# Alternative construction of an outer measure from a premeasure on a ring (or semiring)

In the proof of Carathéodory's extension theorem, an outer measure $\mu^*$ is constructed from a premeasure $\mu_0$ on a ring or semi-ring $A$ over an arbitrary set $X$ as

$$\displaystyle \mu^*(S)\equiv\inf\left\{\sum_{i=1}^\infty \mu_0(S_i): S_i\in A,\ S\subseteq\bigcup_{i=1}^\infty S_i\right\},$$

and then a complete measure space is constructed from the outer measure.

I was wondering if the outer measure can be constructed equivalently as

$$\displaystyle \mu^*_1(S)\equiv\inf\left\{\mu_0(\cup_{i=1}^\infty S_i): S_i\in A,\ S\subseteq \bigcup_{i=1}^\infty S_i \right\}, (1)$$

or $$\displaystyle \mu^*_2(S)\equiv\inf \left\{\sum_{i=1}^\infty \mu_0(S_i): S_i\in A,\ S\subseteq \bigcup_{i=1}^\infty S_i, S_i\text{'s are disjoint} \right\}? (2)$$

I figured out that $\mu^*_1(S) \leq \mu^*(S) \leq \mu^*_2(S)$, where the first inequality is because of $\mu_0(\cup_{i=1}^\infty S_i) \leq \sum_{i=1}^\infty \mu_0(S_i)$. But I am not sure if they are equal.

Thanks and regards!

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Can you assure that $\cup_{i=1}^\infty S_i$ belongs to your ring (semiring)? if not, then you cannot consider $\mu_0(\cup_{i=1}^\infty S_i)$. – Matemáticos Chibchas Jan 20 at 17:38