Dynamical - Homework #4 - Solved

Problem 1.

Suppose we have n > 0 disjoint subsets of N, which we will denote as Ai for each 1 ≤ i ≤ n. For now, suppose each Ai is a finite set. Let us take the union of these sets, A = tni=0Ai. We have that,

X           1
                                                                                          µ(A) =          

2k k∈A

However, since the Ai are all disjoint, we have that each k is in one and only one Ai. Hence, we can re-index this summation as, 1 X µ(A) =     

k∈A1 or k∈A2 or ... or k∈An 2k

Since each of these possibilities are disjoint, let us split it up into separate summations,

                                                                                 X 1            X 1                      X 1

                                                            µ(A) =             +             + ··· +            

                                                                              k∈A1 2k          k∈A2 2k                       k∈An 2k

= µ(A1) + µ(A2) + ··· + µ(An)

So we have covered the case where each Ai is finite. Now suppose at least one is infinite. Then we must also have that A is infinite and so µ(A) = ∞. Observe we also have,

n

X

µ(Ai) = µ(A1) + µ(A2) + ··· + ∞ + ··· + µ(An)

i=1

= ∞

Hence, we have that µ(A) = µ(A1)+µ(A2)+···+µ(An) once again and thus, µ is finitely additive.

Now suppose we have a countable number of sets Ai, where Ai = {i}. Then we have,

X

µ(Ai) = µ(A1) + µ(A2) + ···

= 

2i i=1

= 1

However, if we consider A = tAi, we have that A = N and hence is an infinite set. Thus, µ(A) = ∞ and so,

µ(A) 6= Xµ(Ai)

Thus, µ is not countably additive.

Problem 2.

Since R is a semi-ring, we have that A \ ∅ ∈ R and,

A \ ∅ = A
= tnj=1Ej

for some disjoint sets Ej ∈ R. Moreover, since K = ∪Ki with Ki ∈ R for every i, we can apply Proposition 2.7.1 and get,

K  Ck

where the sets {Ck} are disjoint and in R. Now consider K \ A using these above definitions. We have,

K \ A   \ tnj=1Ej

= t∞k=1Ck \ tnj=1Ej

Observe that for each k, the set Ck \ tnj=1Ej is in R. Hence, we have that

Ck \ tnj=1Ej = tnj=1Ej(k)

Note that for each k, we have that,

Ck \ tnj=1Ej ∩ A = tnj=1Ej(k) ∩ A

= ∅

Moreover, for k1 6= k2, since Ck1 ∩ Ck2 = ∅, then tnj=1Ej(k1) ∩ tjn=1Ej(k2) and so Ejk1 ∩ Eik−2 = ∅ for any i,j.

Lastly, note that,

K \ A t A = t∞k=1 tnj=1Ej(k) t tnj=1Ej

= K

Hence,

µ(K) = µ(K \ A) + µ(A)

   n tj

= X µ 

k=1

∞ n

= XXµ(Ej(k)) + µ(A)

k=1 j=1

Since µ  for all j,k, we must have from the above derivation that µ(A) ≤ µ(K). Moreover, we have that,

K = ∪iKi

= tkCk

So,

µ(K) = µ(∪iKi)

= µ(tkCk)

= Xµ(Ck)

k

Problem 5.

Suppose µ is countably additive. Since finite collections are countable, we must have that for any disjoint sets Ai, 1 ≤ i ≤ n in R, we have that,

n

µ(tni=1Ai) = Xµ(Ai)

i=1

Hence, µ is finitely additive. Moreover if we have a countably infinite collection disjoint of sets Bi where each Bi ∈ R, then since µ is countably additive, we have,

∞

µ(t∞i=1Bi) = Xµ(Bi)

i=1

which satisfies the definition of countable subadditivity.

Now assume µ is additive and countably subadditive.

1.2      Section 3.2 Problem 2.

Let d ∈ {x1.x2x3x4 |x1 ∈ {1,2,...,9} and x2,x3,x4 ∈ {0,1,2,...,9}}. Then d is the set of numbers between 1.000 and 9.999 (inclusive) that have terminating decimal expansion of length 4. Now suppose the decimal representation of 3n starts with d · 103. Then for some integer k ≥ 0,

d · 10k ≤ 3n < (d + 0.001) · 10k

Thus,

log10(d · 10k) ≤ log10 3n < log10((d + 0.001) · 10k

which gives us,

log10 d ≤ nlog10 3 − k < log10(d + 0.001)

and finally,

log10 d ≤ nlog10 3 (mod 1) < log10(d + 0.001)

But this is the same as saying that, letting α = log10 3,

R 

Since 0 ≤ log10 d < 1 based on our definition of d and α is irrational, we can apply

Theorem 3.2.3. Thus, there are infinitely many integers n such that R  0.001)). Hence, there are infinitely many powers of 3 that start with 1984.

1.3      Section 3.4 Problem 1.

We have that the collection of left-closed, right-open dyadic intervals form a sufficient semi-ring for (R,L,λ). Suppose I ∈ C. Then we write I = [k/2i,(k + 1)√ /2i) for integers k,i

 with i,k ∈ Z. We have µ  I       2 . Moreover, we have T−1(x) = x ±               x2 + 4 for x 6= 0 and

 T−1(0) = 0. This gives us, −1                        ii q           2/22i + 4 [

                              T         (I) = k/2 +                      + 4,(k + 1)/2 +        (k + 1)

k/2i − qk2/22i + 4,(k + 1)/2i − q(k + 1)2/22i + 4

T−1(I) is a finite union of intervals and is hence measurable.

  Observe that q(k + 1)2/22i + 4 > q(k + 1)2/22i = (k+1)/2i and qk2/22i + 4 > qk2/22i = k/2i. Hence, T−1(I) is a disjoint union and, µT−1(I) = µk/2i + qk2/22i + 4,(k + 1)/2i + q(k + 1)2/22i + 4+

                                                      µk/2i − qk2/2                                   i q                         2/22i + 4

                                                                                          2i + 4,(k + 1)/2 −         (k + 1)

= (k + 1)/2i + q(k + 1)2/22i + 4 − (k/2i + qk2/22i + 4)+

(k + 1)/2i − q(k + 1)2/22i + 4 − k/2i + qk2/22i + 4

= (k + 1)/2i − k/2i + (k + 1)/2i − k/2i

= 2(k + 1)/2i − 2k/2i

= (k + 1)/2i−1 − k/2i−1

= 1/2i−1

Problem 2.

Suppose (X,S,µ) is a σ-finite measure-space and T : X → X is measure-preserving. Fix X0 ∈ S(X) with T−1(X0) = X0.We want to show that the system (X0,S(X0),µ,T) is a measure-preserving dynamical system. That is (X0,S(X0),µ) is a σ-finite measure space and T : X0 → X0 is a measure preserving transformation.

By Proposition 2.5.1, since X0 ⊂ X is in S, we have that S(X0) = {A : A ⊂ X0 and X0 ∈ S} is a σ-algebra on X0. Since the original measure space was σ-finite, there exist a sequence of measurable sets An of finite measure such that,

∞

X = [ An

n=1

Since X0 is in the collection of measurable sets and is a subset of X, we can remove sets from the sequence Bn, creating a new sequence Bn such that,

∞

X0 = [ Bn

n=1 Hence, the new measure space is σ-finite as well.

Problem 3.

Let (X,S,µ) be a σ-finite measure space and let X0 ∈ S(X) with µ(X\X0) = 0. Suppose there exists a transformation T0 so that (X0,S(X0),µ,T0) is a measure-preserving dynamical system. Since the original measure space is σ-finite

Problem 4.

Suppose (X,S,µ,T) is a measure-preserving dynamical system.