1.1 Section 2.8 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
(A) =
2k kA
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 (A) =
kA1 or kA2 or or kAn 2k
Since each of these possibilities are disjoint, let us split it up into separate summations,
X 1 X 1 X 1
(A) = + +
kA1 2k kA2 2k kAn 2k
= (A1) + (A2) + + (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 (A) = . Observe we also have,
n
X
(Ai) = (A1) + (A2) + + + + (An)
i=1
=
Hence, we have that (A) = (A1)+(A2)++(An) once again and thus, is finitely additive.
Now suppose we have a countable number of sets Ai, where Ai = {i}. Then we have,
X
(Ai) = (A1) + (A2) +
=
2i i=1
= 1
However, if we consider A = tAi, we have that A = N and hence is an infinite set. Thus, (A) = and so,
(A) 6= X(Ai)
Thus, 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
= tk=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 Eik2 = for any i,j.
Lastly, note that,
K A t A = tk=1 tnj=1Ej(k) t tnj=1Ej
= K
Hence,
(K) = (K A) + (A)
n tj
= X
k=1
n
= XX(Ej(k)) + (A)
k=1 j=1
Since for all j,k, we must have from the above derivation that (A) (K). Moreover, we have that,
K = iKi
= tkCk
So,
(K) = (iKi)
= (tkCk)
= X(Ck)
k
Problem 5.
Suppose 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
(tni=1Ai) = X(Ai)
i=1
Hence, is finitely additive. Moreover if we have a countably infinite collection disjoint of sets Bi where each Bi R, then since is countably additive, we have,
(ti=1Bi) = X(Bi)
i=1
which satisfies the definition of countable subadditivity.
Now assume 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 = log10 3,
R
Since 0 log10 d < 1 based on our definition of d and 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,). Suppose I C. Then we write I = [k/2i,(k + 1) /2i) for integers k,i
with i,k Z. We have I 2 . Moreover, we have T1(x) = x x2 + 4 for x 6= 0 and
T1(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
T1(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, T1(I) is a disjoint union and, T1(I) = k/2i + qk2/22i + 4,(k + 1)/2i + q(k + 1)2/22i + 4
k/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)/2i1 k/2i1
= 1/2i1
Problem 2.
Suppose (X,S,) is a -finite measure-space and T : X X is measure-preserving. Fix X0 S(X) with T1(X0) = X0.We want to show that the system (X0,S(X0),,T) is a measure-preserving dynamical system. That is (X0,S(X0),) is a -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 -algebra on X0. Since the original measure space was -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 -finite as well.
Problem 3.
Let (X,S,) be a -finite measure space and let X0 S(X) with (XX0) = 0. Suppose there exists a transformation T0 so that (X0,S(X0),,T0) is a measure-preserving dynamical system. Since the original measure space is -finite
Problem 4.
Suppose (X,S,,T) is a measure-preserving dynamical system.
Reviews
There are no reviews yet.