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Let $\mathfrak R$ be a ring on $M$ and let $\mu: \mathfrak R \to [0,\infty]$ be finitely additive.

Suppose that $\mu$ satisfies countable sub-additivity. Does it follow that $\mu$ is countably additive?

In class, we only showed that $\mu$ being countably additive implies that it is countably sub-additive, but I think that if one assumes that $\mu$ is finitely additive, the other direction also holds:

Let $R_1, R_2,\ldots\in\mathfrak R$ be pairwise disjoint s.t. $\bigcup_{i=1}^\infty R_i \in \mathfrak R$. Then $$ \sum_{i=1}^\infty \mu(R_i) \underset{\text{finite additivity}}{\leq} \mu\left(\bigcup_{i=1}^\infty R_i\right) \underset{\text{countable sub-additivity}}{\leq} \sum_{i=1}^\infty \mu(R_i), $$ so that $\mu\left(\bigcup_{i=1}^\infty R_i\right)=\sum_{i=1}^\infty \mu(R_i)$.

Is there anything wrong with my reasoning?

Thank you!

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You are correct.

Only you are incomplete when it comes to the first inequality. A proof is really needed here.

Start by fixing $n$.

If $R_1,\dots,R_n\in\mathfrak R$ then $\bigcup_{i=1}^nR_i\in\mathfrak R$ because as ring $\mathfrak R$ is closed under finite unions.

If $R:=\bigcup_{i=1}^{\infty}R_i\in\mathfrak R$ then $S:=R-\bigcup_{i=1}^nR_i\in\mathfrak R$ because as ring $\mathfrak R$ is closed under differences.

Then $R=R_1\cup\cdots\cup R_n\cup S$ where the sets mentioned in this union are disjoint, and on base of finite additivity we conclude:

$$\sum_{i=1}^n\mu(R_i)\leq\sum_{i=1}^n\mu(R_i)+\mu(S)=\mu(R)$$

So we have $\sum_{i=1}^n\mu(R_i)\leq\mu(R)$ for every fixed $n$ allowing us to conclude that:$$\sum_{i=1}^{\infty}\mu(R_i)\leq\mu(R)$$

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