for all $E \subset X$ and $\epsilon>0$ there exists $A$ such that $u^*(A) < u^*(E) + \epsilon$ where $u^*$ is an outermeasure

Question

The full problem statement is as follows: Let $\mathcal{A}$ be an algebra on X. let let $\mathcal{A}_{\sigma}$ be the set of countable unions in $\mathcal{A}$. Let $u_{0}$ be a premeasure on $\mathcal{A}$ and let $u^*$ be the induced outer measure. Show that for all $E\in X$ and $\epsilon>0$, there exists $A\in \mathcal{A}_{\sigma}$ so that $u^{*}(A) \leq u^*(E) + \epsilon$

SO my initial thought is lets bring u^*(E) to the other side and use triangle inequality to make $u^*(A \setminus E) <\epsilon$. (where I use $A \setminus E$ to mean $A$ without any element of $E$).

now where I'm stuck on is how can I justify that I can pick A arbitrarily close to but not exactly $E$? is it as simple as letting $A = E \cup{\{e\}}$ where $u^*(e) < \epsilon$ ?

Answer 1

The definition of $\mu^*$ is as follows:$\mu^*(E)=\inf\{\sum_{j=1}^\infty \mu_0(E_j)|A\subseteq \cup_{j=1}^\infty E_j,E_j\in A\}$.

Therefore for any $E$, $\mu^*(E)+\epsilon > \mu^*(E)$. Hence, $\mu^*(E)+\epsilon$ is larger than the infimum. Therefore by the definition of infimum there exists $\{E_j\}_{j\in N}, E_j\in A$ such that $A\subseteq \cup_{j=1}^\infty E_j$ and $\mu^*(E)+\epsilon >\sum_{j=1}^\infty \mu_0(E_j) $.

Note: $\cup_{j=1}^\infty E_j\in A_\sigma$ and $\mu^*(\cup_{j=1}^\infty E_j)\leq \sum_{j=1}^\infty \mu_0(E_j)$.

Hence, $\mu^*(\cup_{j=1}^\infty E_j)<\mu^*(E)+\epsilon$.