Measures

In this section we cover the fundamentals of measures. A measure is mapping from a σ-algebra to $[0,\infty]$ that we think of as assigning a numerical size to each set in the σ-algebra.

Defining measures

Definition

Fix a σ-algebra $\mathscr{B}$ on a set $X$. A measure on $(X,\mathscr{B})$ is a function $\mu : \mathscr{B} \to [0,\infty]$ with the following properties.

$\mu(\emptyset) = 0$
One has \[ \mu \left( \, \bigcup_{n=1}^\infty B_n \right) = \sum_{n=1}^\infty \mu(B_n) \] whenever $B_1,B_2,B_3,\dots$ are pairwise disjoint sets in $\mathscr{B}$.

The second property is called countable additivity. A sequence $B_1,B_2,\dots$ of subsets of $X$ is pairwise disjoint when $B_i \cap B_j = \emptyset$ for all $i \ne j$. We think of the second property as saying that we can calculate the size of a set by breaking it up into infinitely many pieces and summing the sizes.

Note that the definition is plausible: because every σ-algebra contains the empty set we may require a map from $\mathscr{B}$ to $[0,\infty]$ satisfies the first property, and because σ-algebra are closed under countable unions, it makes sense to calculate the left-hand side in the second property whenever $B_1,B_2,\dots$ belong to $\mathscr{B}$.

By a measure space we mean a triple $(X,\B,\mu)$ where $(X,\B)$ is a measurable space and $\mu$ is a measure on $\B$.

Examples

Interesting measures tend to be difficult to construct because which sets belong to a σ-algebra is often difficult to describe. The construction of the most important measure - the Lebesgue measure on the real line - takes up most of the next section. We go here through some examples of measures for which there are explicit formulae.

Counting measure

Fix a set $X$. The counting measure on $X$ is defined on $(X,\P(X))$ by \[ \mu(E) = \sup \{ |A| : A \subset E \textsf{ with } 0 \le |A| \l \infty \} \] and gives the cardinality of $E$ when $E$ is finite, and $\infty$ otherwise. If, for example, we take $X = \R$ then \[ \mu( \{ \sqrt{2}, 4, -\pi \}) = 3 \] and $\mu(\N) = \infty$.

Point measure

Fix a non-empty set $X$ and a point $a \in X$. The point measure at $a$ is defined on $(X,\P(X))$ by \[ \mu(E) = \begin{cases} 1 & a \in E \\ 0 & a \notin E \end{cases} \] and ascribes size according only to whether it contains the point $a$. We sometimes write $\delta_a$ for the point measure at $a$. If, for example, we take $X = \R$ and $a = \sqrt{2}$ then \[ \mu( \{ \sqrt{2}, 4, -\pi \}) = 1 \] and $\mu(\N) = 0$.

First properties

We cover here some fundamental properties of measures. With the perspective of a measure as assigning sizes to subsets of $X$ each should seem reasonable. For example, the first property says that $\mu(B) \le \mu(C)$ whenever $B \subset C$. It simply verifies that larger sets in the sense of set containment cannot have smaller measure.

Theorem

Fix a measure space $(X,\B,\mu)$.

Monotonicity If $B,C \in \B$ with $B \subset C$ then \[ \mu(B) \le \mu(C) \]
Subadditivity If $B_1,B_2,\ldots \in \B$ then \[ \mu \left( \, \bigcup_{n=1}^\infty B_n \right) \le \sum_{n=1}^\infty \mu(B_n) \]

Proof:

Fix a measure space $(X,\B,\mu)$.

Fix $B,C \in \B$ with $B \subset C$. Put $D = C \setminus B$. The sets $B$ and $D$ are disjoint, so we have \[ \mu(C) = \mu(B) + \mu(D) \ge \mu(B) \] because $\mu(D) \ge 0$.
Define inductively a sequence $C_1 = B_1$ and \[ C_{n+1} = B_{n+1} \setminus ( B_1 \cup \cdots \cup B_n) \] for all $n \in \N$. The sets $C_1,C_2,\dots$ all belong to $\B$ and are pairwise disjoint. Moreover \[ \bigcup_{n=1}^\infty C_n = \bigcup_{n=1}^\infty B_n \] and $C_n \subset B_n$ for all $n \in \N$ so \[ \mu \left( \, \bigcup_{n=1}^\infty B_n \right) = \mu \left( \, \bigcup_{n=1}^\infty C_n \right) = \sum_{n=1}^\infty \mu(C_n) \le \sum_{n=1}^\infty \mu(B_n) \] by the definition and part 1. of this theorem.▮

Fix pairwise disjoint sets $B_1,B_2,\ldots$ in $\B$. Countable additivity gives \[ \begin{align*} \mu \left( \, \bigcup_{n=1}^\infty B_n \right) &= \sum_{n=1}^\infty \mu(B_n) \\ &= \lim_{N \to \infty} \sum_{n=1}^N \mu(B_n) = \lim_{N \to \infty} \mu \left( \, \bigcup_{n=1}^N B_n \right) \end{align*} \] which is reminiscent of a continuity property. This is not a precise statement, as we have not said what we mean by placing a limit inside a measure. Nevertheless, limiting properties of measures are important in the development of the theory. The following theorem gives some rigorous examples.

Theorem

Fix a measure $\mu$ on a measurable space $(X,\B)$.

Continuity I If $B_1 \subset B_2 \subset \cdots$ in $\B$ then \[ \mu \left( \, \bigcup_{n=1}^\infty B_n \right) = \lim_{n \to \infty} \mu(B_n) \]
Continuity II If $B_1 \supset B_2 \supset \cdots$ in $\B$ and $\mu(B_1) \l \infty$ then \[ \mu \left( \, \bigcap_{n=1}^\infty B_n \right) = \lim_{n \to \infty} \mu(B_n) \]

Proof:

Fix a measure $\mu$ on a measure space $(X,\B)$.

Define $C_1 = B_1$ and $C_{n+1} = B_{n+1} \setminus B_n$ for all $n \in \N$. The sets $C_1,C_2,\dots$ all belong to $\B$ and are pairwise disjoint. Moreover \[ B_n = C_1 \cup \cdots \cup C_n \] and therefore \[ \mu \left( \, \bigcup_{n=1}^\infty B_n \right) = \lim_{n \to \infty} \sum_{i=1}^n \mu(C_i) = \lim_{n \to \infty} \mu(B_n) \] from the definition of a measure.
Put $C_n = B_1 \setminus B_n$ for all $n \in \N$. Then $C_1 \subset C_2 \subset \cdots$ and therefore \[ \mu \left( \, \bigcup_{n=1}^\infty C_n \right) = \lim_{n \to \infty} \mu(C_n) \] by part 3. of this theorem. Since $\mu(B_1) \l \infty$ we also have $\mu(B_n) \l \infty$ for all $n \in \N$. Thus \[ \mu(C_n) = \mu(B_1) - \mu(B_n) \] and \[ \mu \left( \, \bigcup_{n=1}^\infty C_n \right) = \mu \left( \, B_1 \setminus \bigcap_{n=1}^\infty B_n \right) = \mu(B_1) - \mu \left( \, \bigcap_{n=1}^\infty B_n \right) \] which all combine to give the desired result.▮