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

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.

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.

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

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

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 $(A, \mathcal{A} , m)$.

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