# Induced Outer Measure Restricted to Semiring is Pre-Measure

## Theorem

Let $\mathcal S$ be a semiring over a set $X$.

Let $\mu: \mathcal S \to \overline \R_{\ge 0}$ be a pre-measure on $\mathcal S$, where $\overline \R_{\ge 0}$ denotes the set of positive |extended real numbers.

Let $\mu^*: \mathcal P \left({X}\right) \to \overline \R_{\ge 0}$ be the outer measure induced by $\mu$.

Then:

- $\displaystyle \mu^*\restriction_{\mathcal S} \, = \mu$

where $\restriction$ denotes restriction.

## Proof

Let $S \in \mathcal S$.

It follows immediately from the definition of the induced outer measure that $\mu^* \left({S}\right) \le \mu \left({S}\right)$.

Therefore, it suffices to show that if $\displaystyle \left({A_n}\right)_{n \mathop = 0}^{\infty}$ is a countable cover for $S$, then:

- $\displaystyle \mu \left({S}\right) \le \sum_{n \mathop = 0}^\infty \mu \left({A_n}\right)$

If the above statement is true, then it follows directly from the definition of infimum that $\mu \left({S}\right) \le \mu^* \left({S}\right)$, thus proving the theorem.

Define, for all natural numbers $n \in \N$:

- $\displaystyle B_n = A_n \setminus A_{n-1} \setminus \cdots \setminus A_0$

where $\setminus$ denotes set difference.

We take $B_0 = A_0$.

Using the mathematical induction, we will prove that for all natural numbers $m < n$, $B_{n, m} = A_n \setminus A_{n-1} \setminus \cdots \setminus A_{n-m}$ is the finite union of pairwise disjoint elements of $\mathcal S$.

We take $B_{n, 0} = A_n$.

The base case $m = 0$ is trivial.

Now assume the induction hypothesis that the above statement is true for some natural number $m < n - 1$, and let $D_1, D_2, \ldots, D_N$ be pairwise disjoint elements of $\mathcal S$ such that:

- $\displaystyle B_{n, m} = \bigcup_{k \mathop = 1}^N D_k$

Then:

\(\displaystyle B_{n, m+1}\) | \(=\) | \(\displaystyle B_{n, m} \setminus A_{n-m-1}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({\bigcup_{k \mathop = 1}^N D_k}\right) \setminus A_{n-m-1}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \bigcup_{k \mathop = 1}^N \, \left({D_k \setminus A_{n-m-1} }\right)\) | $\quad$ by Set Difference is Right Distributive over Union | $\quad$ |

By the definition of a semiring, for all natural numbers $k \le N$, $D_k \setminus A_{n-m-1}$ is the finite union of pairwise disjoint elements of $\mathcal S$.

Hence $B_{n, m+1}$ is the finite union of pairwise disjoint elements of $\mathcal S$, completing the induction step.

Therefore, $B_{n, n-1} = B_n$ is the finite union of pairwise disjoint elements of $\mathcal S$, as desired.

Using the above result and applying the axiom of countable choice, we can, for all $n \in \N$, choose a finite set $\mathcal F_n$ of pairwise disjoint elements of $\mathcal S$ for which:

- $\displaystyle B_n = \bigcup \mathcal F_n$

Now, $x \in S$ if and only if there exists an $n \in \N$ such that $x \in S \cap A_n$.

Taking the smallest such $n$, which exists because $\N$ is well-ordered, it follows that $x \notin A_0, A_1, \ldots, A_{n-1}$, and so $x \in S \cap B_n$.

Therefore:

- $\displaystyle S = \bigcup_{n \mathop = 0}^\infty \, \left({S \cap B_n}\right)$

Hence:

\(\displaystyle \mu \left({S}\right)\) | \(=\) | \(\displaystyle \mu \left({\bigcup_{n \mathop = 0}^\infty \, \left({S \cap B_n}\right)}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \mu \left({\bigcup_{n \mathop = 0}^\infty \, \left({S \cap \bigcup_{T \mathop \in \mathcal F_n} T}\right)}\right)\) | $\quad$ by definition of $\mathcal F_n$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \mu \left({\bigcup_{n \mathop = 0}^\infty \, \bigcup_{T \mathop \in \mathcal F_n} \, \left({S \cap T}\right)}\right)\) | $\quad$ by Intersection Distributes over Union | $\quad$ | |||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \, \sum_{T \mathop \in \mathcal F_n} \, \mu \left({S \cap T}\right)\) | $\quad$ because the countable union of finite sets is countable, and by the countable subadditivity of $\mu$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \, \sum_{T \mathop \in \mathcal F_n} \, \mu \left({T}\right)\) | $\quad$ by monotonicity | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \, \mu \left({\bigcup \mathcal F_n}\right)\) | $\quad$ by additivity | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \, \mu \left({B_n}\right)\) | $\quad$ by definition of $\mathcal F_n$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \, \mu \left({A_n}\right)\) | $\quad$ by Set Difference is Subset and monotonicity | $\quad$ |

$\blacksquare$

## Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.