Let $\mathcal{E}$ be a semi-ring of subsets of $\Omega$ and let $\mu:\mathcal{E}\rightarrow [0,\infty]$ be a measure on $\mathcal{E}.$
Define the outer measure $\mu^{*}$ generated by the measure $\mu$ via the formula $$\mu^{*}(A):=\inf \left\{\sum_{n=1}^{\infty}\mu\left(A_{n}\right):\left\{A_{n}\right\} \subset \mathcal{E} \text { and } A \subset \bigcup_{n=1}^{\infty} A_{n}\right\}.$$ Say that a set $A\in \mathcal {P}(\Omega)$ is $\mu^{*}$-measurable if $\mu^{*}(S)=\mu^{*}(S \cap A)+\mu^{*}\left(S \cap A^{c}\right)$ for each subset $S$ of $\Omega$. Then:
- $\mu^{*}(A)=\mu(A)$ for every $A$ belonging to the semi-ring $\mathcal{E}$.
- The collection $\mathcal{U}_{\mu^{*}}$ of $\mu^{*}$-measurable subsets of $\Omega$ is a $\sigma$-algebra which contains $\sigma(\mathcal{E})$, and $\mu^{*}$ is a measure when restricted to $\mathcal{U}_{\mu^{*}}$.
- If $\mu$ is $\sigma$-finite and $\Omega=\bigcup_{i=1}^{\infty} \Omega_{i},\Omega_{i}\in\mathcal{E}$ , then $\mu^{*}$ is the unique extension of $\mu$ to a $\sigma$-finite measure on $\sigma(\mathcal{E})$.
If $\mu$ is finite and $\Omega=\bigcup_{i=1}^{\infty} \Omega_{i},\Omega_{i}\in\mathcal{E}$, Is $\mu^{*}$ the extension of $\mu$ to a finite measure on $\sigma(\mathcal{E})$?
Since $\sigma(\mathcal{E})$ is a $\sigma$-algebra, $\mu^{*}$ is a finite measure on $\sigma(\mathcal{E})\Leftrightarrow \mu^{*}(\Omega)<\infty.$ But how to prove $\mu^{*}(\Omega)<\infty$ under $\mu$ is finite and $\Omega=\bigcup_{i=1}^{\infty} \Omega_{i},\Omega_{i}\in\mathcal{E}?$