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The terminology in this area is somewhat confusing, my question is how to prove:

Given $E \subseteq \mathbb{R}$, there exists a Lebesgue measurable set $A$ such that $E \subseteq A$ and $\lambda^*(E) = \lambda^*(A)$.

Some authors call such $A$ a measurable cover of $E$.

I would appreciate if the proof was as straightforward from the definition of $\lambda^*$ as possible and if no stronger choice principle than CC$(\mathbb{R})$ was used.

Thanks a thousand times,

S.

Edit: You may assume that $\lambda^*(E) < \infty$.

SAP
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If I remember right a definition of Lebesgue outer measure then it seems the following.

For each natural $n$ there exists a cover $\mathcal C_n$ of the set $E$ by open intervals such that $\sum\{\lambda(C):C\in\mathcal C_n\}<\lambda^*(E)+1/n.$ Then a set $A=\bigcap_n\bigcup C_n$ is Borel (hence, Lebesgue measurable), $E\subset A$ and $\lambda^*(E)=\lambda(A).$

Alex Ravsky
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