Measure

Definition

A function $\mu :\mathcal{F}\to \overline{R}$, where $\mathcal{F}$ is a Sigma Field, is a measure, if:

1. $\mu (\varnothing )=0$
2. (countable additivity) For disjoint $\{A_n\in \mathcal{F}{\}}_{n\ge 1}$: $\mu \left(U_{n⩾1}{A}_n\right)={\sum }_{n=1}\mu \left(A_n\right)$.

A measure is called a probability measure if $\mu (\Omega )=1$.

Properties

• (Finite additivity) For disjoint $A,B\in \mathcal{F}$: $\mu (A\cup B)=\mu (A)+\mu (B)$
• For any event $A$, $\mathbb{P}(A^C)=1-\mathbb{P}(A)$
• If $A\subset B$, then $\mu (A)\le \mu (B)$
• (Union bound) For any countable collection of events $\{A_n\}$: $\mu \left({\bigcup }_n{A}_n\right)\le {\sum }_n\mu (A_n)$
• (Inclusion-exclusion formula) For any finite collection of events $\{A_n\}$: $\mathbb{P}\left({\bigcup }_n{A}_n\right)={\sum }_n\mathbb{P}(A_n)-{\sum }_{i<j}\mathbb{P}(A_i\cap A_j)+{\sum }_{i<j<k}\mathbb{P}(A_i\cap A_j\cap A_k)+\dots +(-1{)}^{n-1}\mathbb{P}(A_1\cap \dots \cap A_n)$

Continuity of Probability Measures

A (probability) measure requires countable additivity, which can be difficult to verify. The following theorem provides alternative conditions for establishing the countable additivity from finite additivity.

Continuity of probability measures

Let $\mathcal{F}$ be a $\sigma$-field of subsets of $\Omega$, and suppose that $\mathbb{P}:\mathcal{F}\to [0,1]$ satisfies $\mathbb{P}(\Omega )=1$ as well as the finite additivity property. Then, the following are equivalent:

1. $\mathbb{P}$ is a probability measure (that is, it also satisfies countable additivity).

2. If $\left\{A_i\right\}$ is an increasing sequence of sets in $\mathcal{F}$ (i.e., $A_i\subset A_{i+1}$, for all $i$ ), and $A={\cup }_{i=1}^{\infty }{A}_i$, then ${\lim }_{i\to \infty }\mathbb{P}\left(A_i\right)=\mathbb{P}(A)$.

3. If $\left\{A_i\right\}$ is a decreasing sequence of sets in $\mathcal{F}$ (i.e., $A_i\supset A_{i+1}$, for all $i$ ), and $A={\cap }_{i=1}^{\infty }{A}_i$, then ${\lim }_{i\to \infty }\mathbb{P}\left(A_i\right)=\mathbb{P}(A)$.

4. If $\left\{A_i\right\}$ is a decreasing sequence of sets (i.e., $A_i\supset A_{i+1}$, for all $i$ ) and ${\cap }_{i=1}^{\infty }{A}_i$ is empty, then ${\lim }_{i\to \infty }\mathbb{P}\left(A_i\right)=0$

• 💡 The second statement also holds for general measures, not just probability measures.

• ❗️ The third and fourth statements do not hold for general measures as $\mathbb{P}(A_i)$ is not capped at 1. The statements are true for general measures if at least one $A_i$ has finite measure.

Proof

(a) ⇒ (b). Note that

$$A=\underset{n\ge 1}{⨆}\left(A_n\setminus A_{n-1}\right),$$

where $A_0=\varnothing$. Then by the countable additivity of $\mathbb{P}$, we have

$$\mathbb{P}(A)=\sum _{n\ge 1}\mathbb{P}\left(A_n\setminus A_{n-1}\right)=\underset{N\to \infty }{\lim }\sum _{n=1}^N\left(\mathbb{P}(A_n)-\mathbb{P}(A_{n-1})\right)=\underset{N\to \infty }{\lim }\mathbb{P}(A_N).$$

(b) ⇒ (c). By De Morgan’s law, we have

$$\mathbb{P}(A)=1-\mathbb{P}(A^C)=1-\mathbb{P}\left(\bigcup _{n\ge 1}{A}_n^C\right)=1-\underset{n\to \infty }{\lim }\mathbb{P}(A_n^C)=\underset{n\to \infty }{\lim }\mathbb{P}(A_n).$$

(d) is a special case of (c) by taking $A=\varnothing$.

(c) ⇒ (a). Let $\{B_n\}$ be any sequence of disjoint sets. Let $A_n:={\bigcup }_{i\ge n}{B}_i$. Then clearly $\{A_n\}$ is an decreasing sequence of sets. We claim that ${\cap }_{n=1}^{\infty }{A}_n$ is empty. To see this, suppose $\omega \in {\cap }_{n=1}^{\infty }{A}_n$. Then $\omega \in A_1={\cup }_{i=1}^{\infty }{B}_i$. Then there exists $n_0$ such that $\omega \in B_{n_0}$. Since $\left\{B_n\right\}$ are disjoint, $\omega \notin {\cup }_{i\ge n_0+1}{B}_i=A_{n_0+1}$, which contradicts the assumption that $\omega \in {\cap }_{n=1}^{\infty }{A}_n$. Therefore, by (d), we have ${\lim }_{i\to \infty }{A}_i=0$. Now for any $n$, by the finite additivity of $\mathbb{P}$, we have

$$\mathbb{P}\left(\bigcup _{i=1}^{\infty }{B}_i\right)=\mathbb{P}\left(\bigcup _{i=1}^{n-1}{B}_i+\bigcup _{i\ge n}{B}_i\right)=\sum _{i=1}^{n-1}\mathbb{P}\left(B_i\right)+\mathbb{P}\left(\bigcup _{i\ge n}{B}_i\right)=\sum _{i=1}^{n-1}\mathbb{P}\left(B_i\right)+\mathbb{P}\left(A_n\right).$$

Letting $n\to \infty$ gives

$$\mathbb{P}\left(\bigcup _{i=1}^{\infty }{B}_i\right)=\underset{n\to \infty }{\lim }\sum _{i-1}^{n-1}\mathbb{P}(B_i)+\underset{n\to \infty }{\lim }\mathbb{P}(A_n)=\sum _{i=1}^{\infty }\mathbb{P}(B_i).$$

Complete Measure

A measure space $(\Omega ,\mathcal{F},\mu )$ is said to be complete if for any $A\in \mathcal{F}$ with $\mu (A)=0$, any subset $B\subset A$ is also in $\mathcal{F}$. That is, $\mathcal{F}$ contains all zero-measure sets. A null set in $\mathcal{F}$ is a set $A$ such that $\mu (A)=0$. Let $\mathcal{N}$ be the collection of all null sets. Then the completion of $\mathcal{F}$ is ${\mathcal{F}}^{\ast }=\sigma (\mathcal{F}\cup \mathcal{N})$. The completion of a measure is always possible and unique, by letting ${\mu }^{\ast }(A)=0$ for all $A\subset B\in \mathcal{N}$.

In terms of the Borel Measure, if we refer to the Lebesgue measure as the complete measure ${\mathcal{B}}^{\ast }$, then $\mathcal{B}⫋{\mathcal{B}}^{\ast }$. That is, there exists Lebesgue sets that are not Borel measurable. Moreover, we have $|\mathcal{B}|={\aleph }_1$ and $|{\mathcal{B}}^{\ast }|=2^{{\aleph }_1}={\aleph }_2$, assuming the generalized continuum hypothesis.