If $E\in \mathscr{M}_{\mu^*}$ , then for each $\varepsilon$ exists $A\in \mathscr{A}$ such that $\mu^*(A\triangle E)< \varepsilon$

Let $X$ be a set, $\mathscr{A}$ a ring of subsets of $X$, $\mu ∶ \mathscr{A} \to \overline{\mathbb{R}}_{≥0}$ a premeasure and $\mu^*$ the outer measure generated by $\mu$. (By Caratheodory)

If $E\in \mathscr{M}_{\mu^*}$ and satisfacies $\mu^*(E)<\infty$, then for each $\varepsilon$ existx $A\in \mathscr{A}$ such that $\mu^*(A\triangle E)< \varepsilon$

I try to test this, my first idea was to give a cover of $ E $, I just don't know if I can find said cover in $ \mathscr{A} $, as the measure is not finite, so the extension is not unique, someone can give me a hint how to proceed?

$\textbf{Hint}$:

By definition of the outer measure, there is a countable collection $\{Q_j\}$ of elements in $ \mathscr{A}$ such that $$ \sum_{j=1}^{\infty}\mu(Q_j) \le \mu^*(E) + \varepsilon. $$ Since $\mu^*(E) < \infty$, the series above converges. Hence there is an $N$ so that $$ \sum_{j=N+1}^{\infty} \mu(Q_j) < \frac\varepsilon 2. $$ Consider $A = \bigcup_{j=1}^N Q_j$.