Math 650. Homework 5. Solutions

Problem 1 Let X be an uncountable set. Let R be the collection of all the finite subsets of X. Given A Î R let m(A) be the number of elements in A. Show that R is a ring and m is a measure on R.

[Solution] If A and B are in R, they are finite sets, and so is AÈB and A - B.

If A,B Î R are disjoint, then the number of elements in AÈB equals the number of elements in A plus the number of elements in B, so m(AÈB)=m(A)+m(B).

For m to be a measure, it must be countably additive. Let then {A_i}_i=1^¥ Ì R be a countable collection of mutually disjoint sets such that A=È_i=1^¥ A_i is also in R. Then A and all A_i are finite, and so only finitely many A_i are non-empty. Therefore å_i=1^¥ m(A_i) is a finite sum and

m(A)= ¥ å i=1  m(Ai)

by additivity of m.

Problem 2 Let X be an infinite set and R the collection of all countable subsets of X.

1. Is R a s-ring?
2. Let m be a measure on R. Show that there is a function f:X® [0,¥) such that

   
   

   m(A)= å x Î A  f(x),

   for all A Î R.

3. Show that f in (2) has the following properties: the set { x Î X | f(x) ¹ 0} is countable, and åf(x) < ¥.
4. Conversely, show that if f:X® [0,¥) has these two properties, then the formula

   
   

   m(A)= å x Î A  f(x)

   defines a measure on R.

[Solution] (1) Yes, a countable union of countable sets is a countable set.

(2) Each one-point set {x} Î R. If A Î R, A is countable, and so it can be written as a countable disjoint union of one-point sets: A=È_{x Î A}{x}. Thus, setting f(x)=m({x}), the countably additivity of m gives

m(A)= å x Î A  f(x).

(3) The set { f(x) > 0} is countable. Indeed, let A_n be the interval A_n=([1/(n+1)], [1/n]], n=1,2,¼ and A₀=(1,¥). Then {A_n}_n=0^¥ is a countable partition of (0,¥), and so B_n={ x Î X | f(x) Î A_n} is a countable partition of {f(x) > 0}. If {f(x) > 0} was uncountable, then at least one of the B_n must be also uncountable (for otherwise {f(x) > 0} would be a countable union of countable sets, hence countable). Suppose then that B_n0 is uncountable. Then it contains an infinite countable set C={x₁,x₂,¼}. Since C is countable, C Î R. But its measure

m(C)= ¥ å i=1  f(xi) ³ ¥ å i=1  1 n0+1 =¥,

contradicting the fact that m(A) is a non-negative number for every A Î R.

Once we know that Y={f(x) > 0} is countable, we know that Y Î R and so 0 £ m(Y) < ¥ and

å x Î X  f(x) = å f(x) > 0  f(x) = å x Î Y  f(x) = m(Y) < ¥.

Problem 3 Let X be the real line and R = R_Leb. Given A Î R, let m(A)=1 if, for some positive e, A contains the interval (0,e); otherwise let m(A)=0.

Show that m is an additive set function, but it is not countably additive.

[Solution] We have to show that if A,B Î R are disjoint, then m(AÈB)=m(A)+m(B). A set in R is a finite union of intervals.

Let A=A₁ȼÈA_m and B=B₁ȼÈB_n, where the A_i's and B_j's are intervals. Assume that they are ordered so that the right endpoint of A_i is less than or equal to the left endpoint of A_i+1, i=1,¼, m-1, and similarly for the B_j's. Also assume that the right endpoint of A₁ is smaller that the left endpoint of B₁. (You should verify that this can be done.)

There are two cases to consider.

1. m(AÈB)=1. This means that there exists e > 0 such that (0,e) Ì AÈB = A₁ȼÈA_m ÈB₁ȼB_n. But given that A₁ is the left-most interval in AÈB, this implies that there exists a > 0 such that (0,a) Ì A₁, and hence also that BÇ(0,a)=Æ. It follows that m(A)=1 and m(B)=0.
2. m(AÈB)=0. This means that AÈB does contains no interval of the form (0,e), and since A,B Ì AÈB, the same must be true for A and for B. Thus m(A)=m(B)=0.

To show that m is not countably additive, let A_n=([1/(n+1)], [1/n]], n=1,2,¼, so that m(A_n)=0 for all n. Then A=È_i=1^¥([1/(n+1)],[1/n]] = (0,1] is in R, but

m(A)=1 ¹ ¥ å i=1  m(Ai) = 0+0+0+¼ = 0.

Problem 4 Given any collection C of subsets of X, show that there is a smallest ring R containing C.

[Solution] There is always a ring which contains C, namely the power set ring 2^X. Let R be the intersection of all the rings which contain C. Then R is a non-empty collection of subsets of X.

It is also a ring. For if A,B are in R, then they are in every ring which contains C. If § is a ring which contains C, then the union AÇB and the difference A-B are also in §. Thus AÇB and A-B are in every ring which contains C, that is, they are in R.