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

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 $$u^*(e) < \epsilon$$ ?

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