Dynamical - Homework #3 - Solved

1.1      Section 2.5 Problem 14.

We have that E = {∅} ⊂ X. We also have that {∅} ∈ S. Hence, with A,B = {∅} we trivially have that

A ⊂ E ⊂ B

and µ(B \ A) = 0

Thus, E = {∅} ∈ Sµ and Sµ is non-empty. Now fix E ∈ Sµ and consider EC. Since E ∈ Sµ, we have A,B ∈ S such that

A ⊂ E ⊂ B

and

µ(B \ A) = 0

Applying the complement to the above expression gives us that AC ⊃ EC ⊃ BC. So we want to show that µ(AC \BC) = 0. But note that AC \BC = AC ∩B and B \A = B ∩AC, so we have that,

µ(AC \ BC) = µ(AC ∩ B)

= µ(B \ A)

= 0

Hence, Sµ is closed under complements.

Now let En ∈ Sµ, n ≥ 1 and consider  En. For each n, we have that there exists An,Bn ∈ S such that An ⊂ En ⊂ Bn and µ(Bn \ An) = 0. Since En ⊂ Bn for all n, we must have that ∪En ⊂ ∪Bn. Similarly, we have ∪An ⊂ ∪En.

Now we need to show that µ(∪Bn \ ∪An) = 0. By Corollary 2.4.2 and countable subadditivity, we have,

µ(∪Bn \ ∪An) = µ(∪Bn) − µ(∪An)

≤ µ(B1) + µ(B2) + ··· − µ(A1) − µ(A2) − µ(A3) − ···

= µ(B1) − µ(A1) + µ(B2) − µ(A2) + µ(B3) − µ(A3) + ···

= µ(B1 \ A1) + µ(B2 \ A2) + µ(B3 \ A3) + ···

= 0 + 0 + 0 + ···

= 0

Hence, ∪En ∈ Sµ and so Sµ is closed under countable unions. Hence, Sµ is a σ-algebra.

Now fix E ∈ S. Then if we set A,B = E ∈ S, we have that A ⊂ E ⊂ B and µ(B \ A) = µ(E \ E) = µ(∅) = 0 as required. Thus, every element in S is also in Sµ, and so Sµ is a σ-algebra containing S.

Problem 15.

Define µ on elements of Sµ by µ(E) = µ(A) for any A ∈ S such that there is a B ∈ S with A ⊂ E ⊂ B and µ(B \ A) = 0.

Fix E ∈ Sµ and let A,B ∈ S be any two sets such that A ⊂ E ⊂ B and µ(B \ A) = 0.

Since E ⊂ B, we have that E \ A ⊂ B \ A. Thus,

µ(E \ A) ≤ µ(B \ A) = 0

For any other choice of A0,B0 ∈ S such that A0 ⊂ E ⊂ B0 and µ(B0 \ A0) = 0, we also have that µ(E \ A0) = 0. Since A,A0 ⊂ E, we have,

µ(A04A) ≤ µ(E \ A)

= 0

Hence, we have that µ(A0 \ A) = 0 = µ(A \ A0). Now note that,

µ(A0) = µ(A \ (A \ A0)

= µ(A) − µ(A \ A0)

= µ(A) − 0

= µ(A)

Thus, we have that µ(E) is independent of the choice of A and B.

Now fix E1 ∈ Sµ with µ(E1) = 0 and consider E2 ⊂ E1. We have that there exist A,B ∈ S such that A ⊂ E1 ⊂ B and µ(B \A) = 0. Since µ(E1) = 0, we have that µ(A) = 0 as well. We have that µ(B) ≥ µ(A) = 0, so we can state:

µ(B \ A) = µ(B) − µ(A)

= µ(B)

= 0

Hence, A,B both have measure 0. Let us take A = ∅ ∈ S then and keep B as above. Clearly we have A ⊂ E2 ⊂ E1 ⊂ B and µ(B \ A) = 0. As a result, we get that,

 

µ(E2) = µ(A)

= µ(∅)

= 0

Thus, E2 is measurable and µ is a complete measure on Sµ.

1.2            Section 2.6
Problem 2.

Note that the rational numbers are countable and that for each q ∈ Q, we have that {q} is closed. Thus, {q}C is open for all q ∈ Q. Hence, if we take, \ C

{q} q∈Q

we see that we are taking a countable intersection of open sets. This is precisely the definition of a Gδ set. Moreover, this set is the intersection of the complements of the rational numbers. That is, the only numbers in R not included in this set are the rationals. Hence, we have constructed the irrational numbers.

Note the irrational numbers are not open. If we fix an irrational number x, observe that for any value of  > 0, we will find a rational number q such that q  . This follows directly from the fact that Q is dense in R. Moreover, the irrational numbers are not closed. The irrationals are also dense in R, so by definition we can find a sequence of irrational numbers converging to any real number. If we pick q ∈ Q ⊂ R, then we see that we can construct a sequence of irrational numbers which converge to q 6∈ Tq∈Q{q}C. Thus, the irrational numbers are neither open nor closed as required.

Problem 5.

Let C be the collection of all closed intervals with rational endpoints. That is, intervals of the form hpq11, pq22i with p1,q1,p2,q2 ∈ Z. Note that B contains all the closed sets and open sets. Moreover, since B is a σ-algebra, it must also include all complements and countable unions of the closed and open sets. In particular, all complements and countable unions of the closed intervals with rational endpoints must be included in B. Thus, σ(C) ⊂ B.

Now we need to show that B ⊂ C. Since C contains all closed intervals, any σ algebra containing C must contain the complements of these elements. That is, it must contain all open intervals with rational endpoints. This is true because, for an open interval pq11, pq22, we have that it must equal,

                                                                                                      # "                   !!C

−∞, p 1 [ p2,∞ q1      q2

Note that −∞, pq11i,hpq22,∞ ∈ C and any σ-algebra containing these intervals must contain their union and the complement of this union. Thus, pq11, pq22 must be in any σalgebra containing C for any p1,q1,p2,q2 ∈ Z. We know from the properties of R that any open set can be written as a countable union of disjoint open intervals. Since any σ-algebra containing C must contain all open intervals with rational endpoints and must be closed under countable unions, we have that all open sets must be contained in any σ-algebra containing C. Lastly, since any σ-algebra containing C contains all the open sets and is closed under complements and countable unions, it must contain the σ-algebra generated by the open sets. Thus, we have that,

B ⊂ σ(C)

By the above arguments, we thus have that B = σ(C).

1.3            Section 2.7
Problem 1.

Suppose that C is a semi-ring of subsets of a nonempty set X and ∅ 6= Y ⊂ X. Consider the collection {A ∩ Y : A ∈ C} and suppose that this collection is non-empty.

Suppose C,D ∈ {A∩Y : A ∈ C}. We have C = A∩Y and D = B∩Y for some A,B ∈ C.

Consider the following,

C ∩ D = (A ∩ Y ) ∩ (B ∩ Y )

= (A ∩ B) ∩ Y

Since C is a semi-ring, we have that A ∩ B ∈ C and so C ∩ D = (A ∩ B) ∩ Y is in our collection.

Now consider C \ D. We can express this as the following,

C \ D = (A ∩ Y ) \ (B ∩ Y )

= (A \ B) ∩ Y

Since C is a semi-ring, we have that A \ B can be expressed as

tnj=1Ej

where Ej ∈ C for every j. Thus, we can rewrite the previous statement as,

C \ D = (A \ B) ∩ Y

 Y

 

Since Ej ∈ C, we have that Ej ∩ Y ∈ {A ∩ Y : A ∈ C}. Furthermore, since Ej ∈ C are disjoint, Ej ∩ Y ∈ {A ∩ Y : A ∈ C} are disjoint as well. Therefore, C \ D satisfies all the required properties for a semi-ring.

Now let us extend to the case where C is a ring. We now need to show that our new collection is closed under finite unions. It suffices to show that the collection is closed for a single union (i.e. C ∪ D) because we can then extend this union a finite number of times n. So now let us consider C ∪ D,

C ∪ D = (A ∩ Y ) ∪ (B ∩ Y )

= (A ∪ B) ∩ Y

Since C is a ring, we have that A∪B ∈ C and so (A∪B)∩Y is in our collection. Hence, {A ∩ Y : A ∈ C} is a ring.

Problem 2.

Let (X,L,λ) be a canonical Lebesgue measure space and C a sufficient semi-ring. Now for any nonempty measurable set X0 ⊂ X, consider C ∩ X0 = {C ∩ X0 : C ∈ C}. We want to show that this set is a sufficient semi-ring for (X0,L(X0),λ).

Fix Y ∈ C ∩ X0. Then Y = C ∩ X0 for some C ∈ C and so y ∈ Y ⇐⇒ y ∈ X0 and y ∈ C. Thus, Y ⊂ X0 by definition and, as a result, every set in C ∩ X0 is a subset of X0.

Now since C is a semi-ring, we have ∅ ∈ C. Hence, ∅ ∩ X0 = ∅ ∈ C ∩ X0, and so the collection is non-empty.

Fix A,B ∈ C ∩ X0. Then A = C1 ∩ X0, B = C2 ∩ X0 for some C1,C2 ∈ C. We have,

A ∩ B = (C1 ∩ X0) ∩ (C2 ∩ X0)

= (C1 ∩ C2) ∩ X0

Since C is a sufficient semi-ring, we have that C1∩C2 ∈ C and thus (C1∩C2)∩X0 ∈ C∩X0.

Now consider A \ B. Note that this is given by (C1 ∩ X0) \ (C2 ∩ X0). This is equivalent to (C1 \ C2) ∩ X0. Since C1,C2 ∈ C, we have,

C1 \ C2 = tnj=1Ej

where Ej ∈ C are disjoint. Rewriting our formulation for A \ B yields,

A \ B = (C1 \ C2) ∩ X0

 X0

 

Note that Ej ∩ X0 ∈ C ∩ X0 for every j. Also note that the sets Ej ∩ X0 are disjoint. To prove this, fix i and j and suppose x ∈ Ej ∩ X0 and x ∈ Ei ∩ X0. Then by the definition of intersection, we have that x ∈ Ej, x ∈ Ei, and x ∈ X0. But we know that Ei,Ej are disjoint, so x cannot be in both sets. Thus we have a contradiction and the sets Ej ∩ X0 are disjoint as required.

Hence we have shown that C ∩ X0 is a semi-ring. Now we need to show that it satisfies the sufficient semi-ring property. First we need to show that every set contained in C ∩ X0 has finite measure. Consider A as defined above. We have,

λ(A) = λ(C1 ∩ X0)

≤ λ(C1) < ∞

Hence, A has finite measure. Recall that C is a sufficient semi-ring. Then

 ∞            λ(C1) = inf Xλ(Ij) : C1 ⊂ ∪∞j=1Ij and Ij ∈ C for j ≥ 1

                                                                 j=1                                                                                                  

Applying this to the definition of A yields,

λ(A) = λ(C1 ∩ X0)

                                         ∞                                                                                                                                                  

= inf X X0 and Ij ∩ X0 ∈ C ∩ X0 for j ≥ 1

                                        j=1                                                                                                                                                 

as required.

Problem 6.

Let C be a collection of sets. Denote r(C) as the intersection of all rings containing C. We want to show that this collection is a ring.

Firstly, since C ⊂ R for every ring containing C by definition, we have that C must be contained in their intersection. That is, C ⊂ r(C).

Every ring containing C must also contain the empty set. Hence, ∅ ∈ r(C).

Now fix A,B ∈ r(C). Then A,B are in every ring containing C. Since rings are closed under intersection, we must have that A ∩ B is in every ring containing C as well. Hence, A ∩ B ∈ r(C).

Now consider A \ B. Since A,B ∈ r(C), then A,B are in every ring containing C. Thus, there exist disjoint sets Ej such that

A \ B = tnj=1Ej

Hence, Ej ∈ r(C) for j = 1,...,n.

Lastly, since A,B in every ring containing C and each ring is closed under finite unions, we must have that A ∪ B is in every ring containing C. Hence, A ∪ B ∈ r(C) and we have that r(C) is a ring, as required.

Problem 7.

Suppose that R is a semi-ring. Suppose we take finite unions of all disjoint elements in R. Fix A ∈ R. Since ∅ ∈ R, we can use property (3) of semi-rings to state that, for disjoint sets Ej ∈ R with j = 1,...,n, we have,

A = A \ ∅

= tnj=1Ej

Thus, we can express the union of any two elements A,B ∈ R as follows,

A 

where A = tnj=1EjA and B = tmj=1EjA. Hence, taking finite unions of disjoint sets in R

suffices for expressing finite unions of all sets in R. Thus, this constitutes a ring.

Now suppose r(R) is contained in the ring we have just described. Then there exists at least one element in the above ring which is not in r(R). The only new elements introduced in the above ring were done by taking union of disjoint sets. Thus, this is where the difference must lie. Then there must exist E1,E2 ∈ R such that E1 and E2 are disjoint with E1 tE2 6∈ r(R). However, if E1 t E2 6∈ r(R), then there is sets A ∈ R such that,

A = E1 t E2 tnj=3 E3 6∈ r(R)

Hence, there must exist B ∈ R such that A ∪ B 6∈ r(R). Thus, this would not define a ring and, as a result, the ring which we defined at the start of this problem must be r(R).

1.4            Section 3.3
Problem 1.

Define T : [0,1] → [0,1] by T(x) = 2x if 0 ≤ x ≤ 1/2 and T(x) = 2 − 2x if 1/2 < x < 1. Define S1 : [0,1] → [0,1/2] by S1(y) = y/2 and S2 : [0,1] → [1/2,1] by S2(y) = y/2 + 1/2.

For a measurable set A ⊂ [0,1]. We have that,

T −1(A) = S1(A) t (S2(A \ {0}))

Since we are taking a singleton away from A and S2 is a well-defined function, this will not change the measure of S2(A). Thus, from the above and from the results in Chapter 2, we have,

λ(T −1(A)) = λ(S1(A)) + λ(S2(A \ {0}))

=  A) +  A +  )

=  (A) +  (A)

= λ(A)

Thus, T is measure preserving.

Now, by the explanation on page 76 in the text, we can consider T as operating on the symbolic binary representation of a point x = (a1a2a3 ...). When applying T(x), we get (a2a3 ...). For an irrational number, this sequence never repeats itself. Hence, if x is irrational, we have that√ Tn(x) is distinct for all n ∈ Z≥0. An example in [0,1) would be

 

x0 =        2 − 1.

Since T(x) ∈ [0,1] for all x ∈ [0,1], we have that Tn(x0) ∈ [0,1] for all n.