Construction of Outer Measure

Theorem
Let $X$ be a set.

Let $\powerset X$ be the power set of $X$.

Let $\AA$ be a subset of $\powerset X$ which contains the empty set.

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

Let $\gamma: \AA \to \overline \R_{\ge 0}$ be a mapping such that $\map \gamma \O = 0$.

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


 * $\displaystyle \forall S \in \powerset X: \map {\mu^*} S = \inf \set {\sum_{n \mathop = 0}^\infty \map \gamma {A_n}: \forall n \in \N : A_n \in \AA, \ S \subseteq \bigcup_{n \mathop = 0}^\infty A_n}$

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 $\map \gamma \O = 0$:
 * $\map {\mu^*} \O = 0$

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 $\sequence {S_n}$ 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 $\AA$ for some $n \in \N$.

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

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 $\AA$.

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 $\CC_n \subseteq \AA$ of $S_n$ such that:


 * $\displaystyle \sum_{x \mathop \in \CC_n} \map \gamma x < \map {\mu^*} {S_n} + \frac \epsilon {2^{n + 1} }$

Let $\displaystyle \CC = \bigcup_{n \mathop = 0}^\infty \CC_n$.

Then $\CC$ is a subset of $\AA$ and a cover for $S$.

Furthermore, $\CC$ is the countable union of countable sets.

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

Therefore:

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

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