Construction of Outer Measure

Theorem
Given a set $X$ and a cover $C$ of $X$ with a countable subcover of $X$ and which contains the empty set, suppose there is a function $\gamma : \mathcal P(X) \to \R$ such that


 * $\gamma (A) \in [0, \infty]$ for each $A \in C$, and
 * $\gamma (\varnothing) = 0$.

Then the function $\mu^*: \mathcal P(X) \to \R$ defined by


 * $\displaystyle \mu^*(E) \equiv \inf \left\{{\sum_{n\in\N} \gamma (A_n) : (A_n)_{n\in\N} \text{ is a countable cover of } E \text{ by members of } C}\right\}$

for $E\in\mathcal P(X)$ is an outer measure on $X$.

Proof
We must check each of the criteria of the outer measure in turn.

1. Since $\gamma$ is nonnegative, the sum of any images under $\gamma$ is nonnegative and thus the infimum of any such sums is also nonnegative. Hence $\mu^*(E) \geq 0$ for each $E\in\mathcal P(X)$.

2. By the above, $\mu^* (\varnothing) \geq 0$. Since $\varnothing$ is a cover of itself contained in $C$, and since $\gamma (\varnothing) = 0$ by assumption, $\mu^*(\varnothing) \leq 0$ as well. Hence $\mu^*(\varnothing) = 0.$.

3. For any $B_1\subseteq B_2 \in \mathcal P(X)$, any cover of $B_2$ by sets in $C$ is also a cover of $B_1$. Hence $\mu^*(B_1) \leq \mu^*(B_2)$ and the function is monotonic.

4. Take any countable sequence $(B_n)_{n\in\N}$ of members of $\mathcal P(X)$, and let $B = \bigcup_{n\in\N} B_n$.

Select any $\epsilon > 0\ $. Then by defintion of $\mu^*\ $ as an infimum, for each $k\in \N$, we can select a countable cover $C(k)\subseteq C$ of $B_k\ $ such that:


 * $\displaystyle \sum_{x\in C(k) } \gamma (x) < \mu^*(B_k) + \frac{\epsilon}{2^{k+1}}$.

And since $\displaystyle C^* = \bigcup_{k \in \N}C(k)\subseteq C$ covers $B$, we have:

But since $\epsilon$ was an arbitrary positive number, $\displaystyle \mu^*(B) \leq \sum_{k\in\N}\mu^*(B_k)$.

Consequences
It follows immediately from the theorem that the Lebesgue outer measure is, in fact, an outer measure.