Intro to Probability

5. Measures

5.1 The Notion of a Measure

Definition 5.1 (Measure): Let (Ω,F)(\Omega, \sigmaF) be a measurable space. A function μ:F→R‾+\mu : \sigmaF \to \overline{\R}_{+} is said to be a measure on F\sigmaF if the following two items are satisfied:
  1. μ(∅)=0\mu(\varnothing) = 0
  2. Given any disjoint family

    \qty{A_i : i \in \N} \subset \sigmaF

    ,     we have that

    \mu\qty(\bigcup_{i\in \N} A_i) = \sum_{i \in \N} \mu(A_i)

    .

5.2 Point Measures

Example (Point measure): Let (Ω,F)(\Omega, \sigmaF) be a measurable space and x∈Ωx \in \Omega be a given point of Ω\Omega. Define the function δx(A)={1if x∈A0if x∉A \delta_x(A) = \begin{cases} 1 & \text{if } x \in A \\ 0 & \text{if } x \not \in A \end{cases} Then,

\delta_x : \sigmaF \mapsto \qty{0,1}

is a measure on F\sigmaF.
  1. From the definition of δx\delta_x we immediately see that δx(∅)=0\delta_x (\varnothing) = 0.
  2. Let

    \qty{A_i : i \in \N} \subset \sigmaF

    be disjoint and set A=⋃i∈NAiA = \bigcup_{i \in \N} A_i. If x∉Ax \not \in A, i.e. x∈⋂i∈NAicx \in \bigcap_{i \in \N} A_i^c, there does not exist ii s.t. x∈Aix \in A_i, hence δx(A)=0=∑i∈Nδx(Ai)\delta_x(A) = 0 = \sum_{i \in \N} \delta_x(A_i). Otherwise, if x∈Ax \in A, since

    \qty{A_i : i \in \N}

    is disjoint, there exists a unique j∈Nj \in \N s.t. x∈Ajx \in A_j. Hence also in this case δx(A)=1=δx(Aj)=∑i∈Nδx(Ai)\delta_x(A) = 1 = \delta_x(A_j) = \sum_{i \in \N} \delta_x(A_i).

In general, we state the following result.

Proposition 5.2 (Point measures): Let (Ω,F)(\Omega, \sigmaF) be a measurable space, II be a countable set and

E = \qty{x_i : i \in I} \subset \Omega

be a collection of points in Ω\Omega. Assume that αi\alpha_i, i∈Ii \in I, are s.t. αi∈[0,∞)\alpha_i \in [0, \infty) for any i∈Ii \in I. Define μ:F→R‾+\mu: \sigmaF \to \overline{\R}_{+} as μ(A)=∑i∈Iαiδxi(A)=∑x∈Eαxδx(A) \mu(A) = \sum_{i \in I} \alpha_i \delta_{x_i}(A) = \sum_{x \in E} \alpha_x \delta_x(A) where for any i∈Ii \in I δxi(A)={1if xi∈A0if xi∉A \delta_{x_i}(A) = \begin{cases} 1 & \text{if } x_i \in A \\ 0 & \text{if } x_i \not \in A \end{cases} Then, μ\mu is a measure on F\sigmaF.
Note: Consider the setting of the latter proposition. Clearly, if II is finite, μ(A)<∞\mu(A) < \infty for any A∈FA \in \sigmaF. If II is countably infinite, we that #I=#N\# I = \# \N and hence, for any A∈FA \in \sigmaF, μ(A)=∑n∈Nαnδxn(A)\mu(A) = \sum_{n \in \N} \alpha_n \delta_{x_n}(A). Therefore, since αiδxi(A)≤αi\alpha_i \delta_{x_i}(A) \leq \alpha_i and αi≥0\alpha_i \geq 0, it follows from TODO: ref to series that μ(A)<∞\mu(A) < \infty for any A∈FA \in \sigmaF if ∑i∈Iαi<∞\sum_{i \in I} \alpha_i < \infty.

5.3 Counting Measures

Example (Counting measure): We consider the measurable space (Ω,P(Ω))(\Omega, \powerset(\Omega)) where Ω\Omega is a finite set. Define

\mu: \powerset(\Omega) \to \N \cup \qty{0}

as μ(A)=#A\mu(A) = \# A. Then, μ\mu is a measure on F\sigmaF.
  1. We have μ(∅)=#∅=0\mu(\varnothing) = \# \varnothing = 0.
  2. Let

    \qty{A_i : i \in \N} \subset \powerset(\Omega)

    be disjoint. Naturally, since

    \qty{A_i : i \in \N}

    is a disjoint family of sets, the cardinality of its union is the sum of the individual set cardinalities.
Proposition 5.3 (Counting measures): Consider the measurable space (Ω,P(Ω))(\Omega, \powerset(\Omega)) where Ω\Omega is a countable but not necessarily finite set. Define μ(A)={#Aif A∈P(Ω) s.t. A is finite∞otherwise \mu(A) = \begin{cases}\# A & \text{if } A \in \powerset(\Omega) \text{ s.t. } A \text{ is finite} \\ \infty & \text{otherwise} \end{cases} Then, μ\mu is a measure on P(Ω)\powerset(\Omega).

5.4 Properties of a Measure

Proposition 5.4 (Properties of a measure): Let (Ω,F)(\Omega, \sigmaF) be a measurable space and μ\mu be a measure on F\sigmaF. Then, we have the following properties:
  1. Given n∈Nn \in \N and

    \qty{A_i : 1 \leq i \leq n} \subset \sigmaF

    disjoint, it follows that

    \mu\qty(\bigcup_{i=1}^n A_i) = \sum_{i=1}^n \mu(A_i)

  2. If A,B∈FA, B \in \sigmaF s.t. A⊂BA \subset B, it follows that μ(A)≤μ(B)\mu(A) \leq \mu(B).
  3. If A,B∈FA, B \in \sigmaF s.t. A⊂BA \subset B and μ(A)<∞\mu(A) < \infty, it follows that μ(B∖A)=μ(B)−μ(A)\mu(B \setminus A) = \mu(B) - \mu(A).
  4. Given A,B∈FA, B \in \sigmaF, μ(A)+μ(B)=μ(A∪B)+μ(A∩B)\mu(A) + \mu(B) = \mu(A \cup B) + \mu(A \cap B).
  5. If

    \qty{A_i : i \in \N} \subset \sigmaF

    is s.t. Ai⊂Ai+1A_i \subset A_{i+1}, then

    \mu\qty(\bigcup_{i=1}^n A_i) = \mu(A_n) \uparrow \mu\qty(\bigcup_{i \in \N} A_i)

  6. If

    \qty{A_i : i \in \N} \subset \sigmaF

    is s.t. μ(A1)<∞\mu(A_1) < \infty and Ai+1⊂AiA_{i+1} \subset A_i, then

    \mu\qty(\bigcap_{i=1}^n A_i) = \mu(A_n) \downarrow \mu\qty(\bigcap_{i \in \N} A_i)

  7. If

    \qty{A_i : n \in \N} \subset \sigmaF

    ,     then

    \mu\qty(\bigcup_{i\in \N} A_i) \leq \sum_{i \in \N} \mu(A_i)