c

= X f¹(A)

as required.

Now consider f(A B). Let x f(A B). Then x {y R : f¹(y) A B). Thus, x A or x B, or both. That is, x {y R : f¹(y) A) or x {y R : f¹(y) A) or both. Hence, we have that f(A B) f(A) f(B). Now fix x f(A) f(B). Then f¹(x) A or f¹(x) B. But this is exactly the definition of f(AB) and so x f(AB). 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 I_k for some disjoint open intervals I_k. Now note that disjoint set in a functions 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,

f1(G) = f1 (tk=1Ik)

= tk=1f1(Ik)

Since each I_k is an interval, we have that f¹(I_k) is measurable. And since the countable union of measurable sets is measurable, we have that f¹(I_k) = f¹(G) is measurable, as required.

Now suppose f¹(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) = g¹(f¹(B))

We have that f¹(B) L(R) since f is Lebesgue measurable. We need to use the continuity of f to show that f¹(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 f¹(B) as an expression of open sets through countable unions, countable intersections, and complements.

Hence, f¹(B) is a Borel set and so g¹(f¹(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, g¹(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) = g¹(f¹(B))

We have that f¹(B) L(R) since f is Lebesgue measurable. We need to use the property of g to show that g¹(f¹(B)) is measurable.

Since f¹(B) is measurable, then there exists a G set G and a null set N such that f¹(B) = G N = G N^C. Note that G is a countable intersection of open sets, and hence is Borel. Hence, we have from Problem 5 that,

g1(f1(B)) = g1(G NC)

= g1(G) g1(N)C

Since G is a Borel set, we have that g¹(G) is measurable and since N is a null set, we have that g¹(N)^C is measurable. Thus, we have that g¹(f¹(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 : lim_n f_n(x) > } Then lim_n f_n(x) converges to some number f(x) such that f(x) > . Hence, if we fix > 0, there exists N N such that for all n > N, we have,

|f_n(x) f(x)| < |f_n(x) | <