We are supposed to solve the following problem. I think I managed to solve the first part, but I'm struggling with the second one.
Let $X$ be a set and $\mathcal{E} \subset \mathcal{P}(X)$ with $\varnothing, X \in \mathcal{E}$. Furthermore, let $\rho : \mathcal{E} \rightarrow [0, \infty]$ be a function with $\rho(\varnothing) = 0$. Finally, let $\mu^*$ be the outer measure induced by $\rho$ on $X$.
a) Prove or disprove that $\mu^* \leq \rho$ on $\mathcal{E}$.
b) Prove that for every outer measure $\tau$ on $X$ with $\tau \leq \rho$ on $\mathcal{E}$, it already holds that $\tau \leq \mu^*$.
For a), I think that the statement is true and I managed to prove it:
Let $A \in \mathcal{E}$ and choose $A_1 = A$ and $A_i = \varnothing, \forall i \geq 2$. Then we have $A \in \bigcup_{i = 1}^\infty A_i$ and $\forall i \in \mathbb{N}: A_i \in \mathcal{E}$. Now, we obtain $\rho(A) = \sum_{i = 1}^\infty \rho(A_i) \geq \mu^*(A)$.
As such, $\mu^* \leq \rho$ on $\mathcal{E}$.
But I don't really know how to prove part b). How can I conclude from an inequality that only holds on $\mathcal{E}$, that $\tau \leq \mu^*$?
I would appreciate any help!
In case anyone has the same problem, I think I solved it, so I'm going to post an answer. If there's a mistake, feel free to correct me.
Let $A \in \mathcal{P}(X)$ and $\epsilon > 0$ for the moment. Let $\mu^*(A) < \infty$, else we're finished. Now, there exists a sequence $A_i \in \mathcal{E}$ such that $A \subset \bigcup_{i = 1}^\infty A_i$ and $\sum_{i = 1}^\infty \rho(A_i) \leq \mu^*(A) + \epsilon$.
Because of our assumption, we immediately get that $\sum_{i = 1}^\infty \tau(A_i) \leq \sum_{i = 1}^\infty \rho(A_i) \leq \mu^*(A) + \epsilon$.
Now we'll use the fact that $\tau$ is an outer measure, in particular the monotonicity and $\sigma$-subadditivity: $\tau(A) \leq \tau(\bigcup_{i = 1}^\infty A_i) \leq \sum_{i = 1}^\infty \tau(A_i)$.
As such we obtained the inequality $\tau(A) \leq \mu^*(A) + \epsilon$. Finally, taking the limit $\epsilon \rightarrow 0$ gives us the desired proof.