(clarify this answer?) Prove that for all subsets of the reals A, there exists a Borel set $A \subset B$ such that $\lambda(B) = \lambda^*(A) $

Question

This (or a similar) question has been asked already on stackexchange. I am asking for a variation on the accepted answer.

Prove for each $A\subset \mathbb{R}$ there exists Borel set whose measure is same as outer measure of $A$

how do we relate this to the question above, where in we must consider $\lambda(B)$ instead of $\lambda^*(B)$?

Answer 1

If you analize the definition of $\lambda^*(A)$, you get $\lambda^*(A)=\inf \{\sum_n \lambda(E_n) \mbox{ where } (E_n)_{n \in amthbb{N}} \mbox{ is a countable covering of } A \}$ and thus for all $n \in \mathbb{N}$ there exists $(E_k^n)_k$ such that $\sum_k \lambda (E_k^n) \leq \lambda^*(A)+\frac 1 n$. If you take $E=\bigcap_n\bigcup_k E_k^n$ you have that $E$ is what you are searching for. In fact $\lambda^*(A) \leq \lambda(E) \leq \lambda^*(A)+\frac 1 n$ for all $n \in \mathbb{N}$.

Note that this holds in general for every $(X,\mathcal{A},\mu)$ measure space the "borel" set is replaced by $E \in \mathcal{A}$.