Construction of Outer Measure

Theorem
Let $\mathcal A$ be a subset, containing the empty set, of the power set $\mathcal P \left({X}\right)$ of a set $X$.

Let $\gamma$ be a mapping $\gamma : \mathcal A \to \overline{\R}_{\ge 0}$ such that $\gamma \left({\varnothing}\right) = 0$. Here, $\overline{\R}_{\ge 0}$ denotes the set of positive extended real numbers.

Then the mapping $\mu^*: \mathcal P \left({X}\right) \to \overline{\R}_{\ge 0}$ defined by:


 * $\displaystyle \mu^* \left({S}\right) = \inf \ \left\{{\sum_{n=1}^\infty \gamma \left({A_n}\right) : \forall n \in \N : A_n \in \mathcal A, \ S \subseteq \bigcup_{n=1}^\infty A_n}\right\}$

is an outer measure on $X$.

The infimum of the empty set is taken to be $+\infty$.

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

Because $\gamma \left({\varnothing}\right) = 0$ by assumption, it immediately follows that $\mu^* \left({\varnothing}\right) = 0$.

Any cover of a set is also a cover of any 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 $X$.

Define $\displaystyle S = \bigcup_{n=1}^\infty S_n$.

If, for some $n \in \N$, there does not exist a countable cover for $S_n$ by elements of $\mathcal A$, then there does not exist a countable cover for $S$ by elements of $\mathcal A$. In this case, the claim 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.

Then, by the defintion of $\mu^*$ as an infimum, for each $n \in \N$, there exists a countable cover $\mathcal C_n \subseteq \mathcal A$ of $S_n$ such that:


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

Define $\displaystyle \mathcal C = \bigcup_{n=1}^\infty \mathcal C_n$.

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

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

Therefore:

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

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