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 . Hence, ∪ By ⊂ C , and thus, yC∈
∪ By = C . yC∈
{BxXx : ∈ } By (5), there exists a mapping from 2 onto F . So, if {BxXx : ∈ } is finite, then F ≤ 2{BxXx : ∈ }, i.e., F is finite. But we assume F is not finite, so,
{BxXx : ∈ } is infinite.
∞ This implies that we can find a sequence Bx , B ∈ F , disjoint, ( i )i=1 xi nonempty. Note that for any nonempty I ⊂ N , we know that B is still in ∪ x j jI∈ F because F is close under finite union. Also, for any non-empty and distinct II, ⊂ N , we know that B ≠ B . So, there exists a mapping 12 ∪∪xixi II12 from 2N to F , and thus F ≥ 2N . But 2N is uncountable; hence F is also uncountable, contradicting the initial assumption that F is countably infinite.
• The set of σ-algebra in X is closed under arbitrary intersection, i.e., let Fα be σ-
algebra ∀∈α A where A is some index set, potentially uncountable. Then, ∩ Fα is α∈A still a σ-algebra.
Proof. Let AA,,,12 A …∈ ∩ Fα , then AA,,,12 A … ∈ Fα ∀α ∈ A . Now, because all α∈A ∞ c Fα ’s are σ-algebra, we know that, (1) ∅, (2) A , and (3) ∪ Aj are in Fα j =1 ∞ c ∀∈α A . Hence, ∅, A , and ∪ Aj are in ∩ Fα . j =1 α∈A • If A and B are σ-algebra in X, then, it is not necessary that AB∪ is also a σ- algebra. Proof. Let EE12, ⊂ X distinct, and not complement of one another. Let c Aiii=∅{ ,,E EX ,} . Then, Ei ∈AA1∪ 2 but EE121∪∉∪AA2.
Ω ⎧1, if ω ∈ A • If A∈ 2 , we define the indicator function by 1A ()ω = ⎨ . ⎩0, if ω ∉ A
We often do not explicitly with the ω , and just write 1A .
• Definition: An ∈ A converges to A (we write AAn → ) if ∀ω ∈Ω lim1ω = 1 ω . This is equivalent to AA= limsup= liminf A. AAn ( ) ( ) nn n→∞ n→∞ n→∞ • Let A be a σ -algebra , and A ∞ a sequence of events in A . Then (n )1
• liminf An ∈ A . n→∞ ∞
Proof. liminf An= Bn, where Bn= Am. Note that ∀n Bn ∈ A because A n→∞ ∪ ∩ n=1 mn≥ ∞
is closed under countable intersections. Hence, liminf ABnn=∈A n→∞ ∪ n=1 because A is closed under countable unions.
• limsup An ∈A . n→∞ Proof. Can proof directly by observing that it is just union and intersection of sets in A . c c ∞∞∞ ∞ cc⎛⎞ Alternatively, from liminf ABBnnn=== A m =An, ()n→∞ ∪∩∩∩∩∪⎜⎟ nnnmnnmn===≥=≥111⎝⎠ 1 c we have limsupA = liminf Ac . Now, A is closed under complement nn( ) n→∞ n→∞ c c (*). Hence, ∀n An ∈ A . By the first part liminf An ∈A . Applying (*) n→∞ c again, and we have liminfAAc = limsup ∈A . ( nn) n→∞ n→∞
• An → A ⇒ AAn →∈A
Proof. To say An → A, we have AA= limsupn= liminf An. We already know n→∞ n→∞
that limsup An ∈A (and liminf An ∈ A ) n→∞ n→∞
• Recall that liminfAAnn⊂ limsup . n→∞ n→∞ • The following are properties of σ -algebra which involve inverse image. For the proof, see the notes on measurable function. • Let F be a σ -algebra . M=⊂{SfS : −1 ( ) ∈F} , then M is a σ field on .
• Consider function f : FF12→ . −1 • Let F1 be a σ -algebra on F1 . Then, MF=⊂{FFfF2: ( ) ∈1} is also a σ-
algebra on F2 .
−1 • Let F2 be a σ -algebra on F2 . Then, M={ fBB( ) : ∈F2} is a σ-algebra on
F1 . • Let C ⊂ 2X , the σ-algebra generated by C , σ (C) is ∩ G , i.e., the G is a σ − field CG⊂ intersection of all σ-field containing C . It is the smallest σ-field containing C . • The intersection is well-defined (over a nonempty class) because at least the collection of all subsets of X, i.e., 2X is one such σ-field which contains C . • σ ()C is a σ-field Proof. The set of σ-algebra in X is closed under arbitrary intersection.
Proof. Let ABA,,12 , A ,…∈σ (C), then ABA,,12 , A ,… ∈ every G . All G is σ- field; thus, 1) ∅ in every G . 2) Ac also in every G . 3) A ∪ B also in ∞ every g. 4) ∪ Aj also in every G . j =1 • Is the smallest σ-field containing C in the sense that if H is a σ-field and CH⊂ , then σ ()CH⊂ . Proof. H is one of the G ’s. • By definition CC⊂ σ (). • σσ()()CC= σ() Proof. By definition σσσ()C⊂ ( (C)) . Now, σσ(CC) ⊂ ( ), so, σ ()C is a σ- algebra containing σ (C) . But σσ( (C )) is the smallest σ-algebra containing σ ()C . Hence, σσ( (CC)) ⊂ σ( ) . • C is a σ -algebra if and only if CC= σ ( ) . Proof. “⇒” If C is already a σ -algebra , then C itself is a σ -algebra containing C . So, σ ()C⊂ C. Combining with C⊂ σ (C) , we have CC= σ ( ) . “⇐” σ ()C is a σ -algebra ; so if CC= σ ( ) , then C is also a σ -algebra . • Properties of σ (C) : • σ ()CG= ∩ G is a σ − field CG⊂ • CC⊂ σ () • σσ()()CC= σ() • If H is a σ-field and CH⊂ , then σ (CH) ⊂ . • σσ(){∅=} (){XX} =∅{ , } • σ ({A}) =∅{ ,,AAc , X} for A ⊂ X • σσ()AA=∪∅=∪=∪∅() {} σ( A{XX}) σ( A{ , }) • [PS0] AB⊂⇒σσ() A ⊂( B) • σσ()A , ()BABA⊂∪⊂∪ σ () σ( ) σ( B) • The following properties are useful for proving σσ(AB) = ( )
• [PS1] Let Ci be a class of subsets of X. The following statements are equivalent:
(1) σσ()CC21⊂ ()
(2) ∀∈C C2 C can be generated by elements of C1
(3) CC21⊂ σ ()
(4) σσ()CC21⊂⊂ () A for any σ-algebra A containing C1 .
(5) σ ()C2 ⊂ A for any σ-algebra A containing C1 . • BA⊂ σ ( ) ⇒ σσσ()BAA⊂=∪( ) ( B) • If σσ()A = ()B , then σσσ( A) ==∪(BA) ( B) . • AB⊂⊂σ ( A) ⇒ σσ()BA= ( ) . • σσ()AA=∪∅=∪=∪∅() {} σ( A{XX}) σ( A{ , })
• Suppose we have (possibly uncountable) collection of classes Ai 's. Then ⎧⎫⊂ ⎪⎪⎛⎞⎛ ⎞⎧⊂ ⎛⎞ ⎫. σσ()AAii⊂⊂ ()⎨⎬⎧⎫σσσσAAi=() i⎜⎟ ⎨AA i: i∈ σ()Ai⎬ ∪∪A ⎜⎟⎜∪ ⎟ ∪ ii⎪⎪⎨⎬∪ AAii: ∈σ () i⊂ ⎝⎠⎝i ⎠⎩= ⎝⎠i ⎭ ⎩⎩ i ⎭ ⎭ Those in blue are only true when there are countable number of collections. • σσσσσ()ABABABAB, ()⊂∪⊂∪=∪ () ( ) ( ) σσσ( ( ) ( )) ⎛⎞ • [PS2] σσσ()AAii⊂⊂∪∪ () ⎜⎟Ai ii⎝⎠
• Suppose ABi⊂ σ ()i (which is equivalent to σσ(Ai) ⊂ (Bi)), then ⎛⎞⎛⎞ σσ⎜⎟⎜∪∪ABii⊂ ⎟. ⎝⎠⎝ii⎠ ⎛⎞⎛⎞ • Suppose σσ()Ai= ()Bi. Then, σσ⎜⎟⎜∪∪ABii= ⎟. ⎝⎠⎝ii⎠ • σσ(){∅=} (){XX} =∅{ , }. • If A ⊂ X , then σ (){A} =∅{ ,,AAc , X} . • A ⊂ B ⇒ σσ()A ⊂ ()B . Proof. By definition, B⊂ σ (B) . From A ⊂ B , we have A ⊂⊂Bσ ()B. So, σ (B) is a σ-algebra containing A. σ ( A) is the smallest σ-algebra containing A. Hence, σσ()A ⊂ ()B . Proof. Let A be a set of σ-algebra which contains A, and B be the set of σ-algebra which contains B. Because A ⊂ B , any σ-algebra in B also contains A and hence in A . So, BA⊂ . Therefore, σσ()A =∈⊂∈=∩∩{SS::AB} { SS} ( B) because σ ()A have more intersections. • Let A be a class of subsets of X. We will say that a set C can be generated by elements of C if CA∈σ ().
• [PS1] Let Ci be a class of subsets of X. The following statements are equivalent:
(1) σσ()CC21⊂ ()
(2) ∀∈C C2 C can be generated by elements of C1
(3) CC21⊂ σ ()
(4) σσ()CC21⊂⊂ () A for any σ-algebra A containing C1 .
(5) σ ()C2 ⊂ A for any σ-algebra A containing C1 . Proof. (2) and (3) are equivalent by definition.
“(3) ⇒ (1)” because CC21⊂ σ ( ) ⇒ σσσσ(CC21) ⊂=( ( )) ()C1.
“(1) ⇒ (3)” because CC2⊂ σ ( 2) . Therefore, σσ(CC21) ⊂ ( ) ⇒ CC21⊂ σ () .
“(1) ⇒ (4)” by definition of σ (C1 ) . “(4) ⇒ (5)” because (5) is expressed in
(4). “(5) ⇒ (1)” because σ (C1 ) is also a σ-algebra containing C1 .
• Remark: we can’t prove that σσ(C1) ⊂ (C2) . For example, let Xab= { , } ,
C1 = {{ab},{ }}, C2 = {{ab, }} . Then, σ (C1 ) =∅{ ,,,,{abab} { } { }} and
σ ()C2 =∅{ ,, {ab}}. So, σσ(CC21) ⊂ ( ) but σσ(CC12) ⊄ ( ) . • If B⊂ σ ()A, then σσσ()BAA⊂=∪( ) ( B) . Proof. Because A ⊂∪AB, we have σσ( A) ⊂∪( AB) . Now, from B⊂ σ ()A and A ⊂ σ ()A , we also have A ∪⊂Bσ ( A) , which implies σσ()A ∪⊂BA( ) . • If σσ()A = ()B , then σσσ( A) ==∪(BA) ( B) . • AB⊂⊂σ () A ⇒ σσ()BA= ( ) . Proof. 1) AB⊂⇒σσ() A ⊂( B) . 2) BA⊂⇒⊂σσσ( ) ( B) ( A) .
• Suppose we have (possibly uncountable) collection of classes Ai 's. Then ⎛⎞ • σσσ()AAii⊂⊂∪∪ () ⎜⎟Ai. ii⎝⎠ ⎛⎞ Proof. Ai⊂ ∪Ai which implies σσ()Ai⊂ ⎜∪Ai⎟. Because this is true ∀i , i ⎝⎠i ⎛⎞ we also have ∪∪σσ()AAii⊂ ⎜⎟. ii⎝⎠ ⎛⎞⎛ ⎞ • σσσ⎜⎟⎜∪∪AAii= ()⎟ ⎝⎠⎝ii⎠ ⎛⎞⎛ ⎞ Proof. “⇒” AAii⊂ σ () ⇒ ∪∪AAii⊂ σ ( ) ⇒ σσσ⎜⎟⎜∪∪AAii⊂ ()⎟. ii ⎝⎠⎝ii⎠ ⎛⎞ “⇐” We already know that ∪σσ()Aii⊂ ⎜∪A⎟; hence, ii⎝⎠ ⎛⎞⎛⎞ σσ⎜⎟⎜∪∪()AAii⊂ σ ⎟. ⎝⎠⎝ii⎠
• Suppose ABii⊂⊂σ ( A i) . Then, ⎛⎞⎛⎞⎛ ⎞⎛ ⎞ σσσσσσ⎜⎟⎜⎟⎜∪∪∪ABii==() B i ⎟⎜ = ∪ (Ai )⎟. ⎝⎠⎝⎠⎝iii ⎠⎝ i⎠
Proof. ABii⊂⊂σ () A i ⇒ σσ(BAi) = ( i) . The middle two terms are sandwiched. ⎧ ⎫ • Lemma: If ∅∈Ai ∀i , then ∪∪AAiii⊂∈⎨ AA: i⎬. ii⎩⎭ Proof. A∈ A ⇒ A∈ A . Take AA= ∈ A and A = ∅∈A ∀≠ii. Then, ∪ i i0 ii00ii0 i ⎧⎫ ∪∪AAiii⊂∈⎨⎬AA: i. ii⎩⎭ ⎛⎞⎛⎞ ⎧ ⎫ • If ∅∈Ai ∀i , then σσ⎜⎟∪∪AAii⊂∈⎜⎟⎨ AA: ii⎬ ⎝⎠ii⎝⎠ ⎩ ⎭ • Because ∅∈σ ()Ai automatically, we have ⎧⎫ • ∪∪σσ()AAiii⊂∈⎨⎬AA: ()i ii⎩⎭ ⎛⎞⎧⎛⎞⎫ • ⇒ σσ⎜⎟∪∪()AAii⊂∈ σ⎜⎟⎨ AA: i σ()i⎬ ⎝⎠⎩ii⎝⎠⎭ ⎛⎞⎛ ⎞⎛⎞ ⎧ ⎫ • σσσσ⎜⎟⎜∪∪AAiii=⊂∈() ⎟⎜⎟⎨ ∪AA: i σ()Ai⎬ . ⎝⎠⎝ii ⎠⎝⎠ ⎩ i ⎭ • Corollary: σσσσσ()ABABABAB, ()⊂∪⊂∪=∪( ) ( ) ( ) σσσ( ( )()) Intuitively, • The first inclusion is obvious because union gives bigger set. • The second inclusion comes from the fact that set in σσ(A)(∪ B) is either generated from A or generated from B . However, the set in σ ()AB∪ can also mix elements of A and B . • For the third equality, we know that there are more generating elements in σσ(()AB∪ σ ()). However, all of them can already be generated by elements in σ ()AB∪ .
• Suppose we have (finite or) countable collection of classes Ai 's. Then ⎛ ⎧⎫⎛⎞ ⎞ • σσ⎜⎟⎨⎬∪∪AAii: ∈AA i⊂ ⎜⎟i ⎝⎠⎩⎭⎝ii⎠
Proof. Let B= ∪Ai . Suppose Aii∈AB⊂ . Then, Ai ∈B . So, Ai ∈⊂BBσ (). i
σ (B) is a σ-algebra; hence, ∪ Ai ∈σ (B) . Therefore, i ⎧⎫⎛⎞ ⎨⎬∪∪AAii: ∈⊂AA i σ ⎜⎟i. ⎩⎭⎝ii⎠ ⎧⎫⎛⎞⎛⎞ • ⎨⎬∪∪AAii: ∈=σσ()AA i⊂ ⎜⎟⎜i σσ∪()Ai⎟. ⎩⎭ii⎝⎠⎝i⎠
Proof. A= ∪ Ai with Aii∈σ (A ) ⇒ Ai∈∪σ (Ai) ⇒ the countable union i i ⎛⎞ ∪∪Aii∈σσ⎜⎟()A . ii⎝⎠ ⎧⎫⎧⎫⎛⎞⎛⎞⎛⎞ • ⎨⎬⎨⎬∪∪∪AAii::∈∈=σσ()AAA i⊂ ⎜⎟AAii σ() i ⊂ σ⎜⎟⎜i σσ∪()Ai⎟. ⎩⎭⎩⎭⎝iii⎝⎠⎠ ⎝⎠i Proof. If CD⊂ and D is a σ-algebra, then CC⊂⊂σ ( ) D because of the minimality of σ (C) . • σσ()AA=∪∅=∪=∪∅() {} σ() A{XX} σ( A{ , }) Proof. BA=∪∅{ }. Then, AB⊂ . Suppose B ∈B , then 1) B ∈⊂AAσ (), or 2) B =∅ which automatically in σ (A) . So, BA⊂ σ ( ) . Because AB⊂⊂σ () A, we conclude that σσ(BA) = ( ) . For BA=∪{X} , replace ∅ in the above proof by X. The rest is the same. For BA=∪∅{ , X} , can combine the argument of the above two parts or set CA=∪{X} . Then, the second part gives, σσ(A) = (C) , and the first part gives σσ(CC){=∪∅( }).
• Suppose ABi⊂ σ ()i (which is equivalent to σσ(Ai) ⊂ (Bi)), then ⎛⎞⎛⎞ σσ⎜⎟⎜∪∪ABii⊂ ⎟. ⎝⎠⎝ii⎠ [PS2] ⎛⎞ Proof. ∪∪AAii⊂⊂⊂σσσ() ∪ () B i⎜ ∪Bi⎟. Hence, by [PS0], ii i ⎝⎠i ⎛⎞⎛⎞ σσ⎜⎟⎜∪∪ABii⊂ ⎟. ⎝⎠⎝ii⎠ ⎛⎞⎛⎞ • Suppose σσ()Ai= ()Bi. Then, σσ⎜⎟⎜∪∪ABii= ⎟. ⎝⎠⎝ii⎠
Proof. σσ()Ai= ()Bi ⇒ (1) σσ(Ai) ⊂ (Bi) and (2) σσ(Ai) ⊃ (Bi) . Apply the above result. ⎧⎫ • Suppose AA12,,… is a partition of X . Then, σ (){}AA12,,… = ⎨⎬∪ Ai : I⊂∪∅{}. ⎩⎭iI∈ ⎧⎫ Proof. First we will show that G = ⎨⎬∪ AIi : ⊂∪∅ {} is a σ -algebra . 1) ⎩⎭iI∈ c Trivially, by definition, ∅ ∈G . 2) B ∈G ⇒ B = ∪ Ai ⇒ B = ∪ Ai . iI∈ iI∈ \ c Because \ I ⊂ , B ∈G . 3) B j ∈G ⇒ Bji= ∪ A ⇒ ∪∪Bj = Ai . iI∈ j j iI∈∪ j j
I j ⊂ ⇒ ∪I j ⊂ ⇒ ∪Bj ∈G . j j
{AA12,,…} ⊂ G because we can set I = {i} . Suppose F is another σ -algebra containing {AA12,,…}. Then, ∅∈F and
∀⊂I ∪ Ai ∈ F because F is closed under countable union. Therefore, iI∈
GF⊂ . Hence, G is the smallest σ -algebra containing {AA12,,…}.
• In particular, for Xxi=∈{ i : } , where xi ’s are distinct. If we want a σ -algebra X which contains {xi} ∀i , then, the smallest one is 2 which happens to be the biggest one. So, 2X is the only σ -algebra which is “reasonable” to use. • Extended σ-algebra:
Let I be an index set, possibly uncountable. Suppose ∀i ∈ I , Fi is a σ-algebra on ⎧ ⎫ disjoint X i . Let X = ∪ X i , and F=∈⎨∪ EEii: F i⎬ . Then, F is a σ-algebra on i∈I ⎩⎭i∈I X .
Proof. 1) ∅∈F because ∅∈Fi ; so ∪∅ =∅∈F . 2) Let E ∈F . Then, i∈I
c ⎛⎞ EE= ∪ i , Ei ∈ F . Hence, EX= \\⎜⎟∪∪ Eiii=() XE∈F . (Note that i∈I ⎝⎠ii∈∈II∈Fi
this step need X i ’s to be disjoint.)
X
X i
Ei
3) Let Ai ∈ F . Then, Ai= ∪ Eij, where Eij, ∈ F j. Then, j∈I ∞∞ ∞ ∪∪∪∪∪AEii==,,ji Ej∈F . iijji==∈∈=11II 1 ∈Fj
• FFj ⊂ ∀j
Proof. Set Ei =∅ for ij≠ .
Remark: That the X i ’s are disjoint is necessary. • Let X12== X{ abc,,} . F1 =∅{ ,,,,,,{a} { bc} { abc}} , c F2 =∅{ ,,,,,,{bacabc} { } { }}. Then, {ab, } = { a} ∪∈{ b} F but {}ab, =∉{} c F .
• Let X1 = { abc,,}, Xbc2 = { , } , F1 =∅{ ,,,{ab} { c} ,,,{ abc}} , F2 =∅{ ,,{bd}}. Then, F =∅{ ,,,{ab} { c} ,,,,,{ abc} { bd} ,,,,,,,,,,{ abd} { bcd} { abcd}} . Note that {ab, }∈ F but {}ab,,c =∉{} cd F . • Restricted σ-algebra: Let F be σ-algebra in X, and E ⊂ X . Then,
FE =∩{FEF: ∈F} is a σ-algebra in X ∩ EE= . Proof. ∀∈F F FX⊂ . Therefore, FEXE∩ ⊂∩. 1) ∅∈F ⇒
∅∩E =∅∈FE . 2) Suppose A∈ FE , then, A = FE∩ for some F ∈ F . Hence, AXEAXEFEXFEc =∩\\ =∩ ∩= \ ∩∈F . 3) ()( ) ( ) ( ) E ∈F
Ai∈ FE ⇒ AFEii=∩ for some Fi ∈ F ⇒ ∞∞ ⎛⎞∞ ∪∪AFEFEii=∩=∩∈()⎜⎟ ∪ i FE. ii==11 ⎝⎠ i=1 ∈F Let FF=∈FF: ⊂E, then FF= . • E { } E E Proof. (1) A F F F A = FE∩ A F and A ⊂ E A∈F . ∈ E ⇒ ∃∈ ⇒ ∈ ⇔ E
(2) A∈F ⇒ A∈F and A ⊂ E ⇒ A = AE∩ ⇒ A∈ FE . E ∈F
• Restricted σ-algebra: Let X 0 ⊂ X and A is a class of sets in X , then
if X0 ∈σ X (A) σσAA∩=X ∩XF = ∈ σ A: F ⊂X, XX0 ()()00{}X() 0
i.e. we just collect the members of σ X (A) which are inside X 0 .
• In the above theorem, we define FF∩=∩XFXF00{ : ∈} . • Hence, for example, for X ⊂ R , we have
BBX = {A∈⊂=∩RR: AX} B X. Proof
• If F is a σ-field in X, then FF∩=∩XFXF00{ : ∈} is a σ-field in X 0 .
Proof. 1) ∅∈F and ∅∩X 0 =∅; hence, ∅ ∈∩F X 0 . 2) Suppose
B ∈∩F X 0 ; then ∃F ∈ F such that B = FX∩ 0 . (Also, B ⊂ X 0 .) Now, cc BXBXFX==00\\( ∩==∩ 000) XFXF \ ,
c c where FXF=∈\ F . So, B ∈∩F X 0 . X
X 0
Bc B
F
c c Note the difference in the definition between B = XB0 \ and FX= \ F which resulted from different spaces under consideration.
3) Suppose Bn ∈∩F X 0 , then ∀n ∃Fn ∈F such that Bnn=∩FX0 . ⎛⎞ Hence, ∪∪Bnn=∩=∩()FX0⎜⎟ ∪ F n X0. F is a σ-field; so, nn ⎝⎠n
∪ Fn ∈ F , and therefore, ∪ Bn ∈∩F X 0 . n n
• Let F0 be a σ-field in X 0 . Let G=⊂{GXGX: ∩∈00F} . Then, G is a σ-field in X.
Proof. 1) ∅⊂X , ∅∩X 0 =∅, and ∅ ∈ F0 (because F0 is a σ-field). Hence, ∅∈G .
2) Suppose G ∈G . Then GX∩ 00∈ F . F0 is a σ-field in X 0 ⇒
XGX00−∩() ∈F0 also. Note that XGXX000− ( ∩=−) G. Now, c c GX∩=−∩=−00() XGXXG0; hence, GX∩ 0∈ F0. Therefore, Gc ∈G .
3) Suppose Gn ∈G . Then, GXn ∩ 0∈ F0. Because F0 is a σ-field, we ⎛⎞ have ∪∪()GXnn∩=00⎜⎟ G ∩∈ XF0. Hence, ∪Gn ∈G . nn⎝⎠ n
• Let F0 be a σ-field in X 0 which contain A ∩ X 0 . Then, σ X (AF) ∩⊂X 00.
Proof. Let G=⊂{GXGX: ∩∈00F} . Then, we want to show σ X (AG) ⊂ . To do this we will show that G is a σ-field in X, and AG⊂ . From above, we
already have the first part. By definition of F0 , we have AF∩⊂X 00; so AG⊂ .
• If FA= σ X () (i.e. A generates the σ-field F in X), then FA∩=XXσ ∩ . 00X0 () In summary: σσAA∩=XX ∩ . XX() 000 ( ) Proof. 1) FA= σ X () ⇒ AF⊂ ⇒ AF∩ X 0⊂∩X 0. 2) Because F is a σ-
field in X, we know that F ∩ X 0 is a σ-field in X 0 . From (1) and (2) , we know that σ AF∩⊂∩XX. (This is where the “smallest” is X0 ()00
necessary; we can’t prove FF00⊂∩X for any F0 defined above.) FA∩⊂Xσ ∩X follows from above because σ A ∩ X is a 0X0 (0) X0 ( 0 )
σ-field in X 0 which contain A ∩ X 0 .
• Now, if X 0 ∈σ X ()A , then σσXX(AA) ∩=∈XF00{ ( ) : FX ⊂} (because
σ X ()A is a field) i.e. we just collect the members of σ X (A) which are inside
X 0 .
Proof. Let DA=∈{D σ X ( ) : DX ⊂0}. Then,
1) EX∈∩σ X ()A 0 ⇒ EX⊂ 0 . Because X 0 ∈σ X (A) , we have
E ∈σ X ()A . So, E ∈ D .
2) D ∈ D ⇒ D ∈σ X (A) and DX⊂ 0 . Hence,
DDX=∩∈∩00σ X (A) X. ∈σ X ()A Borel σ-algebra
• If X = , the Borel σ-algebra or Borel algebra B is the σ-algebra generated by the open sets (or by the closed sets, which is equivalent). • The members of B are called the Borel sets of X. • On R , the σ-algebra generated by any of the followings are Borel σ-algebra: • Open sets • Closed sets • Intervals • Open intervals • Closed intervals • Intervals of the form (−∞,a] , where a ∈ • Can replace by • Can replace (−∞,a] by (−∞,a) , [a,+∞) , or (a,+∞) . • Let Aa=+∞∈{(),: a } , Bb=+∞∈{[ ,:) b } , Ccb=−∞{( ,:] ∈ } , Ddb=−∞{(),: ∈ }. Then, σσσσ( ABCD) ===( ) ( ) ( ) =∪=∪σσ()(AD BC) =∪∪∪σ ()ABCD Proof. σσ( A) = (C) because E ∈ A iff E c ∈C . So, elements in A can be generated by elements in C and vice versa. By the same reason, σσ(BD)(= ). Element [b,+∞) in B can be generated from elements in A ⎛⎞1 by ∩ ⎜⎟bb−+∞=+∞,,[ ) . Element (a,+∞) in A can be generated from n∈ ⎝⎠n ⎡⎞1 element in B by ∪ ⎢aa+ ,,+∞⎟ =( +∞). So, σσ( A) = (B). n∈ ⎣⎠n
{(abˆ, ]} {(ab, ˆ)}
()⋅ c {(ab, ]} {(−∞,b]} {(a,+∞)} {(ab, )}
1 1 aˆ,+∞ −∞,bˆ ± ± {( )} {( )} n n
∩ to limit both endpoints ()⋅ c ab, {[ab, )} {(−∞,b)} {[a,+∞)} {[ ]} lim n = ∞ to change finite to infinite n→∞ ∪ to include infinity
{(abˆ, )} ⎡ab, ˆ {⎣ )} Already include as special case ab, ∈
abˆ, ˆ∈ ∪±∞{} {(abˆ, ˆ)}
• σσ()A = (){ (a,b )} Proof. “ ⊂ ”: ()−∞,a , (a,+∞) are of the form (a,b) . So, AD∪⊂σ ({()a,b }) which implies σσ()(AD∪⊂({ a,b)}) . Use σσ( A) =∪( AD) .
c “ ⊃ ”: ()(ab,,,=−∞∪() a][ b ∞) . So, {(a,b)} ⊂∪=σσ(BC)( A). • Every open set in is the countable union of open intervals (with rational centers). Proof. Structure theorem. Proof. Let A be an open set in . Let I be the set of all rational numbers contained in A. Then, I ⊂ and hence I is countable. Because A is open, ∀∈iA, ∃>ε 0
such that (ii−+⊂εε, )A. Let εεεεi =−+⊂sup{ :(ii , ) A}. (Can use max.)
If ∃i such that εi =∞, then A = which is an open interval, and we are
done. If not, let Niii=−( ε , i +εi) . Clearly, because NAi ⊂ , ∪ NAi ⊂ . i Now, for any x∈ A , ∃ε > 0 ( x − εε, x+⊂) A. Because is dense in , ε ⎛⎞εε ∃∈q such that xqx<<+. Note that ⎜⎟qq− , +⊂A, 2 ⎝⎠22 ⎛⎞⎛⎞εε ⎜⎟∀∈yq⎜⎟ −,, q + yxyqqx − < − + − =ε , which implies ⎝⎠⎝⎠22 ⎛⎞εε ⎛⎞ε ε ⎜⎟qq−+⊂, Nq . Also, xq∈−⎜⎟, q + . Hence, x ∈ Nq . ⎝⎠22 ⎝⎠22 • σσ(){}open= ({() a,b }): The σ-algebra generated by open sets is the same as the σ- algebra generated by open interval. Proof. Every open set in is the countable union of open intervals. Moreover, any open interval is an open set. • σσ({open}) = ({ closed}) : The σ-algebra generated by open sets is the same as the σ-algebra generated by closed sets. Proof. Any closed set F can be generated from an open set GF= c . Similarly, any open set G can be generated from a closed set FG= c . • {()a,b } ⊂−∞∈σ ({( ,,cc] }): Let D be the set of all intervals of the form (−∞,c], where c ∈ . Let (ab, ) be an open interval in . Then, (ab, ) can be generated by elements in D . Proof. Let a ∞ be a sequence of rationals decreasing to a and b ∞ be a ()n n=1 ()n n=1 sequence of rationals increasing strictly to b. (Can do this because the ∞ rationals are dense in .) Then, ()ab,,= ∪( ann b]. But n=1 c (abnn,,\,,,] =−∞( bn] ( −∞ a n] =−∞( bn] ∩−∞( an] . • {(−∞,cc] , ∈ } ⊂{ closed} ⇒σσ{( −∞ ,cc] , ∈} ⊂ {closed} • For , let
BB1 =∪−∞∈{EE{ } : } ,
BB2 =∪+∞∈{EE{ } : } ,
BB3 ={EE ∪{ −∞,: +∞} ∈ } .
The extended Borel algebra BBB=∪∪13 B ∪ B3. • B is a σ-algebra. [see extended • Example of Borel sets subset of B • open sets (by definition) • closed sets (because closed set is a complement of open set which is ∈B by definition) • countable unions of closed sets • countable intersections of open sets. • Hausdorff’s notation
• Fσ set is a countable unions of closed sets
• Gδ set is a countable intersection of open sets • Ex. open sets, intervals, countable intersection of countable union of open sets.
• is not Gδ .
• Gδ is not a field; thus, not a σ-field.
• Let interval X ⊂ . BX = σ-field generated by the field of finite unions of subintervals of X.
• Call BX the σ-field of Borel subsets of X or σ-field of Borel sets on X.
• Call set B ∈ BX Borel set. • Is the smallest σ-field containing finite unions of subintervals of X.
• We have already shown above that BBX = {A∈⊂=∩RR: AX} B X, where
BBRR∩=XAXA{ ∩: ∈}. • Borel σ-algebra on the extended real line • is
• {ABA∪∈:,,,,,B B ∈∅−∞∞−∞∞{ {}{}{ }}} • is generated by, for example
• A ∪−∞∞{{ },{ }} where σ (AB) = ⎡⎡⎤abˆˆˆ,,,,,ˆˆ ab abˆ⎤ • {⎣⎣⎦)} { } {( ⎦} • {[−∞,,bbaa]} {[ −∞ ,,,)} {( +∞]} ,,{[ +∞]}
⎡⎤⎡,,bbaaˆˆ ,,,ˆˆ ,, • {⎣⎦⎣−∞} { −∞)} {( +∞]} {[] +∞ } Here, ab, ∈ and abˆ, ˆ∈∪±∞ { } • Remark: • Also see the section on extended σ -algebra . • We can’t generate ∞ by union real numbers. So, having {∞} or some set that contains ∞ is crucial.
• {()ab,:, ab∈∪−∞∞ { ,}} can not generate B ∪±∞{}. In fact,
σ {()ab,:, ab∈∪−∞∞= { ,}} B . • For α,β ∈ , σασασασα(){( ,,+∞]} =(){[ +∞]} =({[ −∞ ,,]}) =({[ −∞ )}) =+∞+∞−∞−∞σα(){}[],,,,,,,( α][ α ][ α) σαβσαβσαβσαβσα(){[ ,,]} {() ,,,,,} {( ]} {[ )} ⊂+∞({( ,]}) ⎛⎤1 ⎡ 1 ⎤ Proof. []α,,+∞ =∩⎜b + +∞⎥ , and (α,,+∞=] ∪⎢b − +∞⎥ . Also, n≥1⎝⎦n n≥1 ⎣ n ⎦ (α,,+∞] =[ −∞ α ]c , and [α,+∞] =[ −∞,α )c . Hence, σασασασα(){( ,,+∞]} =({[ +∞]}) =({[ −∞ ,,]}) =({[ −∞ )})
=+∞+∞−∞−∞σα(){}[],,,,,,,( α][ α ][ α) Note that we can throw in {∞},{−∞} because {∞} =∩(n,∞], and n≥1 c ⎛⎞ {}−∞ =⎜⎟∪∩[] −nn,, ∞ = [ −∞ − ). ⎝⎠nn≥≥11 c Now, []αβ,,,=−∞∪([ α)( β ∞]) ; hence in σα({( ,+∞]}) . Similarly, {()αβ,,,,,} {( αβ]} {[ αβ)} ⊂+∞ σ({( α ,]}) . Remark: {[−∞,,:ααα)} ={[ −∞) ∈ } , {[αβ,,:,]} =∈{[ αβ] αβ } . Hence, [α,,∞∉] {[αβ]} . • Borel σ-algebra in Rk is generated by • the class of open sets in Rk • the class of closed sets in Rk • the class of bounded semi-open rectangles (cells) I of the form k k I =∈xaxbiR :,1,iii <≤ =…,k. Note that I =⊗(abii, ] where ⊗ denotes the { } i=1 Cartesian product ×. k • the class of “southwest regions” Sx of points “southwest” to x ∈ R , i.e. k Syxi=∈{ R :,1,, yxik ≤i =… } .