Intro to Probability
5. Measures
5.1 The Notion of a Measure
Definition 5.1 (Measure): Let $(\Omega, \sigmaF)$ be a measurable space. A function $\mu : \sigmaF \to \overline{\R}_{+}$ is said to be a measure on $\sigmaF$ if the following two items are satisfied:
- $\mu(\varnothing) = 0$
- 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 $(\Omega, \sigmaF)$ be a measurable space and $x \in \Omega$ be a given point of $\Omega. $ Define the function
$$
\delta_x(A) = \begin{cases} 1 & \text{if } x \in A \\ 0 & \text{if } x \not \in A \end{cases}
$$
Then, $\delta_x : \sigmaF \mapsto \qty{0,1}$ is a measure on $\sigmaF. $
- From the definition of $\delta_x$ we immediately see that $\delta_x (\varnothing) = 0
. $ - Let $\qty{A_i : i \in \N} \subset \sigmaF$ be disjoint and set $A = \bigcup_{i \in \N} A_i
. $ If $x \not \in A, $ i.e. $x \in \bigcap_{i \in \N} A_i^c, $ there does not exist $i$ s.t. $x \in A_i, $ hence $\delta_x(A) = 0 = \sum_{i \in \N} \delta_x(A_i). $ Otherwise, if $x \in A, $ since $\qty{A_i : i \in \N}$ is disjoint, there exists a unique $j \in \N$ s.t. $x \in A_j. $ Hence also in this case $\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 $(\Omega, \sigmaF)$ be a measurable space, $I$ be a countable set and $E = \qty{x_i : i \in I} \subset \Omega$ be a collection of points in $\Omega. $ Assume that $\alpha_i, $ $i \in I, $ are s.t. $\alpha_i \in [0, \infty)$ for any $i \in I. $ Define $\mu: \sigmaF \to \overline{\R}_{+}$ as
$$
\mu(A) = \sum_{i \in I} \alpha_i \delta_{x_i}(A) = \sum_{x \in E} \alpha_x \delta_x(A)
$$
where for any $i \in I$
$$
\delta_{x_i}(A) = \begin{cases} 1 & \text{if } x_i \in A \\ 0 & \text{if } x_i \not \in A \end{cases}
$$
Then, $\mu$ is a measure on $\sigmaF. $
Note: Consider the setting of the latter proposition. Clearly, if $I$ is finite, $\mu(A) < \infty$ for any $A \in \sigmaF. $ If $I$ is countably infinite, we that $\# I = \# \N$ and hence, for any $A \in \sigmaF, $ $\mu(A) = \sum_{n \in \N} \alpha_n \delta_{x_n}(A). $ Therefore, since $\alpha_i \delta_{x_i}(A) \leq \alpha_i$ and $\alpha_i \geq 0, $ it follows from TODO: ref to series that $\mu(A) < \infty$ for any $A \in \sigmaF$ if $\sum_{i \in I} \alpha_i < \infty. $
5.3 Counting Measures
Example (Counting measure): We consider the measurable space $(\Omega, \powerset(\Omega))$ where $\Omega$ is a finite set. Define $\mu: \powerset(\Omega) \to \N \cup \qty{0}$ as $\mu(A) = \# A. $ Then, $\mu$ is a measure on $\sigmaF. $
- We have $\mu(\varnothing) = \# \varnothing = 0
. $ - 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 $(\Omega, \powerset(\Omega))$ where $\Omega$ is a countable but not necessarily finite set. Define
$$
\mu(A) = \begin{cases}\# A & \text{if } A \in \powerset(\Omega) \text{ s.t. } A \text{ is finite} \\ \infty & \text{otherwise} \end{cases}
$$
Then, $\mu$ is a measure on $\powerset(\Omega). $
5.4 Properties of a Measure
Proposition 5.4 (Properties of a measure): Let $(\Omega, \sigmaF)$ be a measurable space and $\mu$ be a measure on $\sigmaF. $ Then, we have the following properties:
- Given $n \in \N$ and $\qty{A_i : 1 \leq i \leq n} \subset \sigmaF$ disjoint, it follows that $$ \mu\qty(\bigcup_{i=1}^n A_i) = \sum_{i=1}^n \mu(A_i) $$
- If $A, B \in \sigmaF$ s.t. $A \subset B
, $ it follows that $\mu(A) \leq \mu(B). $ - If $A, B \in \sigmaF$ s.t. $A \subset B$ and $\mu(A) < \infty
, $ it follows that $\mu(B \setminus A) = \mu(B) - \mu(A). $ - Given $A, B \in \sigmaF
, $ $\mu(A) + \mu(B) = \mu(A \cup B) + \mu(A \cap B). $ - If $\qty{A_i : i \in \N} \subset \sigmaF$ is s.t. $A_i \subset A_{i+1}
, $ then $$ \mu\qty(\bigcup_{i=1}^n A_i) = \mu(A_n) \uparrow \mu\qty(\bigcup_{i \in \N} A_i) $$ - If $\qty{A_i : i \in \N} \subset \sigmaF$ is s.t. $\mu(A_1) < \infty$ and $A_{i+1} \subset A_i
, $ then $$ \mu\qty(\bigcap_{i=1}^n A_i) = \mu(A_n) \downarrow \mu\qty(\bigcap_{i \in \N} A_i) $$ - If $\qty{A_i : n \in \N} \subset \sigmaF
, $ then $$ \mu\qty(\bigcup_{i\in \N} A_i) \leq \sum_{i \in \N} \mu(A_i) $$