Measures

Introduction to Measures

Recommended Prerequesites

1. Probability
2. Sigma Algebra

Definition

A measure can be thought of as a function that assigns a number to a set.
More formally, let $\mathcal{X}$ be a set and $\mathcal{F}$ be a sigma algebra on $\mathcal{X}$, then a measure, $\mu$ on $(\mathcal{X},\mathcal{F})$ is a function $\mu \to [0,\mathrm{\infty }]$ with the following properties:

1. Non-negativity: For $A\in \mathcal{F},\mu (A)\ge 0.$
2. $\mu (\mathrm{\varnothing })=0$
3. Sigma-additivity: if $\{A_n\}$ is a countable collection of disjoint sets in \mathcal{F} (i.e., $A_i\cap A_j=\mathrm{\varnothing }$ for $i\ne j$, then
   $\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (A_n).$

Properties of a Measure

From the 3 above properties, what are some implied ones?

Monotonicity

If $A\subseteq B$ and $A,B\in \mathcal{F}$, $\mu (A)\le \mu (B)$
Why might this be the case?
Recall that a measure of a set is non-negative, so the contribution to the "size" of B from the non-A elements cannot lower the "size".

Proof

If $A\subseteq B$, then we can decompose B into two subsets, A and the set $B\setminus A$. Since these are disjoint sets by construction, by the property of sigma-additivity we have: $$\mu (A)+\mu (B\setminus A)=\mu (B)$$ Do to non-negativity, $\mu (B\setminus A)\ge 0$, thus we get the result $$A\subseteq B⟹\mu (A)\le \mu (B)$$ And if $B\setminus A=\mathrm{\varnothing }$, recall that $\mu (\mathrm{\varnothing })=0$

Sub-Additivity

$$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)\le \sum _{n=1}^{\mathrm{\infty }}\mu (A_n)$$

Proof

Let $\{\{A_n{\}}_{n=1}^{\mathrm{\infty }}\}$ be a countable collection of sets in a sigma algebra, $\mathcal{F}$ and let $\mu$ be a measure on $\mathcal{F}$. These sets $\{B_n\}$ are disjoint by construction and satisfy $$B_1=A_1,B_n=A_n\setminus \bigcup _{k=1}^{n-1}{A}_k\text{ for }n\ge 2.$$ $$\bigcup _{n=1}^{\mathrm{\infty }}{A}_n=\bigcup _{n=1}^{\mathrm{\infty }}{B}_n$$ Since $\{B_n\}$ is a disjoint collection, we can apply sigma-additivity: $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (B_n).$$ Since $B_n\subseteq A_n$ for each n, and by the previously proven monotonoicity of measures, we have $$\mu (B_n)\le \mu (A_n)\text{for each }n$$ $$\sum _{n=1}^{\mathrm{\infty }}\mu (B_n)\le \sum _{n=1}^{\mathrm{\infty }}\mu (A_n)$$ $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)=\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\le \sum _{n=1}^{\mathrm{\infty }}\mu (A_n)$$ $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)\le \sum _{n=1}^{\mathrm{\infty }}\mu (A_n)$$

Completeness

If A is a set of measure zero and $B\subseteq A$, then $\mu (B)=0$.

Examples of Measures

Lebesgue Measure

The Lebesgue measure is the analogue of length. Let $\lambda$ be the Lebesgue measure on $\mathcal{B}(\mathbb{R})$. Then $\lambda ((a,b))=b-a$

Probability Measure

A measure, $\mathbb{P}$, on set X with sigma algebra $\mathcal{F}$, if $\mathbb{P}(X)=1$.

Counting Measure

Let X be a set and $\mathcal{P}(X)$ be its power set. Let ${\mu }_{\text{count}}$ be the counting measure on that sigma algebra. $${\mu }_{\text{count}}(A)=\text{ the number of elements in A}$$ for $A\subseteq X$.

Measure Practice Problems

1. Explain why the Counting Measure satisfies the criterium for a measure.
2. Let X be the set of outcomes of rolling a fair die once. Let $\mathcal{F}=\mathcal{P}(X)$. Let $\mathbb{P}$ be the probability measure on X and $\mathcal{F}$. Let $A_1$ be the rolling a 3. Let $A_2$ be rolling an odd number. Verbally explain monotonicity and sub-additivity as they relate to this example.