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Field ( -Algebra) σ-field (σ-algebra) We will shown later that the class 2X of all subsets can be too large for our quantities of interest to hold, with consistency, across all member of the class. In this section, we present smaller classes which have several “nice” properties. • Let X be a (nonempty) set (universal), and F is a family/set of subsets of X. F is called a field (or algebra ) (of sets) provided that 1) ∅∈F (or X ∈F or F is non-empty.) 2) AA∈⇒FFc ∈. 3) AB, ∈⇒∪∈FF A B . • F ⊂ 2X 3* ) AB, ∈⇒∩∈FF A B c Proof. AB∩=()cc A ∪ B • AB,\∈⇒FF A B ∈ Proof. A \ BA=∩c B. • Property (1) can be replaced by X ∈F because they are equivalent under property (2). • Property (1) can be replaced by (*) F is non-empty. This is because, (2) and (3) implies that (1) is equivalent to (*). Proof. “⇒” Trivial. ∅∈F ⇒ F is non-empty. “⇐” If F is non-empty, then there exists A∈ F. Now, by (2) , we have Ac is also in F. Hence, by (3), XAA=∪c ∈F . By (2), ∅ =∈X c F . • Property (3) can be replaced by (3* ) : AB, ∈ FF⇒∩∈ A B . Proof. Under (2), we will show that (3) is equivalent to (3* ) . Suppose AB, ∈ F ; by (2), we have ABcc, ∈ F . Now, suppose we have (3), then c ABcc∪∈F . Apply (2) again, we have AB∩ =∪( Acc B) ∈F . Similarly, suppose we have (3* ) . Then, ABcc∩ ∈ F . Apply (2) and we c then have AB∪=( Acc ∩ B) ∈F . • AB,\∈⇒Δ=FF A B() A B ∪ B\ A ∈. • An algebra F, is closed under finite unions and finite intersections: If n n AA12,,,… An ∈ F , then ∪ Ai ∈ F , and ∩ Ai ∈ F . i=1 i=1 Proof. Use induction. Hence, a field is closed under finite set-theoretic operations. • Example of filed • X = any fixed interval of . F = {finite unions of intervals contained in X}. • X = (0,1]. B0 = the collection of all finite unions of intervals of the form (ab,0,] ⊂ ( 1] . ∞ ⎛⎤11 • Not a σ -field . Consider the set ∪⎜ , ⎥ . i=1 ⎝⎦212ii+ • Let F consist of finite and the cofinite sets (A being cofinite if Ac is finite. Then F is a field. Proof. 1) ∅∈F because ∅ = 0 is finite. 2) If A∈F , then A or Ac is finite. c Now, for Ac to be in F , we need Ac or ( Ac ) = A to be finite which is exactly what we have. So, Ac ∈ F . 3) If AB, ∈ F , then ABcc∩≤min( AB cc , ) is guaranteed to be finite except when both Acc, B are infinite. However, in that case, A , B are finite because AB, ∈ F . So, A ∪≤BAB + is finite. • If X itself is finite, then all subsets of X is finite and hence F = 2X and F is a σ -field also. • If X is infinite, then F is not a σ -field . Consider the sequence xx12,,… of distinct elements in X. Let . Then Ai ∈ F ∞ because it is finite. However, BAxi= ∪ 22ij =∈∉{}: F because B is j=1 c clearly not finite, and B contains the infinite set {xii : odd} . • Suppose X 0 ∈ F , a field (in X). Then FF∩=∈XF00{ : FX ⊂}, i.e. we just collect the members of F which are inside X 0 . Proof. Let EF=∈{EEX: ⊂0}. 1) Suppose E ∈E , then E = EX∩∈∩0F X0. Hence, EF⊂∩X 0 . ∈F • (1) says that any set in F which is already inside X 0 is good. 2) Suppose A =∩FX0 . Then, AFX= ∩∈0 F (FF∩⊂X 0 ) , and ∈F ∈F ∈F AF=∩ X0 ⊂ X0. So, FE∩ X 0 ⊂ . • (2) says that we don’t have to care about set which is not in F (there is nothing new) or is not contained in X 0 . • A field F (on X) is called a σ-field provided that ∞ 3′ If A1, A2, … is a countable sequence of sets in F, then ∪ Aj ∈ F . j =1 (closed under countable union) ∞ ′′ 3 ) ∩ Aj ∈ F . (closed under countable intersection.) j =1 c ∞∞⎛⎞ c ′ Proof. ∩∪Aj= ⎜Aj⎟ which is in F by applying (2), (3 ) , and then (2). jj==11⎝⎠ • (3′) ⇒ (3); hence, any σ-algebra is an algebra. Also, necessary and sufficient condition for being a σ-algebra is then (1), (2), and (3′). ∞ Proof. Let the sequence Ai ’s be A,,,,,BBBB…. Then, ∪ AABj = ∪∈F . j =1 • ()3′ can be replaced by ()3′′ . Proof. Use (2) and DeMorgan’s law. • A σ-field is closed under countable set-theoretic operations. • ()3′ can be replaced by ()33+ () , where ()3 : AAAn ⇒∈F ∈F Proof. Start with An ∈ F . “⇒” is obvious because A= ∪ An . “⇐” Define n n Bn= ∪ Ai. Then, by (3), Bn ∈ F . Also, Bnnn ∪∪AB= = A. Hence, i=1 nn by (3), A∈F . • ()3′ can be replaced by ()33+ (ˆ) or (33* ) + (ˆ) , where ˆ ()3: AAAn ⇒∈F ∈F Proof. Under (2), we know that • ()3 and (3* ) are equivalent. • ()3′ (countable union) is equivalent to (3′′) (countable intersection). Now, start with An ∈ F . “⇒” is obvious because (3ˆ) (monotone intersection) is a special case of n ′′ * ()3 (countable intersection). “⇐” Define Bni= ∩ A . By ()3 , Bn ∈ F . i=1 ˆ Also, Bnnn ∩∩AB= = A. Hence, by (3,) A∈F . nn • A field is sometimes called a finitely additive field to stress that it need not be a σ -field . • Definition: A class M of subsets of X is monotone if it satisfies (3) and (3.ˆ) • A monotone field is equivalent to σ -field Proof. (1) + (2) + (3′) ≡ (1) + (2) + (3) + (3) or (3.ˆ) • Example of σ-algebra: • A =∅{ , X} (the trivial σ-algebra. This is the smallest σ-algebra in X. • 2X : Set of all subsets of X. This is the largest σ-algebra in X. • {∅,,A AXc , } for A ⊂ X . • F = collection of subsets of X which are either countable or have countable complements (i.e. F consist of the countable and the cocountable sets) Proof. 1) ∅ is countable ⇒ ∅∈F . 2) Suppose A∈F , then A or Ac is countable, and hence (by symmetry of the definition) Ac ∈ F . 3) Suppose c ∞ ∞∞ ∞ ⎛⎞ c c Ai ∈ F . Consider ∪ Ai and ⎜⎟∪∩Ai= Ai. (3.1) If ∩ Ai is countable, i=1 ⎝⎠ii==11 i=1 ∞ c c then we are done. (3.2) Suppose ∩ Ai is uncountable, then all Ai ’s are i=1 uncountable (because the uncountable intersection is in it). Because c Ai ∈ F , and Ai is uncountable, we have Ai are all countable. Countable ∞ union of countable sets are countable; hence, ∪ Ai is countable. i=1 • Let X be an uncountable set. Then, by axiom of choice, X contains a set A such that A and Ac are both uncountable. Such a set is not in F . So, • The σ-algebra may not contain all the subsets of X. • The σ-algebra may not be closed under the formation of arbitrary unions. (Here, all singletons {x}∈ F because they are all finite and hence countable. However, A is an uncountable union of some {x}∈ F , but A∉F . • The Borel σ-algebra defined below. • Unfortunately, there is no constructive means of describing the σ -algebra generated by a class of sets. That is, we cannot give a prescription of adding all countable unions, then all complements, and so on, and be ensured of thereby giving an algorithm for obtaining all σ -algebra members as sequences of set theoretic operations on members of the original collection. [Gray p. 16] • Properties of σ-algebra F : • Nonempty: ∅∈F , X ∈F • F ⊂ 2X • AA∈⇒FFc ∈ • AB, ∈ F ⇒ AB∪∈F , AB∩ ∈ F , AB\ ∈F n n • AA12,,,… An ∈ F ⇒ ∪ Ai ∈ F , and ∩ Ai ∈ F i=1 i=1 ∞ ∞ • A1, A2, … ∈ F ⇒ ∪ Aj ∈ F , and ∩ Aj ∈ F . j =1 j =1 • A σ-field is closed under countable set-theoretic operations. • The set of σ-algebra in X is closed under arbitrary intersection. • X 0 ∈ F ⇒ FF∩=∈XF00{ : FX ⊂}. • If F is a σ-algebra in X, then ( X ,F ) is called a measurable space, and the members of F are called the F − measurable sets in X. When the σ-algebra F is fixed, the set will usually be said to be measurable. • Example of algebras that are not σ-algebras. • Suppose X is an infinite set, and F=⊂{SXS: or Sc is finite}. Then F is an algebra but not a σ-algebra. • An infinite σ − algebra F on X is uncountable i.e. σ − algebra is either finite or uncountable. Proof. Suppose F is countably infinite. Define SAx =∈{ F : xA ∈} . Then, Sx is at most countable because Sx ⊂ F . Let Bx = ∩Sx (intersection of all the sets in Sx ). Then, 1) Bx is still in F because each set A in Sx is in F and σ − algebra is close under countable intersection. 2) xA∈∈F ⇒ AS⊂ x ⇒ Bx ⊂ A . 3) For any C ∈F and x ∈ X , either Bx ⊂ C or Bx ⊂ XC\ . This follows directly from above (i) x ∈CBC⇒⊂x , and (ii) x ∉CBXC⇒⊂x \ . 4) Bx ≠ By ⇒ BBxy∩=∅. Recall that Bx = By iff BBxy\ =∅ and BByx\ =∅. Hence, Bx ≠ By iff BBxy\ ≠ ∅ or BByx\ ≠∅. Without loss of generality, assume BByx\ ≠ ∅ . Then, ∃A0 ∈ F such that x ∈ A0 but y ∉ A0 ( y ∈ XA\ 0 .) Otherwise, if ∀A∈ F x ∈ Ay⇒∈A, then, SSx ⊂ y and hence By ⊂ Bx which disagrees with BBxy\ ≠∅. Now, because xA∈∈0 F and yXA∈ \ 0 ∈ F we have Bx ⊂ A0 and By ⊂ XA\ 0 . Hence, BBxy∩ =∅. 5) ∀∈C F C is a (disjoint) union of some Bx ’s. This is because CB⊂ ∪ y yC∈ by for yC∈ we also have (by (2)) By ⊂ C .
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