2  Probability measure

Reference: Chapter 2. Resnick (2005).

A probability space is a triple $(\mathrm{\Omega },\mathcal{B},\mathbb{P})$ where

1. $\mathrm{\Omega }$, the sample space.
2. $\mathcal{B}$, a $\sigma$-field of subsets of $\mathrm{\Omega }$ where each element is called event.
3. $\mathbb{P}$ is a probability measure.

Definition 2.1 ($\mathbf{\text{Probability measure}}$)
A probability measure $\mathbb{P}$ is any function $\mathbb{P}:\mathcal{B}\to [0,1]$ such that

1. $\mathbb{P}(A)\ge 0$ for all sets $A\in \mathcal{B}$.
2. $\mathbb{P}(\mathrm{\Omega })=1$.
3. $\mathbb{P}$ is $\sigma$-additive: if $\{A_n{\}}_{n\ge 1}$ is a sequence of disjoint events in $\mathcal{B}$, then: $$\begin{array}{}\text{(2.1)}& \mathbb{P}\left(\stackrel{\mathrm{\infty }}{\underset{n=1}{⨆}}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mathbb{P}(A_n)\text{.}\end{array}$$

In general, a probability measure $\mathbb{P}$ is a function that always goes from a $\sigma$-field of subsets of $\mathrm{\Omega }$ to $[0,1]$.

2.1 Consequences of the axioms

Proposition 2.1 ($\mathbf{\text{Probability of the complement}}$)
The probability of the complement of a set $A$ reads $$\begin{array}{}\text{(2.2)}& \mathbb{P}(A^{\mathsf{c}})=1-\mathbb{P}(A)\text{.}\end{array}$$

Proof: Proposition 2.1

Proof. Since it is possible to write $\mathrm{\Omega }=A\cup A^{\mathsf{c}}$ as the union of disjoint set, we can apply $\sigma$-additivity (Equation 2.1) to obtain: $$\begin{array}{rl}\mathrm{\Omega }=A\bigsqcup A^{\mathsf{c}}& \stackrel{\mathbb{P}}{⟶}\mathbb{P}(\mathrm{\Omega })=\mathbb{P}(A)+\mathbb{P}(A^{\mathsf{c}})\\ & ⟹1=\mathbb{P}(A)+\mathbb{P}(A^{\mathsf{c}})\\ & ⟹\mathbb{P}(A^{\mathsf{c}})=1-\mathbb{P}(A)\end{array}$$

Proposition 2.2 ($\mathbf{\text{Probability of the empty set}}$)
The probability of the empty set $\mathrm{\varnothing }$ is zero, i.e. $\mathbb{P}(\mathrm{\varnothing })=0$.

Proof: Proposition 2.2

Proof. Using the fact that $\mathbb{P}(\mathrm{\Omega })=1$ by assumption and applying Equation 2.2: $$\mathbb{P}(\mathrm{\varnothing })=1-\mathbb{P}({\mathrm{\varnothing }}^c)=1-\mathbb{P}(\mathrm{\Omega })=0\text{.}$$

Proposition 2.3 ($\mathbf{\text{Probability of the union}}$)
The Probability of the union of two sets: $$\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)\text{.}$$

Proof: Proposition 2.3

Proof. Let’s write the sets $A$ and $B$ in terms of union of disjoint events (Equation 1.9) and apply $\mathbb{P}$ on both side and $\sigma$-additivity (Equation 2.1). $$\begin{array}{}\text{(2.3)}& \begin{array}{rl}& \mathbb{P}(A)=\mathbb{P}(A\cap B)+\mathbb{P}(A\cap B^c)⟹\mathbb{P}(A\cap B^c)=\mathbb{P}(A)-\mathbb{P}(A\cap B)\\ & \mathbb{P}(B)=\mathbb{P}(A\cap B)+\mathbb{P}(B\cap A^c)⟹\mathbb{P}(B\cap A^c)=\mathbb{P}(B)-\mathbb{P}(A\cap B)\end{array}\end{array}$$ Let’s now decompose $A\cup B$ in the disjoint union of 3 events (Equation 1.8) and again, apply $\mathbb{P}$ on both side and $\sigma$-additivity: $$\mathbb{P}(A\cup B)=\mathbb{P}(A\cap B)+\mathbb{P}(A\cap B^{\mathsf{c}})+\mathbb{P}(A^{\mathsf{c}}\cap B)\text{.}$$ Substituting $\mathbb{P}(A\cap B^c)$ and $\mathbb{P}(A\cap B^c)$ from Equation 2.3 gives the result: $$\begin{array}{rl}\mathbb{P}(A\cup B)& =\mathbb{P}(A\cap B)+\mathbb{P}(B)-\mathbb{P}(A\cap B)+\mathbb{P}(A)-\mathbb{P}(A\cap B)=\\ & =\mathbb{P}(B)+\mathbb{P}(A)-\mathbb{P}(A\cap B)\end{array}$$

Proposition 2.4 ($\mathbf{\text{Monotonicity of probability measure}}$)
The probability measure $\mathbb{P}$ is non-decreasing, in the sense that given two events $A$ and $B$, then $$A\subset B⟹\mathbb{P}(A)\le \mathbb{P}(B)\text{.}$$

Proof: Proposition 2.4

Proof. The proof of the statements follows once the set $B$ is written as disjoint union of subsets of $A$ and $B$ (Equation 1.9). Then, applying the probability $\mathbb{P}$ and $\sigma$-additivity on both sides one obtain: $$\mathbb{P}(B)=\mathbb{P}(A)+\mathbb{P}(B-A)\ge \mathbb{P}(A)\text{.}$$

Further properties are:

1. Subadditivity: the measure $\mathbb{P}$ is $\sigma$-subadditive. For a sequence of events $\{A_n{\}}_{n\ge 1}$ in $\mathcal{B}$ then: $$\begin{array}{}\text{(2.4)}& \mathbb{P}\left(\stackrel{\mathrm{\infty }}{\bigcup _{n=1}}{A}_n\right)\le \sum _{n=1}^{\mathrm{\infty }}\mathbb{P}(A_n)\text{.}\end{array}$$

2. Continuity: the measure $\mathbb{P}$ is continuous for a monotone sequence of sets $A_n\in \mathcal{B}$, i.e.  $$\begin{array}{}\text{(2.5)}& A_n\uparrow A⟹\mathbb{P}(A_n)\uparrow \mathbb{P}(A)\text{,}{A}_n\downarrow A⟹\mathbb{P}(A_n)\downarrow \mathbb{P}(A)\text{.}\end{array}$$

3. Fatou’s lemma: consider a sequence of events $\{A_n{\}}_{n\ge 1}$ in $\mathcal{B}$, then we have the following result: $$\begin{array}{}\text{(2.6)}& \mathbb{P}(\underset{n\to \mathrm{\infty }}{liminf}{A}_n)\le \underset{n\to \mathrm{\infty }}{liminf}\mathbb{P}(A_n)\le \underset{n\to \mathrm{\infty }}{limsup}\mathbb{P}(A_n)\le \mathbb{P}(\underset{n\to \mathrm{\infty }}{limsup}{A}_n)\text{.}\end{array}$$

Resnick, Sidney I. 2005. A Probability Path. Birkhauser. https://link.springer.com/book/10.1007/978-0-8176-8409-9.