What is a Measure? (Measure Theory)

In layman’s terms, “measures” are functions that are intended to represent ideas of length, area, mass, etc. The inputs for the measure functions would be sets, and the output would be a real value, possibly including infinity.

It would be desirable to attach the value 0 to the empty set $\emptyset$ and measures should be additive over disjoint sets in X.

Definition (from Bartle): A measure is an extended real-valued function $\mu$ defined on a $\sigma$ -algebra X of subsets of X such that
(i) $\mu (\emptyset)=0$
(ii) $\mu (E) \geq 0$ for all $E\in \mathbf{X}$
(iii) $\mu$ is countably additive in the sense that if $(E_n)$ is any disjoint sequence ( $E_n \cap E_m =\emptyset\ \text{if }n\neq m$ ) of sets in X, then

$\displaystyle \mu(\bigcup_{n=1}^\infty E_n )=\sum_{n=1}^\infty \mu (E_n)$ .

If a measure does not take on $+\infty$ , we say it is finite. More generally, if there exists a sequence $(E_n)$ of sets in X with $X=\cup E_n$ and such that $\mu (E_n) <+\infty$ for all n, then we say that $\mu$ is $\sigma$ -finite. We see that if a measure is finite implies it is $\sigma$ -finite, but not necessarily the other way around.

Examples of measures

(a) Let X be any nonempty set and let X be the $\sigma$ -algebra of all subsets of X. Let $\mu_1$ be definied on X by $\mu_1 (E)=0$ , for all $E\in\mathbf{X}$ . We can see that $\mu_1$ is finite and thus also $\sigma$ -finite.

Let $\mu_2$ be defined by $\mu_2 (\emptyset) =0$ , $\mu_2 (E)=+\infty$ if $E\neq \emptyset$ . $\mu_2$ is an example of a measure that is neither finite nor $\sigma$ -finite.

The most famous measure is definitely the Lebesgue measure. If X=R, and X=B, the Borel algebra, then (shown in Bartle’s Chapter 9) there exists a unique measure $\lambda$ defined on B which coincides with length on open intervals. I.e. if E is the nonempty interval (a,b), then $\lambda (E)=b-a$ . This measure is usually called Lebesgue measure (or sometimes Borel measure). It is not a finite measure since $\lambda (\mathbb{R})=\infty$ . But it is $\sigma$ -finite since any interval can be broken down into a sequence of sets ( $E_n$ ) such that $\mu (E_n)<\infty$ for all n.