Let $H$ be a semiring over $\Omega$ and let $\mu$ and $\nu$ be two measures on $H$. Show that $(\mu + \nu)^* = \mu^* + \nu^*$ with the induced outer measures $\mu^*$ and $\nu^*.$
We defined the induced outer measure as $\mu^*(A) := \inf \left\{ \sum_{n=1}^\infty \mu(E_n) \;|\; A \subset \bigcup_{n=1}^\infty E_n \land \forall n \in \mathbb{N}:(E_n \in H) \right\}$ with $A \in 2^\Omega$. $\nu^*$ is defined analogously.
Now, my idea was to show that $(\mu + \nu)^* \leq \mu^* + \nu^*$ and $(\mu + \nu)^* \geq \mu^* + \nu^*$. The latter inequality is obvious, since $\inf \{\mu(A) + \nu(A) : A\in H\} \geq \inf \{\mu(A) : A \in H\} + \inf \{\nu(A) : A \in H\}$.
Then, for $(\mu + \nu)^* \leq \mu^* + \nu^*$ my idea was to take two covers $E_n^\mu$ and $E_n^\nu$ made from sets in H, such that for an $A \in 2^\Omega$ we have $\mu^*(A) \leq \sum_{n=1}^\infty \mu(E_n^\mu)$ and $\nu^*(A) \leq \sum_{n=1}^\infty \nu(E_n^\nu)$. Then, with the cover $E_n := E_n^\mu \cap \bigcup_{m=1}^\infty E_m^\nu$ we would have, if each $E_n \in H$, that $(\mu + \nu)^*(A) \leq \sum_{n=1}^\infty \mu(E_n) + \nu(E_n) \leq \mu^*(A) + \nu^*(A).$
My problem is that I don't know how to reason that each $E_n \in H$. I tried with the continuation of $\mu$ and $\nu$ on the ring generated by $H$, but didn't have any luck with that.
I would appreciate any hints and help. Thanks.
You should have $\mu^*(A) \geq \sum_{n=1}^\infty \mu(E_n^\mu) - \varepsilon$ and $\nu^*(A) \geq \sum_{n=1}^\infty \nu(E_n^\nu) - \varepsilon$. To solve your problems, just note that by Caratheodory's theorem, $\mu^*$ is a measure on $\sigma(H)$ (and agrees with $\mu$ on $H$). To carry out your idea, you may also want to assume that the covers consist of disjoint sets (you can assume this because $H$ is a semi-ring).