$B$ is a Borel envelope of $A$ iff any measurable subset of $B\setminus A$ has Lebesgue measure 0 Suppose that $A\subseteq R^{m}$, let $B(R^{m})$ be the set of all Borel subsets of $R^{m}$. We say that $B\subseteq R^{m}$ is a Borel envelope of $A$ if $B\in B(R^{m})$ and for every $L^{m}$ measurable set (here $L^{m}$ means Lebesgue outer measure) $F$ in $R^{m}$ one has $L^{m}(A\cap F)=L^{m}(B\cap F)$. Then I want to show the following: Let $B\in B(R^{m})$ and $A\subseteqq B$. Prove that $B$ is a Borel envelope of A iff $L^{m}(S)=0$ whenever $S$ is an $L^{m}$-measurable subset of $B \setminus A$. I want to use the fact that there is a Borel set $A'$ such that $L^{m}(A)=L^{m}(A')$ but I don't know what to do next.
 A: $\newcommand{\de}{\delta}\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}$To simplify the writing, let us first work locally, say with subsets of $(0,1]^m$, instead of $\R^m$, so that to avoid infinite values of $L^m$. (Going back to $\R^m$ is then straightforward: using Carathéodory's criterion, for any set $A\subseteq\R^m$ we "collect" its "finite pieces" by the formula $L^m(A)=\sum_{z\in\Z^m}L^m(A\cap(z+(0,1]^m))$.)
Then it is easy to show (see Lemma 1 at the end of this answer) that the definition of a Borel envelope simplifies as follows: a Borel set $B$ is a Borel envelope of $A$ if $B\supseteq A$ and $L^{m}(A)=L^{m}(B)$.
We want to show the following: Suppose that $B$ is Borel and $A\subseteq B$; then $B$ is a Borel envelope of $A$ iff $L^{m}(S)=0$ whenever $S$ is an $L^{m}$-measurable subset of $B\setminus A$.
The only-if part: Suppose that $B$ is a Borel envelope of $A$. To obtain a contradiction, suppose that $L^{m}(S)>0$ for some $L^{m}$-measurable $S\subseteq B\setminus A$. Then $S\subseteq B$ and for the $L^{m}$-measurable set $S_1:=B\setminus S$ we have $A\subseteq S_1$ and hence $L^{m}(A)\le L^{m}(S_1)=L^{m}(B)-L^{m}(S)<L^{m}(B)$, which contradicts the condition $L^{m}(A)=L^{m}(B)$, as desired.
The if part: Suppose that $L^{m}(S)=0$ whenever $S$ is an $L^{m}$-measurable subset of $B\setminus A$. To obtain a contradiction, suppose that $B$ is not a Borel envelope of $A$ -- that is, $L^{m}(A)<L^{m}(B)$.
Note that $L^{m}(A)=\inf\{L^{m}(C)\colon C\supseteq A, C\text{ Borel}\}$. So,
$L^{m}(B)>L^{m}(C)$ for some Borel set $C$ such that $C\supseteq A$, whence $L^{m}(B)> L^{m}(C\cap B)$ and $L^{m}(B\setminus C)=L^{m}(B)-L^{m}(C\cap B)>0$, whereas the set $S:=B\setminus C$ is $L^m$-measurable and contained in $B\setminus A$ -- a desired contradiction.
To complete the proof, we only to need to state and prove

Lemma 1: A Borel set $B\subseteq(0,1]^m$ is a Borel envelope of a set $A\subseteq(0,1]^m$ iff $B\supseteq A$ and $L^{m}(A)=L^{m}(B)$.

Proof of Lemma 1: The only-if part of Lemma 1 is trivial: just take, for instance, $F=(0,1]^m$ or $F=\R^m$.
Let us now consider the if part of Lemma 1. Here we suppose that $A\subseteq B\subseteq(0,1]^m$, $B$ is Borel, and $L^{m}(A)=L^{m}(B)$. To obtain a contradiction, suppose that $B$ is not a Borel envelope of $A$. Then $L^{m}(A\cap F)\ne L^{m}(B\cap F)$ and hence $L^{m}(A\cap F)<L^{m}(B\cap F)$ for some $L^{m}$-measurable set $F\subseteq\R^m$. Also, obviously, $L^{m}(A\cap F^c)\le L^{m}(B\cap F^c)$, where $F^c:=\R^m\setminus F$. Adding the latter two inequalities and using Carathéodory's criterion again (together with the upper bound $L^{m}(A)\le L^{m}((0,1]^m)\le1<\infty$), we get
$L^{m}(A)=L^{m}(A\cap F)+L^{m}(A\cap F^c)<L^{m}(B\cap F)+L^{m}(B\cap F^c)
=L^{m}(B)$, which does contradict the condition $L^{m}(A)=L^{m}(B)$.
This completes the proof of Lemma 1 and thus the entire proof of the desired characterization of the Borel envelope.
