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Measures

Why

We extend our notion of length, area, and volume beyond the Lebesgue measure on the product spaces of real numbers.

Definition

Suppose $\mathcal{A} $ is an algebra of sets. A function $f: \mathcal{A} \to \Rbar_+$ is finitely additive if

\[ f(\cup_{i = 1}^{n} A_i) = \sum_{i = 1}^{n} f(A_i) \quad \text{for all disjoint } A_1, \dots , A_n \in \mathcal{A} \]

Similarly, suppose $\mathcal{F} $ is a $\sigma $-algebra. Then $f$ is countably additive if

\[ f(\cup_{i = 1}^{\infty} F_i) = \sum_{i =1 }^{\infty} f(F_i) \quad \text{for all disjoint sequences } \set{F_i}_{i \in \N } \text{ in } \mathcal{F} \]

If, in addition, $f(\varnothing) = 0$, then $f$ is called a finitely additive measure or countably additive measure respectively. Since a countably additive measure is finitely additive (the converse is false!), when we speak of a measure we mean a countable additive one.

When $(X, \mathcal{F} )$ is a countably unitable subset algebra and $\mu : \mathcal{F} \to \Rbar_+$, then we call $(X, \mathcal{F} )$ a measurable space and call $(X, \mathcal{F} , \mu )$ a measure space. The word “space” is natural, since the notion of a measure generalized the notion of volume in real space (see Real Space and N-Dimensional Space). We often call $\mathcal{F} $ the measurable sets. In other words, a measure space is a triple: a base set, a sigma algebra, and a measure.

Notation

We often use $\mu $ for a measure since it is a mnemonic for “measure”. We often also us $\nu $ and $\lambda $ since these letters are near $\mu $ in the Greek alphabet.

Examples

Let $(A, \mathcal{A} )$ a measurable space. Let $\mu : \mathcal{A} \to [0, +\infty]$ such that $\mu (A)$ is $\card{A}$ if $A$ is finite and $\mu (A)$ is $+\infty$ otherwise. Then $\mu $ is a measure. We call $\mu $ the counting measure.
Let $(A, \mathcal{A} )$ measurable. Fix $a \in A$. Let $\mu : \mathcal{A} \to [0, +\infty]$ such that $\mu (A)$ is $1$ if $a \in A$ and $\mu (A)$ is $0$ otherwise. Then $\mu $ is a measure. We call $\mu $ the point mass concentrated at $a$.
The Lebesgue measure on the measurable space $(\R , \mathcal{B} (\R ))$ is a measure.
Let $\mathcal{A} $ the co-finite algebra on $N$. Let $\mu : \mathcal{A} \to [0, +\infty]$ be such that $\mu (A)$ is 1 if $A$ is infinite or 0 otherwise. Then $\mu $ is a finitely additive measure. However it is impossible to extend $\mu $ to be a countably additive measure. Observe that if $A_n = \set{n}$ the $\mu (\cup_{n} A_n) = 1$ but $\sum_{n} \mu (A_n) = 0$.

Let $(A, \mathcal{A} )$ a measurable space. Let $\mu : \mathcal{A} \to [0, +\infty]$ be $0$ if $A = \varnothing$ and $\mu (A)$ is $+\infty$ otherwise. Then $\mu $ is a measure.

Let $A$ be set with at least two elements ($\card{A} \geq 2$). Let $\mathcal{A} = \powerset{A}$. Let $\mu : \mathcal{A} \to [0, +\infty]$ such that $\mu (A)$ is $0$ if $A = \varnothing$ and $\mu (A) = 1$ otherwise. Then $\mu $ is not a measure, nor is $\mu $ finitely additive.
Let $B, C \in \mathcal{A} $, $B \cap C = \varnothing$ then using finite additivity We obtain a contradiction

\[ 1 = \mu (B \cup C) \neq \mu (B) + \mu (C) = 2 \]

Properties

Suppose $(A, \mathcal{A} , \mu )$ is measure space. Then

\[ \mu (B) \leq \mu (C) \quad \text{for all } B \subset C \subset A \]

Suppose $(A, \mathcal{A} , m)$ is a measure space and $\set{A_n} \subset \mathcal{A} $ is a countable family. Then $m(\cup A_n) \leq \sum_{i} m(A_i)$.
For a measure space $(A, \mathcal{A} , m)$.

\[ m(\cup_{n = 1}^{\infty} A_i) = \lim_{n \to \infty} m(A_i) \]

For a measure space $(A, \mathcal{A} , m)$.

\[ m(\cap _{n = 1}^{\infty} A_i) = \lim_{n \to \infty} m(A_i) \]

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