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 \(\mu\rightarrow [0,\infty]\) with the following properties:

1. Non-negativity: For \(A\in \mathcal{F}, \mu(A)\geq 0.\)
2. \(\mu(\emptyset)=0\)
3. Sigma-additivity: if \(\{A_n\}\) is a countable collection of disjoint sets in \mathcal{F} (i.e., \(A_i \cap A_j=\emptyset\) for \(i\neq j\), then
   \(\mu\left( \bigcup_{n=1}^{\infty} A_n \right) = \sum_{n=1}^{\infty} \mu(A_n).\)

Properties of a Measure

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

Monotonicity

If \(A\subseteq B\) and \(A, B \in \mathcal{F}\), \(\mu(A) \leq \mu(B)\)
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, \(\mu(B\setminus A)\geq 0\), thus we get the result $$A \subseteq B \implies \mu(A) \leq \mu(B)$$ And if \(B\setminus A=\emptyset\), recall that \(\mu(\emptyset)=0\)

Sub-Additivity

$$\mu\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty} \mu(A_n)$$

Proof

Let \(\{\{A_n\}_{n=1}^\infty\}\) 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 $$B_1 = A_1, \quad B_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k \text{ for } n \geq 2.$$ $$\bigcup_{n=1}^{\infty} A_n=\bigcup_{n=1}^{\infty} B_n$$ Since \(\{B_n\}\) is a disjoint collection, we can apply sigma-additivity: $$\mu\left(\bigcup_{n=1}^{\infty} B_n\right) = \sum_{n=1}^{\infty} \mu(B_n).$$ Since \(B_n \subseteq A_n\) for each n, and by the previously proven monotonoicity of measures, we have $$\mu(B_n) \leq \mu(A_n) \quad \text{for each } n$$ $$\sum_{n=1}^{\infty} \mu(B_n) \leq \sum_{n=1}^{\infty} \mu(A_n)$$ $$\mu\left(\bigcup_{n=1}^{\infty} A_n\right)=\mu\left(\bigcup_{n=1}^{\infty} B_n\right) = \leq \sum_{n=1}^{\infty} \mu(A_n)$$ $$\mu\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty} \mu(A_n)$$

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.