Approximating an arbitrary set by a Borel set

Question

Q) Show that for each (not necessarily measurable) subset $A\subset \mathbb{R}$, there exists a Borel subset $B\subset \mathbb{R}$ such that $A\subset B$ and $\lambda^*(A) = \lambda(B)$ where $\lambda^*$ is the outer measure.

I know that for a Lebesgue measurable $A\subset \mathbb{R}$, there exists a Borel set $B\subset \mathbb{R}$ s.t. $B\supset A$ and $\lambda(B-A)=0$. How can I go from there?

Answer 1

For each $n$ there exist open interval $U_k$ with $A \subset \bigcup_k U_k$ and $\lambda (B_n) \leq \lambda^{*} (A)+\frac 1 n$ where $B_n=\bigcup_k U_k$. Take $B=\bigcap B_n$. Then $A \subseteq B$, $B$ is a Borel set and $\lambda (B) \leq \lambda^{*} (A)$. The reverse inequality is obvious.