Measures

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

Defining measures

Definition

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

1. $\mu (\varnothing )=0$
2. One has $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (B_n)$$ whenever $B_1,B_2,B_3,\dots$ are pairwise disjoint sets in $\mathcal{B}$.

The second property is called countable additivity. A sequence $B_1,B_2,\dots$ of subsets of $X$ is pairwise disjoint when $B_i\cap B_j=\varnothing$ 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 $\mathcal{B}$ to $[0,\mathrm{\infty }]$ 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 $\mathcal{B}$.

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

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 $(X,\mathcal{P}(X))$ by $$\mu (E)=sup\{|A|:A\subset E\mathsf{\text{ with }}0\le |A|<\mathrm{\infty }\}$$ and gives the cardinality of $E$ when $E$ is finite, and $\mathrm{\infty }$ otherwise. If, for example, we take $X=\mathbb{R}$ then $$\mu (\{\sqrt{2},4,-\pi \})=3$$ and $\mu (\mathbb{N})=\mathrm{\infty }$.

Point measure

Fix a non-empty set $X$ and a point $a\in X$. The point measure at $a$ is defined on $(X,\mathcal{P}(X))$ by $$\mu (E)=\{\begin{array}{ll}1& a\in E\\ 0& a\notin E\end{array}$$ 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 $X=\mathbb{R}$ and $a=\sqrt{2}$ then $$\mu (\{\sqrt{2},4,-\pi \})=1$$ and $\mu (\mathbb{N})=0$.

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 $(X,\mathcal{B},\mu )$.

1. Monotonicity If $B,C\in \mathcal{B}$ with $B\subset C$ then $$\mu (B)\le \mu (C)$$
2. Subadditivity If $B_1,B_2,\dots \in \mathcal{B}$ then $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)\le \sum _{n=1}^{\mathrm{\infty }}\mu (B_n)$$

Proof:

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

1. Fix $B,C\in \mathcal{B}$ 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 $\mu (D)\ge 0$.
2. 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 $n\in \mathbb{N}$. The sets $C_1,C_2,\dots$ all belong to $\mathcal{B}$ and are pairwise disjoint. Moreover $$\bigcup _{n=1}^{\mathrm{\infty }}{C}_n=\bigcup _{n=1}^{\mathrm{\infty }}{B}_n$$ and $C_n\subset B_n$ for all $n\in \mathbb{N}$ so $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{C}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (C_n)\le \sum _{n=1}^{\mathrm{\infty }}\mu (B_n)$$ by the definition and part 1. of this theorem.▮

Fix pairwise disjoint sets $B_1,B_2,\dots$ in $\mathcal{B}$. Countable additivity gives $$\begin{array}{rl}\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)& =\sum _{n=1}^{\mathrm{\infty }}\mu (B_n)\\ & =\underset{N\to \mathrm{\infty }}{lim}\sum _{n=1}^N\mu (B_n)=\underset{N\to \mathrm{\infty }}{lim}\mu \left(\bigcup _{n=1}^N{B}_n\right)\end{array}$$ 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 $(X,\mathcal{B})$.

1. Continuity I If $B_1\subset B_2\subset \cdots$ in $\mathcal{B}$ then $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\underset{n\to \mathrm{\infty }}{lim}\mu (B_n)$$
2. Continuity II If $B_1\supset B_2\supset \cdots$ in $\mathcal{B}$ and $\mu (B_1)<\mathrm{\infty }$ then $$\mu \left(\bigcap _{n=1}^{\mathrm{\infty }}{B}_n\right)=\underset{n\to \mathrm{\infty }}{lim}\mu (B_n)$$

Proof:

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

1. Define $C_1=B_1$ and $C_{n+1}=B_{n+1}\setminus B_n$ for all $n\in \mathbb{N}$. The sets $C_1,C_2,\dots$ all belong to $\mathcal{B}$ and are pairwise disjoint. Moreover $$B_n=C_1\cup \cdots \cup C_n$$ and therefore $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n\mu (C_i)=\underset{n\to \mathrm{\infty }}{lim}\mu (B_n)$$ from the definition of a measure.
2. Put $C_n=B_1\setminus B_n$ for all $n\in \mathbb{N}$. Then $C_1\subset C_2\subset \cdots$ and therefore $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{C}_n\right)=\underset{n\to \mathrm{\infty }}{lim}\mu (C_n)$$ by part 3. of this theorem. Since $\mu (B_1)<\mathrm{\infty }$ we also have $\mu (B_n)<\mathrm{\infty }$ for all $n\in \mathbb{N}$. Thus $$\mu (C_n)=\mu (B_1)-\mu (B_n)$$ and $$\mu \left(\bigcup _{n=1}^{\mathrm{\infty }}{C}_n\right)=\mu (B_1\setminus \bigcap _{n=1}^{\mathrm{\infty }}{B}_n)=\mu (B_1)-\mu \left(\bigcap _{n=1}^{\mathrm{\infty }}{B}_n\right)$$ which all combine to give the desired result.▮