[Math] How do i prove that “countably monotone + Finitely Additive” implies “Premeasure” on a semiring

measure-theory

Let $S$ be a semi-ring of subsets of $X$

Let $\mu:S \rightarrow [0,\infty]$ be a set function on $X$

If $\mu$ is countably monotone and finitely additive, then $\mu$ is a premeasure.

I know that this must be true, since $\mu$ can be extended to $\mu^*$ on a ring generated by $S$ which is countably monotone and finitely additive.

However, i don't know how to prove this directly.

Help..

Best Answer

I'm the OP and since i have a low bounty, i cannot comment on Karene's question, so i'm writing this as an answer. The set $S$ Karene constructed is not a semi-ring on $X$. Since $\mathbb{N} \setminus \{a\}$ cannot be a finite union of elements of $S$.

I have proved the following:

Let $X$ be a set.

Let $S$ be a semi-ring on $X$.

Let $\mu:S\rightarrow [0,\infty]$ be a function such that $\mu(\emptyset)=0$.

Then, $\mu$ is countably additive if and only if it is countably monotone and finitely additive.

The proof for if part does not require any choice, but the proof for only if part requires choice.

The key idea for the proof is that $\{A\subset X: A \text{ is a finite disjoint union of elements in } S\}$ is indeed a ring on $X$.

Finite additivity makes it possible to construct a function $\mu^*$ on this ring such that $\mu^* (A) = \sum_{i=1}^n \mu(A_i)$ whenever $A$ is a disjoint union of a finite sequence $A_i$ in $S$.

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