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2 Probability, Random Elements, Random Sets Tel Aviv University, 2012 Measurability and continuity 25 2 Probability, random elements, random sets 2a Probability space, measure algebra . 25 2b Standard models . 30 2c Random elements . 33 2d Random sets . 37 Hints to exercises ....................... 44 Index ............................. 45 After introducing probability measures we discuss random objects of two kinds. 2a Probability space, measure algebra We know that algebras are easy but σ-algebras are not. Surprisingly, a mea- sure makes σ-algebras easy by neglecting null sets. In particular, the Lebesgue σ-algebra is simpler than the Borel σ-algebra. 2a1 Definition. (a) A probability measure on a measurable space (X; A) is a map µ : A! [0; 1] such that µ(X) = 1 and µ(A1 [ A2 [ ::: ) = µ(A1) + µ(A2) + ::: whenever sets A1;A2; · · · 2 A are pairwise disjoint. (b) A probability space is a triple (X; A; µ) such that (X; A) is a measur- able space and µ is a probability measure on (X; A). 2a2 Example. One and only one probability measure µ on ([0; 1]; B[0; 1]) satisfies µ([0; x]) = x for all x 2 [0; 1]. This µ is often called \Lebesgue measure on [0; 1]", but I prefer to add \restricted to the Borel σ-algebra". 2a3 Definition. Let (X; A; µ) be a probability space, and A ⊂ X. ∗ (a) Inner measure µ∗(A) and outer measure µ (A) of A are defined by ∗ µ∗(A) = maxfµ(B): B 2 A;B ⊂ Ag ; µ (A) = minfµ(B): B 2 A;B ⊃ Ag ; (b) A is a null (or negligible) set1 if µ∗(A) = 0; ∗ (c) A is a µ-measurable set (symbolically, A 2 Aµ) if µ∗(A) = µ (A). 1Terence Tao calls it \sub-null set", requiring A 2 A for null sets. For some authors, \null" means just \empty". Tel Aviv University, 2012 Measurability and continuity 26 In contrast, sets of A will be called A-measurable. Clearly, A ⊂ Aµ; that is, A-measurability implies µ-measurability. For µ of 2a2, Aµ is the Lebesgue σ-algebra on [0; 1]. 2a4 Core exercise. Prove that (a) the maximum and minimum in 2a3(a) ∗ are reached; and (b) µ∗(A) = 1 − µ (X n A). 2a5 Core exercise. Prove that µ∗ : 2X ! [0; 1] satisfies (a) µ∗(;) = 0, (b) µ∗(A) ≤ µ∗(B) whenever A ⊂ B ⊂ X (\monotonicity"); ∗ ∗ ∗ (c) µ (A1 [ A2 [ ::: ) ≤ µ (A1) + µ (A2) + ::: for all A1;A2; · · · ⊂ X (\countable subadditivity"). Deduce that the null sets are a σ-ideal, that is, every subset of a null set is a null set, and a countable union of null sets is a null set. If A is a null set then its complement X n A is called a set of full measure (or conegligible), and one says that x2 = A for almost all x, in other words, almost everywhere or almost surely (\a.s."). 2a6 Definition. Let (X; A; µ) be a probability space, and A; B ⊂ X. (a) The distance between A and B is1 dist(A; B) = µ∗(A4B); here the symmetric difference A4B = (A n B) [ (B n A) = fx : 1A(x) 6= 1B(x)g. (b) Sets A; B are equivalent or almost equal (symbolically, A ∼µ B) if dist(A; B) = 0, that is, their symmetric difference is a null set. Note that (2a7) dist(A; B) = minfµ∗(C): A \ (X n C) = B \ (X n C)g : (\By throwing C away we get A = B". ) 2a8 Core exercise. Prove that dist : 2X × 2X ! [0; 1] is a pseudometric, that is, (a) dist(A; A) = 0, (b) dist(A; B) = dist(B; A) (\symmetry"), (c) dist(A; C) ≤ dist(A; B) + dist(B; C) (\triangle inequality"). 1Not at all the distance between their closest points. Tel Aviv University, 2012 Measurability and continuity 27 2a9 Core exercise. Let An;Bn ⊂ X, dist(An;Bn) = "n, then (a) dist(A1 \ A2;B1 \ B2) ≤ "1 + "2, (b) dist(A1 [ A2;B1 [ B2) ≤ "1 + "2, (c) dist(A1 n A2;B1 n B2) ≤ "1 + "2, (d) dist(A14A2;B14B2) ≤ "1 + "2. Prove it. Generalize (a), (b) to finite and countable operations. Deduce that µ µ µ µ 8n An ∼ Bn implies A1 \ A2 ∼ B1 \ B2, A1 [ A2 ∼ B1 [ B2, A1 n A2 ∼ µ B1 nB2 and A14A2 ∼ B14B2. Generalize the first two relations to countable operations. 2a10 Core exercise. (a) A set is µ-measurable if and only if it is equivalent to some A-measurable set. (b) Aµ is a σ-algebra. (c) there exists one and only one probability measureµ ¯ on (X; Aµ) such thatµ ¯jA = µ. Prove it. The probability space (X; Aµ; µ¯) is called the completion of (X; A; µ). It is complete; that is, the given σ-algebra contains all null sets. Sometimes I'll write µ(A) (instead ofµ ¯(A)) for A 2 Aµ. Consider the set 2X =∼µ of all equivalence classes. According to 2a9 we may apply finite and countable (but not uncountable) operations to equivalence classes; and the usual properties (such as (A [ B) \ C = (A \ C) [ (B \ C)) hold. The relation A ⊂ B for A; B 2 2X =∼µ may be defined by µ∗(AnB) = 0, and is equivalent to A = A \ B (as well as B = A [ B). Also the distance is well-defined on 2X =∼µ , and is a metric; that is, a pseudometric satisfying dist(A; B) = 0 =) A = B for A; B 2 2X =∼µ : ∗ X µ Note that µ∗; µ : 2 =∼ ! [0; 1] are well-defined. The set of all equivalence classes of measurable sets,1 µ µ A=∼ = Aµ=∼ ; is called the measure algebra of (X; A; µ). It is closed under the finite and countable operations.2 Note that µ : A=∼µ ! [0; 1] is well-defined. 1\Many of the difficulties of measure theory and all the pathology of the subject arise from the existence of sets of measure zero. The algebraic treatment gets rid of this source of unpleasantness by refusing to consider sets at all; it considers sets modulo sets of measure zero instead." (P.R. Halmos, \Lectures on ergodic theory", Math. Soc. Japan 1956, p. 42.) 2In fact, A=∼µ is a complete Boolean algebra, and is topologically closed in 2X =∼µ . Tel Aviv University, 2012 Measurability and continuity 28 X Recall ∼E; Ed; Es; Eδ; Eσ introduced in Sect. 1(a,b) for arbitrary E ⊂ 2 . Now they are also well-defined for arbitrary E ⊂ 2X =∼µ . Similarly to 1a10 X µ and 1b8, a set E ⊂ 2 =∼ is called an algebra if it contains ∼E, Ed, Es, and σ-algebra, if it contains ∼E, Eδ, Eσ. 2a11 Core exercise. Measure subalgebras, that is, sub-σ-algebras of A=∼µ , are in a natural bijective correspondence with sub-σ-algebras of Aµ that contain all null sets.1 Formulate it accurately, and prove. We know (recall 1a21 and 1c18 or the paragraph before 1b8) that Eds = Esd X but generally Eδσ 6= Eσδ for E ⊂ 2 . µ 2a12 Proposition. For every algebra E ⊂ A=∼ the set Eδσ is a σ-algebra, and Eδσ = Eσδ. A surprise: no more Eσδσ and the like! In particular, for an arbitrary probability measure on the Cantor set we µ µ have Fσ=∼ = Gδ=∼ (recall 1b7). 2a13 Extra exercise. Does the equality Eδσ = Eσδ hold for arbitrary sets (not just algebras) E ⊂ A=∼µ ? The \sandwich" below implies 2a12 and gives substantially more detailed information. 2a14 Proposition. Let E ⊂ A be an algebra and B 2 σ(E); then there µ µ exist A 2 Eδσ and C 2 Eσδ such that A ⊂ B ⊂ C and A ∼ B ∼ C. For an arbitrary probability measure on the Cantor set we get A 2 Fσ, C 2 Gδ. In particular, every null set is contained in a Gδ null set. In order to get the same for a probability measure on Rd we generalize 2a14. Given a set F ⊂ A, we introduce µ∗F (A) = supfµ(B): B 2 F;B ⊂ Ag for A ⊂ X: 2a15 Proposition. If F ⊂ A satisfies Fs ⊂ F and Fδ ⊂ F then the set f A 2 Aµ : µ∗F (A) = µ(A) ^ µ∗F (X n A) = µ(X n A) g is a σ-algebra. Let us denote this σ-algebra by Sandwich(F). Clearly, the set F of all closed (or only compact) subsets of Rd satisfies Fs ⊂ F and Fδ ⊂ F. 1That is, all subsets of all measure 0 sets of the whole σ-algebra A. Tel Aviv University, 2012 Measurability and continuity 29 2a16 Core exercise. Let X = Rd, A = B(Rd), F be the set of all closed subsets of Rd, and µ a probability measure on (X; A). Then (a) A ⊂ Sandwich(F). (b) Aµ = Sandwich(F). Deduce it from 2a15. Do the same for compact (rather than closed) subsets of Rd. 2a17 Core exercise. If Fδ ⊂ F then the set E = fA 2 Aµ : µ∗F (A) = µ(A)g satisfies Eδ = E. Prove it. 2a18 Core exercise. If F ⊂ A satisfies Fs ⊂ F then µ fA 2 Aµ : µ∗F (A) = µ(A)g = fA 2 Aµ : 9B 2 Fσ ( B ⊂ A ^ B ∼ A ) g : Prove it. 2a19 Core exercise. For arbitrary F ⊂ A the set E = fA 2 Aµ : 9B 2 µ Fσ ( B ⊂ A ^ B ∼ A ) g satisfies Eσ ⊂ E. Prove it. 2a20 Core exercise. Prove Prop. 2a15. 2a21 Core exercise. Prove Prop. 2a14. Here is another useful implication of the sandwich. 2a22 Proposition. If two probability measures on (X; A) coincide on an algebra E ⊂ A then they coincide on σ(E). Proof.
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