Lecture 2:Measures 1 of 17 Course: Theory of Probability I Term: Fall 2013 Instructor: Gordan Zitkovic Lecture 2 Measures Measure spaces Definition 2.1 (Measure). Let (S, S) be a measurable space. A map- ping m : S! [0, ¥] is called a (positive) measure if 1. m(Æ) = 0, and 2. m([n An) = ∑n2N m(An), for all pairwise disjoint fAngn2N in S. A triple (S, S, m) consisting of a non-empty set, a s-algebra S on it and a measure m on S is called a measure space. Remark 2.2. 1. A mapping whose domain is some nonempty set A of subsets of some set S is sometimes called a set function. 2. If the requirement 2. in the definition of the measure is weakened so that it is only required that m(A1 [···[ An) = m(A1) + ··· + m(An), for n 2 N, and pairwise disjoint A1,..., An, we say that the mapping m is a finitely-additive measure. If we want to stress that a mapping m satisfies the original requirement 2. for sequences of sets, we say that m is s-additive (countably additive). Definition 2.3 (Terminology). A measure m on the measurable space (S, S) is called 1.a probability measure, if m(S) = 1, 2.a finite measure, if m(S) < ¥, 3.a s-finite measure, if there exists a sequence fAngn2N in S such that [n An = S and m(An) < ¥, 4. diffuse or atom-free, if m(fxg) = 0, whenever x 2 S and fxg 2 S. A set N 2 S is said to be null if m(N) = 0. Last Updated: September 19, 2013 Lecture 2:Measures 2 of 17 Example 2.4 (Examples of measures). Let S be a non-empty set, and let S be a s-algebra on S. 1. Measures on countable sets. Suppose that S is a finite or countable set. Then each measure m on S = 2S is of the form m(A) = ∑ p(x), x2A for some function p : S ! [0, ¥] (why?). In particular, for a finite set S with N elements, if p(x) = 1/N then m is a probability measure called the uniform measure1 on S. 1 In the finite case, it has the well-known #A property that m(A) = #S , where # de- 2. Dirac measure. For x 2 S, we define the set function dx on S by notes the cardinality (number of ele- 8 ments). <1, x 2 A, dx(A) = :0, x 62 A. It is easy to check that dx is indeed a measure on S. Alternatively, dx is called the point mass at x (or an atom on x, or the Dirac function, even though it is not really a function). Moreover, dx is a probability measure and, therefore, a finite and a s-finite measure. It is atom free only if fxg 62 S. 3. Counting Measure. Define a set function m : S! [0, ¥] by 8 <#A, A is finite, m(A) = :¥, A is infinite, where, as above, #A denotes the number of elements in the set A. Again, it is not hard to check that m is a measure - it is called the counting measure. Clearly, m is a finite measure if and only is S is a finite set. m could be s-finite, though, even without S being finite. Simply take S = N, S = 2N. In that case m(S) = ¥, but for An = fng, n 2 N, we have m(An) = 1, and S = [n An. Finally, m is never atom-free and it is a probability measure only if #S = 1. Example 2.5 (A finitely-additive set function which is not a measure). Let S = N, and S = 2S. For A 2 S define m(A) = 0 if A is finite and m(A) = ¥, otherwise. For A1,..., An ⊆ S, either n 1. all Ai is finite, for i = 1, . , n. Then [i=1 Ai is also finite and so n n 0 = m([i=1 Ai) = ∑ m(Ai), or i=1 n 2. at least one Ai is infinite. Then [i=1 Ai is also infinite and so n n ¥ = m([i=1 Ai) = ∑ m(Ai). i=1 Last Updated: September 19, 2013 Lecture 2:Measures 3 of 17 On the other hand, take Ai = fig, for i 2 N. Then m(Ai) = 0, for Note: It is possible to construct very simple-looking finite-additive measures each i 2 N, and, so, ∑ 2N m(Ai) = 0, but m([i Ai) = m(N) = ¥. i which are not s-additive. For example, there exist f0, 1g-valued finitely-additive Proposition 2.6 (First properties of measures). Let (S, S, m) be a measure measures on all subsets of N, which are not s-additive. Such objects are called space. ultrafilters and their existence is equiva- 1. For A ,..., A 2 S with A \ A = Æ, for i 6= j, we have lent to a certain version of the Axiom of 1 n i j Choice. n n ∑ m(Ai) = m([i=1 Ai) (Finite additivity) i=1 2. If A, B 2 S,A ⊆ B, then m(A) ≤ m(B) (Monotonicity of measures) 3. If fAngn2N in S is increasing, then m([n An) = lim m(An) = sup m(An). n n (Continuity with respect to increasing sequences) 4. If fAngn2N in S is decreasing and m(A1) < ¥, then m(\n An) = lim m(An) = inf m(An). n n (Continuity with respect to decreasing sequences) 5. For a sequence fAngn2N in S, we have m([n An) ≤ ∑ m(An). (Subadditivity) n2N Proof. 1. Note that the sequence A1, A2,..., An, Æ, Æ, . is pairwise disjoint, and so, by s-additivity, n m([i=1 Ai) = m([i2N Ai) = ∑ m(Ai) i2N n ¥ n = ∑ m(Ai) + ∑ m(Æ) = ∑ m(Ai). i=1 i=n+1 i=1 2. Write B as a disjoint union A [ (B n A) of elements of S. By (1) above, m(B) = m(A) + m(B n A) ≥ m(A). 3. Define B1 = A1, Bn = An n An−1 for n > 1. Then fBngn2N is a n pairwise disjoint sequence in S with [k=1Bk = An for each n 2 N (why?). By s-additivity we have n m([n An) = m([nBn) = m(Bn) = lim m(Bk) ∑ n ∑ n2N k=1 n = lim m([ Bk) = lim m(An). n k=1 n Last Updated: September 19, 2013 Lecture 2:Measures 4 of 17 4. Consider the increasing sequence fBngn2N in S given by Bn = A1 n An. By De Morgan laws, finiteness of m(A1) and (3) above, we have m(A1) − m(\n An) = m(A1 n (\n An)) = m([nBn) = lim m(Bn) n = lim m(A1 n An) = m(A1) − lim m(An). n n Subtracting both sides from m(A1) < ¥ produces the statement. 5. We start from the observation that for A1, A1 2 S the set A1 [ A2 can be written as a disjoint union A1 [ A2 = (A1 n A2) [ (A2 n A1) [ (A1 \ A2), so that m(A1 [ A2) = m(A1 n A2) + m(A2 n A1) + m(A1 \ A2). On the other hand, m(A1) + m(A2) = (m(A1 n A2) + m(A1 \ A2)) + m(A2 n A1) + m(A1 \ A2) = m(A1 n A2) + m(A2 n A1) + 2m(A1 \ A2), and so m(A1) + m(A2) − m(A1 [ A2) = m(A1 \ A2) ≥ 0. Induction can be used to show that n m(A1 [···[ An) ≤ ∑ m(Ak). k=1 Since all m(An) are nonnegative, we now have m(A1 [···[ An) ≤ a, for each n 2 N, where a = ∑ m(An). n2N n The sequence fBngn2N given by Bn = [k=1 Ak is increasing, so the continuity of measure with respect to increasing sequences implies that m([n An) = m([nBn) = lim m(Bn) = lim m(A1 [···[ An) ≤ a. n n Remark 2.7. The condition m(A1) < ¥ in the part (4) of Proposition 2.6 cannot be significantly relaxed. Indeed, let m be the counting mea- sure on N, and let An = fn, n + 1, . g. Then m(An) = ¥ and, so limn m(An) = ¥. On the other hand, \An = Æ, so m(\n An) = 0. Last Updated: September 19, 2013 Lecture 2:Measures 5 of 17 In addition to unions and intersections, one can produce other im- portant new sets from sequences of old ones. More specifically, let fAngn2N be a sequence of subsets of S. The subset lim infn An of S, defined by lim inf An = [nBn, where Bn = \k≥n Ak, n is called the limit inferior of the sequence An. It is also denoted by 2 2 limn An or fAn, ev.g (ev. stands for eventually ). the reason for the use of the word even- tually is the following: lim infn An is the Similarly, the subset lim supn An of S, defined by set of all x 2 S which belong to An for all but finitely many values n A = \ B B = [ A of the index , i.e., lim sup n n n, where n k≥n k, from some value of the index n onwards. n is called the limit superior of the sequence An. It is also denoted by 3 3 limn An or fAn, i.o.g (i.o. stands for infinitely often ). Clearly, we have in words, lim supn An is the set of all x 2 S which belong An for infinitely lim inf An ⊆ lim sup An. many values of n. n n Problem 2.1. Let (S, S, m) be a finite measure space.