Intro to Probability

4. Measurable Spaces

4.1 σ\sigma-Algebras and Measurable Spaces

Definition 4.1 (σ\sigma-algebra): Let Ω\Omega be a non-empty set. A family of subsets F\sigmaF of Ω\Omega is called a σ\sigma-algebra on Ω\Omega if
  1. Ω∈F\Omega \in \sigmaF
  2. A∈F  ⟹  Ac∈F\evA \in \sigmaF \implies \evA^c \in \sigmaF
  3. (An)n∈N\sequence{\evA} with ∀n∈N:An∈F  ⟹  ⋃n∈NAn∈F\forall n \in \N : \evA_n \in \sigmaF \implies \bigcupN \evA_n \in \sigmaF
Note: Elements of F\sigmaF are called events and denoted by A\evA, B\evB, C\evC, etc.
Example (Trivial σ\sigma-algebra): Let Ω≠∅\Omega \neq \varnothing be an arbitrary set. Let F={∅,Ω}\sigmaF = \set{\varnothing, \Omega}. Then, F\sigmaF is a σ\sigma-algebra on Ω\Omega.
  1. Clearly, Ω∈F\Omega \in \sigmaF.
  2. Let A∈F\evA \in \sigmaF. Then, there are only two cases, either A=∅\evA = \varnothing or A=Ω\evA = \Omega. In each case, Ac∈F\evA^c \in \sigmaF.
  3. Consider a countable collection {Ai | i∈N}⊆F\set{\evA_i \mid i \in \N} \subset \sigmaF. This collection is composed only of the sets Ai=∅\evA_i = \varnothing or Ai=Ω\evA_i = \Omega, i∈Ni \in \N. Thus ⋃n∈NAi={Ωif ∃i s.t. Ai=Ω∅otherwise \bigcupN \evA_i = \begin{cases} \Omega & \text{if } \exists i \text{ s.t. } \evA_i = \Omega \\ \varnothing & \text{otherwise} \end{cases} and ⋃n∈NAi⊆F\bigcupN \evA_i \subset \sigmaF.
Example (Power set): Let Ω≠∅\Omega \neq \varnothing be an arbitrary set. Let F\sigmaF be the family which consists of all possible subsets of Ω\Omega, i.e. F={A | A⊆Ω}\sigmaF = \set{\evA \mid \evA \subset \Omega}. This F\sigmaF, called power set and denoted with P(Ω)\powerset(\Omega) or 2Ω2^{\Omega}, is a σ\sigma-algebra on Ω\Omega.
  1. Since Ω⊆Ω\Omega \subset \Omega, Ω∈F\Omega \in \sigmaF.
  2. Let A∈F\evA \in \sigmaF, then by definition Ac=Ω∖A⊆Ω\evA^c = \Omega \setminus \evA \subset \Omega and hence Ac∈F\evA^c \in \sigmaF.
  3. Let {Ai | i∈N}⊆F\set{\evA_i \mid i \in \N} \subset \sigmaF. Then, by definition ⋃n∈NAi={ω∈Ω | ∃i s.t. ω∈Ai}⊆Ω \bigcupN \evA_i = \set{\omega \in \Omega \mid \exists i \text{ s.t. } \omega \in \evA_i} \subset \Omega and ⋃n∈NAi⊆F\bigcupN \evA_i \subset \sigmaF.
Note: The trivial σ\sigma-algebra is the smallest possible σ\sigma-algebra on Ω\Omega. The power set P(Ω)\powerset(\Omega) is the largest possible σ\sigma-algebra on Ω\Omega.
Example (Uncountable Ω\Omega, countable A\evA or Ac\evA^c): Let Ω\Omega be an uncountable set. We consider the family F={A | A⊆Ω s.t. A is countable or Ac is countable} \sigmaF = \set{\evA \mid \evA \subset \Omega \text{ s.t. } \evA \text{ is countable or } \evA^c \text{ is countable}} Then, F\sigmaF is a σ\sigma-algebra on Ω\Omega.
  1. We have that Ωc=∅\Omega^c = \varnothing. We know that #∅=0\#\varnothing = 0, in particular ∅\varnothing is countable. Thus, Ω∈F\Omega \in \sigmaF.
  2. Let A∈F\evA \in \sigmaF. Thus, A\evA is countable or Ac\evA^c is countable. Since (Ac)c=A(\evA^c)^c = \evA, Ac∈F\evA^c \in \sigmaF.
  3. Let {Ai | i∈N}⊆F\set{\evA_i \mid i \in \N} \subset \sigmaF. If there exists j∈Nj \in \N s.t. Aj\evA_j is uncountable, we have that {⋃n∈NAi}c=⋂i∈NAic=⋂i∈N, i≠jAic∩Ajc⊆Ajc \set{\bigcupN \evA_i}^c = \bigcap_{i \in \N} \evA_i^c = \bigcap_{i \in \N, \ i \neq j} \evA_i^c \cap \evA_j^c \subset \evA_j^c hence, using the fact that Ajc\evA_j^c must be countable, we have #{⋃n∈NAi}c≤#Ajc≤#N \#\set{\bigcupN A_i}^c \leq \# \evA_j^c \leq \# \N and ⋃n∈NAi∈F\bigcupN \evA_i \in \sigmaF. If for all i∈Ni \in \N we have that Ai\evA_i is countable we rely on TODO: reference Proposition (Union of countable sets) and conclude that ⋃n∈NAi\bigcupN \evA_i must be countable as well and hence ⋃n∈NAi∈F\bigcupN \evA_i \in \sigmaF.
Note: Notice for the aforementioned σ\sigma-algebra that since Ω\Omega is uncountable it is not true that F=P(Ω)\sigmaF = \powerset(\Omega). As an example consider Ω=[0,1)\Omega = [0,1), then both A=[0,0.5)\evA = [0,0.5) and Ac=[0.5,1)\evA^c = [0.5,1) are not countable.
Example (σ\sigma-algebra restricted on subset): Let Ω\Omega be a non empty set and F\sigmaF be a σ\sigma-algebra on Ω\Omega. Let Ω0⊆Ω\Omega_0 \subset \Omega s.t. Ω0≠∅\Omega_0 \neq \varnothing. Then, the collection F∩Ω0={A∩Ω0 | A∈F}\sigmaF \cap \Omega_0 = \set{\evA \cap \Omega_0 \mid \evA \in \sigmaF} is a σ\sigma-algebra on Ω0\Omega_0.
  1. Clearly, Ω0=Ω∩Ω0∈F∩Ω0\Omega_0 = \Omega \cap \Omega_0 \in \sigmaF \cap \Omega_0.
  2. If B∈F∩Ω0\evB \in \sigmaF \cap \Omega_0, then B=A∩Ω0\evB = \evA \cap \Omega_0 for some A∈F\evA \in \sigmaF and Ω0∖B=(Ω0∖A)∪(Ω0∖Ω0)=Ω0∖A=Ac∩Ω0∈F∩Ω0\Omega_0 \setminus \evB = (\Omega_0 \setminus \evA) \cup (\Omega_0 \setminus \Omega_0) = \Omega_0 \setminus \evA = \evA^c \cap \Omega_0 \in \sigmaF \cap \Omega_0.
  3. Suppose that {Bi | i∈N}⊆F∩Ω0\set{\evB_i \mid i \in \N} \subset \sigmaF \cap \Omega_0. Therefore, for any i∈Ni \in \N, Bi=Ai∩Ω0\evB_i = \evA_i \cap \Omega_0 for some Ai∈F\evA_i \in \sigmaF. Then, since ⋃n∈NBi=(⋃i∈NAi)∩Ω0\bigcupN \evB_i = (\bigcup_{i\in\N} \evA_i) \cap \Omega_0 and ⋃i∈NAi∈F\bigcup_{i\in\N} \evA_i \in \sigmaF, it follows that ⋃n∈NBi∈F∩Ω0\bigcupN \evB_i \in \sigmaF \cap \Omega_0.
Example (Infinte Ω\Omega, finite A,Ac\setA, \setA^c): Let Ω\Omega be an infinite set. Define the family G={A | A⊆Ω s.t. A is finite or Ac is finite} \setG = \set{A \mid A \subset \Omega \text{ s.t. } \setA \text{ is finite or } \setA^c \text{ is finite}} Then, G\setG is not a σ\sigma-algebra on Ω\Omega. To see this, let {ωi | i∈N}\set{\omega_i \mid i \in \N} be a countably infinite sequence of distinct points of Ω\Omega and define the set A={ω2i | i∈N}\setA = \set{\omega_{2i} \mid i \in \N}. We note that both A\setA and, since {ω2i+1 | i∈N}⊆Ac\set{\omega_{2i + 1} \mid i \in \N} \subset \setA^c, Ac\setA^c are not finite. Let Ai={ω2i}\setA_i = \set{\omega_{2i}} s.t. A=⋃n∈NAi\setA = \bigcupN \setA_i. It is clear that Ai∈G\setA_i \in \setG but A∉G\setA \not\in \setG, hence G\setG is not a σ\sigma-algebra on Ω\Omega.
Example (Left-open real intervals): Let Ω=R\Omega = \R and define the family R={A=(a,b] | a,b∈R}∪{∅} \setR = \set{\setA = (a,b] \mid a,b \in \R} \cup \set{\varnothing} Then, R\setR is not a σ\sigma-algebra on Ω\Omega. To see this, let A=(a,b]∈R\setA = (a,b] \in \setR with A≠∅\setA \neq \varnothing, then Ac=(−∞,a]∪(b,∞)∉R\setA^c = (-\infty,a] \cup (b,\infty) \not \in \setR.
Definition 4.2 (Sub-σ\sigma-algebra): Let F\sigmaF be a σ\sigma-algebra on Ω\Omega. If A⊆F\setA \subset \sigmaF and A\setA is a σ\sigma-algebra on Ω0⊆Ω\Omega_0 \subset \Omega, then A\setA is referred to as a sub-σ\sigma-algebra of F\sigmaF.
Example (σ\sigma-algebra restricted on subset continued): We show Ω0∈F\Omega_0 \in \sigmaF if and only if F∩Ω0\sigmaF \cap \Omega_0 is a sub-σ\sigma-algebra of F\sigmaF. We prove the direction “Ω0∈F\Omega_0 \in \sigmaF”   ⟹  \implies “F∩Ω0\sigmaF \cap \Omega_0 is a sub-σ\sigma-algebra of F\sigmaF” first. We know that F∩Ω0={A∩Ω0 | A∈F}\sigmaF \cap \Omega_0 = \set{\evA \cap \Omega_0 \mid \evA \in \sigmaF} and from an exercise (TODO, make proposition) we know that if A∈F\evA \in \sigmaF and Ω0∈F\Omega_0 \in \sigmaF, then A∩Ω0∈F\evA \cap \Omega_0 \in \sigmaF, thus F∩Ω0⊆F\sigmaF \cap \Omega_0 \subset \sigmaF. We already proved before that F∩Ω0\sigmaF \cap \Omega_0 is a σ\sigma-algebra. For the direction “F∩Ω0\sigmaF \cap \Omega_0 is a sub-σ\sigma-algebra of F\sigmaF”   ⟹  \implies “Ω0∈F\Omega_0 \in \sigmaF” we note that Ω0∈F∩Ω0\Omega_0 \in \sigmaF \cap \Omega_0 as Ω∩Ω0=Ω0\Omega \cap \Omega_0 = \Omega_0. Hence Ω0∈F\Omega_0 \in \sigmaF follows from F∩Ω0⊆F\sigmaF \cap \Omega_0 \subset \sigmaF.
Definition 4.3 (Measurable space): Let Ω≠∅\Omega \neq \varnothing and F\sigmaF be a σ\sigma-algebra on Ω\Omega. The pair (Ω,F)(\Omega, \sigmaF) is referred to as a measurable space.
Note: If A∈F\evA \in \sigmaF, then A\evA is said to be measurable, i.e. each event is measurable.

4.2 σ\sigma-Algebras generated by Families of Sets

The next result is of general importance as it shows that even though a family of subsets G\setG might not be a σ\sigma-algebra, one can always find a σ\sigma-algebra which is the smallest possible σ\sigma-algebra that contains G\setG.

Proposition 4.4 (σ\sigma-algebra generated by G\setG): Let Ω≠0\Omega \neq 0 and G\setG be a family of subsets of Ω\Omega. Then, there exists a σ\sigma-algebra σ(G)\sigma(\setG) which satisfies:
  1. G⊆σ(G)\setG \subset \sigma(\setG)
  2. If G⊆U\setG \subset \setU and U\setU is a σ\sigma-algebra, then σ(G)⊆U\sigma(\setG) \subset \setU.
Note: The second property means that σ(G)\sigma(\setG) is a sub-σ\sigma-algebra of any other σ\sigma-algebra U\setU on Ω\Omega.

To prove proposition (TODO: ref) we rely on the following result.

Proposition 4.5 (Union of σ\sigma-algebras is a σ\sigma-algebra): Let Ω≠∅\Omega \neq \varnothing be a set. Let Fi\sigmaF_i, i∈Ii \in \setI be a collection of σ\sigma-algebras on Ω\Omega over an arbitrary set I\setI. Then F=⋂i∈IFi\sigmaF = \bigcap_{i \in \setI} \sigmaF_i is a σ\sigma-algebra on Ω\Omega.

TODO: PROOF

The σ\sigma-algebra σ(G)\sigma(\setG) is referred to as the σ\sigma-algebra generated by G\setG.

Proposition 4.6 (Properties of σ\sigma-algebra generated by G\setG): Let σ(G)\sigma(\setG) be the σ\sigma-algebra generated by a family of subsets G\setG of Ω\Omega. Let A\setA be another family of subsets of Ω\Omega. Then,
  1. if A\setA is a σ\sigma-algebra s.t. G⊆A\setG \subset \setA and A⊆σ(G)\setA \subset \sigma(\setG), then A=σ(G)\setA = \sigma(\setG)
  2. A⊆G  ⟹  σ(A)⊆σ(G)\setA \subset \setG \implies \sigma(\setA) \subset \sigma(\setG)
  3. A⊆G⊆σ(A)  ⟹  σ(A)=σ(G)\setA \subset \setG \subset \sigma(\setA) \implies \sigma(\setA) = \sigma(\setG)
Example (G={∅}\setG = \set{\varnothing}): Let Ω≠∅\Omega \neq \varnothing and let G={∅}\setG = \set{\varnothing}. Then σ(G)={∅,Ω}\sigma(\setG) = \set{\varnothing, \Omega}, i.e. the trivial σ\sigma-algebra on Ω\Omega. By 1. of Proposition 4.6 it is enough to show that {∅,Ω}⊆σ(G)\set{\varnothing, \Omega} \subset \sigma(\setG) since {∅,Ω}\set{\varnothing, \Omega} is a σ\sigma-algebra that contains G\setG. It is clear that {∅,Ω}⊆σ(G)\set{\varnothing, \Omega} \subset \sigma(\setG) is a σ\sigma-algebra, hence it must contain both, ∅\varnothing and ∅c=Ω\varnothing^c = \Omega.
Example (G={{1}}\setG = \set{\set{1}}): Let Ω={1,2,3}\Omega = \set{1,2,3} and define G={{1}}\setG = \set{\set{1}}. Clearly, σ(G)\sigma(\setG) must contain {1}\set{1}, {1}c={2,3}\set{1}^c = \set{2,3}, Ω\Omega and ∅\varnothing. Thus, we claim σ(G)={∅,{1},{2,3},Ω}\sigma(\setG) = \set{\varnothing, \set{1}, \set{2,3}, \Omega}. More generally, if A⊆Ω\evA \subset \Omega, then σ(A)={∅,A,Ac,Ω}\sigma(\setA) = \set{\varnothing, \evA, \evA^c, \Omega}.
Example (G={{ω} | ω∈Ω}\setG = \set{\set{\omega} \mid \omega \in \Omega}): Let Ω\Omega be an uncountable set and F={A | A⊆Ω s.t. A is countable or Ac is countable} \sigmaF = \set{\evA \mid \evA \subset \Omega \text{ s.t. } \evA \text{ is countable or } \evA^c \text{ is countable}} be the σ\sigma-algebra introduced in a previous example. We show that F=σ(G)\sigmaF = \sigma(\setG) where G={{ω} | ω∈Ω}\setG = \set{\set{\omega} \mid \omega \in \Omega}. Clearly G⊆F\setG \subset \sigmaF since each set {ω}\set{\omega} has cardinality one. Thus, it remains to show that F⊆σ(G)\sigmaF \subset \sigma(\setG). Let A∈F\evA \in \sigmaF. Then either A\evA or Ac\evA^c is countable. Suppose that A\evA is countable, then A={ωi | i∈N}\evA = \set{\omega_i \mid i \in \N} for some collection of singletons ωi∈Ω\omega_i \in \Omega. Therefore, A=⋃n∈N{ωi}\evA = \bigcupN\set{\omega_i} and thus A∈σ(G)\evA \in \sigma(\setG) since {ωi}∈σ(G)\set{\omega_i} \in \sigma(\setG). If Ac\evA^c is countable, by the latter argument, Ac∈σ(G)\evA^c \in \sigma(\setG) and hence also (Ac)c=A∈σ(G)(\evA^c)^c = \evA \in \sigma(\setG).
Note: The set G={{ω} | ω∈Ω}\setG = \set{\set{\omega} \mid \omega \in \Omega} is referred to as the point-partition on Ω\Omega.
Proposition 4.7 (Generating sub-σ\sigma-algebra): Let Ω\Omega and Ω0⊆Ω\Omega_0 \subset \Omega be non-empty sets and let G\setG be a family of subsets of Ω\Omega. Then σ(G)∩Ω0=σ({A∩Ω0 | A∈G})\sigma(\setG) \cap \Omega_0 = \sigma(\set{\setA \cap \Omega_0 \mid \setA \in \setG}).

4.3 Borel Sets of Real Coordinate Spaces

Definition 4.8 (Borel σ\sigma-algebra on the real numbers): Let Ω=R\Omega = \R and R={(−∞,x] | x∈R}∪{∅} \setR = \set{(-\infty,x] \mid x \in \R} \cup \set{\varnothing} be the family of left open intervals with the empty set adjoined. The σ\sigma-algebra B(R)=σ(R)\borelB(\R) = \sigma(\setR) is referred to as the Borel σ\sigma-algebra on R\R.
Definition 4.9 (Borel σ\sigma-algebra on real coordinate spaces): Let Ω=Rk\Omega = \R^k, k∈Nk \in \N. Define Rk={A | A=⨂i=1k(ai,bi], ai,bi∈R, i=1,…,k}∪{∅} \setR_k = \set{\setA \mid \setA = \bigtimes_{i=1}^k (a_i,b_i], \ a_i, b_i \in \R, \ i = 1,\ldots,k} \cup \set{\varnothing} i.e. the family of rectangles in Rk\R^k. Then, the σ\sigma-algebra B(Rk)=σ(Rk)\borelB(\R^k) = \sigma(\setR_k) is referred to as the Borel σ\sigma-algebra on Rk\R^k.
Definition 4.10 (Borel σ\sigma-algebra on real coordinate spaces subset): Let E⊆Rk\setE \subset \R^k s.t. E≠∅\setE \neq \varnothing. The Borel σ\sigma-algebra on E\setE is defined by B(E)=B(Rk)∩E={A∩E | A∈B(Rk)} \borelB(\setE) = \borelB(\R^k) \cap \setE = \set{\evA \cap \setE \mid \evA \in \borelB(\R^k)}

Upon Proposition 4.7, the following result is obtained.

Proposition 4.11 (Alternative formulation of B(E)\borelB(\setE)): Given any non-empty E⊆Rk\setE \subset \R^k, we have B(E)=σ({G∩E | G∈Rk})\borelB(\setE) = \sigma(\set{\setG \cap \setE \mid \setG \in \setR_k}).