Let $A$ be a borel measurable set. I need to prove the existence of $B$ in the Borelians such that $A\subset B$ and $l^*(B\setminus A)=0$.
Because of the way the outer measure $l^*(A)$ is defined (as the infimum of a sum in the algebra), we can take a covering such that $A\subset \cup_i A_i$ and $l^*(A)\leq l^*(\cup_i A_i)\leq \sum_i l^*(A_i)\leq l^*(A)+1/n$. If $l^*(A)<\infty$ this means $l^*(\cup A_i\setminus A)=l^*(\cup A_i)-l^*(A)<1/n$, where $\cup A_i= E_{1/n}$ but this is not good enough and we need to trim the excess sets. That is why I tried taking intersection of these elements which surely remains within the borelians:
$$A\subset\cap_n E_{1/n}\quad \quad l^*(\cap_n E_{1/n}\setminus A) \leq l^*( E_{1/n}\setminus A)\leq \frac{1}{n} $$
Because this holds for every $n$, then we have the desired condition that $l^*(\cap E_{1/n}\setminus A)=0$.
When $l^*(A)=\infty$, we could try to proceed similarly with: $A\subset \cup_i A_i$ and $\infty=l^*(A)\leq l^*(\cup_i A_i)\leq \sum_i l^*(A_i)$. However, taking intersection is dangerous, because we have no guarantee it will be countable and remain in the Borelians. Any help is appreciated.
I was able to finish things off. The case $l^*(A)<\infty$ is proven. Now let $I_n=(n,n+1]$ and $A_n=I_n\cap A$. Clearly, $A_n$ has finite measure and frow what we have seen, there is a $B_n$ in the borelians such that $A_n\subset B_n$ and $l^*(B_n\setminus A_n)=0$. Let us define the following countable union which is in the Borelians $\cup_{n\in \mathbb{Z}}B_n$. Clearly $A\subset \cup_{n\in \mathbb{Z}}B_n=B$ and because $l^*$ is a measure in Lebesgue measurable sets:
$$l^*(B\setminus A)=l^*({\cup}(B_n\setminus A))\leq l^*({\cup}(B_n\setminus A_n))\leq \sum l^*(B_n\setminus A_n)=\sum 0=0$$