Construction of Outer Measure

Theorem
Let $X$ be a set.

Let $\mathcal P \left({X}\right)$ be the power set of $X$.

Let $\mathcal A$ be a subset of $\mathcal P \left({X}\right)$ which contains the empty set.

Let $\overline \R_{\ge 0}$ denote the set of positive extended real numbers.

Let $\gamma: \mathcal A \to \overline \R_{\ge 0}$ be a mapping such that $\gamma \left({\varnothing}\right) = 0$.

Let $\mu^*: \mathcal A \to \overline \R_{\ge 0}$ be the mapping defined as:


 * $\displaystyle \forall S \in \mathcal P \left({X}\right): \mu^* \left({S}\right) = \inf \ \left\{ {\sum_{n \mathop = 0}^\infty \gamma \left({A_n}\right) : \forall n \in \N : A_n \in \mathcal A, \ S \subseteq \bigcup_{n \mathop = 0}^\infty A_n}\right\}$

Then $\mu^*$ is an outer measure on $X$.

The infimum of the empty set is the greatest element, $+\infty$.

Proof
We check each of the criteria for an outer measure.

From the assumption $\gamma \left({\varnothing}\right) = 0$:
 * $\mu^* \left({\varnothing}\right) = 0$

Any cover of a set is also a cover of a subset of it.

Thus $\mu^*$ is monotone by Infimum of Subset.

It remains to be shown that $\mu^*$ is countably subadditive.

Let $\left\langle{S_n}\right\rangle$ be a sequence of subsets of $X$.

Let $\displaystyle S := \bigcup_{n \mathop = 0}^\infty S_n$.

Suppose there does not exist a countable cover for $S_n$ by elements of $\mathcal A$ for some $n \in \N$.

Then there does not exist a countable cover for $S$ by elements of $\mathcal A$.

In this case, the result follows immediately.

Now suppose that for each $n \in \N$, there exists a countable cover for $S_n$ by elements of $\mathcal A$.

Let $\epsilon \in \R_{>0}$ be an arbitrary (strictly) positive real number.

By definition of infimum, for each $n \in \N$, we can apply the axiom of countable choice to choose a countable cover $\mathcal C_n \subseteq \mathcal A$ of $S_n$ such that:


 * $\displaystyle \sum_{x \mathop \in \mathcal C_n} \gamma \left({x}\right) < \mu^* \left({S_n}\right) + \frac \epsilon {2^{n+1}}$

Let $\displaystyle \mathcal C = \bigcup_{n \mathop = 0}^\infty \mathcal C_n$.

Then $\mathcal C$ is a subset of $\mathcal A$ and a cover for $S$.

Furthermore, $\mathcal C$ is the countable union of countable sets.

By Countable Union of Countable Sets is Countable, $\mathcal C$ is itself countable.

Therefore:

Since $\epsilon$ was arbitrary, the result follows.

Consequences
It follows immediately that the induced outer measure is an outer measure.