Approximate measures of sets with measures of borel subsets. Show that for each subset $A$  of $\mathbb{R}$ there is a Borel subset of $B$ of $\mathbb{R}$ that includes $A$ such that $ \lambda (B) = \lambda ^*(A)$
If A is Borel it is evident? So we need to approximate A with some Borel subset, which is just a "little" bit bigger?
How is this done? I cannot see this when for example A is some strange uncountable set.
Thanks.
 A: Of course if $A$ is Borel you take $B=A$. I assume $\lambda^*$ denotes the Lebesgue outer measure, i.e. 
$$
\lambda^*(A) = \inf \left \{ \left. \sum_{n \ge 1} (b_n - a_n) \, \right| \, A \subseteq \bigcup_{n \ge 1} ]a_n,b_n[ \right \}.
$$
If $\lambda^*(A) = \infty$, just choose $B = \mathbb R$. Otherwise, choose $a_n^m, b_n^m$ such that $\sum_{n \ge 1} (b_n^m - a_n^m) \le \lambda^*(A) + \frac 1m$ and $A \subseteq \bigcup_{n \ge 1} ]a_n^m,b_n^m[$. Let $\mathcal O_m = \bigcup_{n \ge 1} ]a_n^m,b_n^m[$. Then let
$$
B = \bigcap_{m \ge 1} \mathcal O_m.
$$
It is clear that $B$ is Borel. Since $A \subseteq \mathcal O_m$ for every $m$, $A \subseteq B$. By monotonicity of the outer measure, $\lambda^*(A) \le \lambda^*(B) = \lambda(B)$. But by monotonicity of the measure $\lambda$, we have $B \subseteq \mathcal O_m$ for every $m$, hence
$$
\lambda(B) \le \lambda(\mathcal O_m) \le \lambda^*(A) + \frac 1m
$$
which implies $\lambda(B) \le \lambda^*(A)$, hence $\lambda(B) = \lambda^*(A)$. 
Hope that helps,
