The following theorem is used in measure theory to construct outer measures on a set $X$ , starting with a non-negative function on a collection of subsets of $X$ . For example, if we take $X$ to be the real numbers, $\mathcal{C}$ to be the collection of bounded open intervals of $\mathbb{R}$ and define $p$ by $p((a,b))=b-a$ for real numbers $a<b$ , then the Lebesgue outer measure is obtained.

Proof. The definition of $\mu^*$ immediately gives $\mu^*(A)\le\mu^*(B)$ for sets $A\subseteq B$ , and if $A=\emptyset$ then we can take $A_i=\emptyset$ in () to obtain $\mu^*(\emptyset)\le\sum_ip(\emptyset)=0$ , giving $\mu^*(\emptyset)=0$ . Only the countable subadditivity of $\mu^*$ remains to be shown. That is, if $A_i$ is a sequence in $\mathcal{P}(X)$ then \begin{equation}\label{eq:2} \mu^*\left(\bigcup_i A_i\right)\le\sum_i\mu^*(A_i). \end{equation}To prove this inequality, we may restrict to the case where $\mu^*(A_i)<\infty$ for each $i$ so that, choosing any $\epsilon>0$ , equation () says that there exists a sequence $A_{i,j}\in \mathcal{C}$ such that $A_i\subseteq\bigcup_j A_{i,j}$ and, \begin{equation*} \sum_{j=1}^\infty p(A_{i,j})\le\mu^*(A_i)+2^{-i}\epsilon. \end{equation*}As $\bigcup_iA_i\subseteq\bigcup_{i,j}A_{i,j}$ , equation () defining $\mu^*$ gives \begin{equation*} \mu^*\left(\bigcup_iA_i\right)\le\sum_{i,j}p(A_{i,j})=\sum_i\sum_jp(A_{i,j})\le\sum_i(\mu^*(A_i)+2^{-i}\epsilon)=\sum_i\mu^*(A_i)+\epsilon. \end{equation*}As $\epsilon>0$ is arbitrary, this proves subadditivity ().

Although this result is rather general, placing few restrictions on the function $p$ , there is no guarantee that the outer measure $\mu^*$ will agree with $p$ for the sets in $\mathcal{C}$ nor that $\mathcal{C}$ will consist of $\mu^*$ -measurable sets.

For example, if $X=\mathbb{R}$ , $\mathcal{C}$ consists of the bounded open intervals, and $p((a,b))=(b-a)^2$ for real numbers $a<b$ , then $\mu^*((a,b))=0\not=p((a,b))$ .

Alternatively if $p((a,b))=\sqrt{b-a}$ for all $a<b$ then it follows that $\mu^*((a,b))=\sqrt{b-a}$ so \begin{equation*} \mu^*((0,1))+\mu^*([1,2))=1+1\not=\mu^*((0,2))=\sqrt{2}, \end{equation*}and $(0,1)$ is not $\mu^*$ -measurable.