Construction of Outer Measure

Theorem
Given a set $$X$$ and a covering $$C$$ of $$X$$ with a countable subcovering 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


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

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 covering 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 covering of $$B_2$$ by sets in $$C$$ is also a covering 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 covering $$C(k)\subseteq C$$ of $$B_k\ $$ such that


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

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

$$ $$ $$ $$

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

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