So this is a problem about Borel sets that says:
Prove that a set $A \subset \mathbb{R}^n$ is measurable $\Longleftrightarrow$ there exist a set $B$ wich is an $F_{\sigma}$ and a set $C$ which is a $G_{\delta}$ such that $B \subset A \subset C$ and $C \sim B$ is a null set.
(Here, $F_{\sigma}$ is a countable union of closed sets and $G_{\delta}$ is a countable intersection of open sets. )
There's an easy direction ($\Rightarrow$), so if I have a measurable set $A$, by approximation property I know that for every $k \in \mathbb{N},$ there exist a closed set $F_{k}$ and an open set $G_{k}$ such that $F_{k} \subset A \subset G_{k}$ and $\lambda (G_{k} \sim F_{k}) < k^{-1}$ then we can set $B= \bigcup_{k} F_{k}$ and $C=\bigcap_{k} G_{k}$ and we're done with that.
Now, let's see that $C \sim B$ is a null set:
\begin{eqnarray*} C \sim B & = & (\bigcap_{k} G_{k}) \sim (\bigcup_{k} F_{k}) \\[6pt] &=& (\bigcap_{k} G_{k}) \cap (\bigcup_{k} F_{k})^{c}\\[6pt] &=& (\bigcap_{k} G_{k}) \cap (\bigcap_{k} F_{k}^{c}) \\[6pt] &=& \bigcap_{k} (G_{k} \cap F_{k}^{c})\\[6pt] &=& \bigcap_{k} (G_{k} \sim F_{k}) \\[6pt] \end{eqnarray*}
Then $C \sim B \subset G_{k} \sim F_{k},$ for every $k$, therefore
$$\lambda(C \sim B) < \lambda( G_{k} \sim F_{k})< k^{-1}, \forall k.$$
Consequently, $\lambda( C \sim B)=0$ which means that $C \sim B$ is a null set.
I know have to prove the other direction ($\Leftarrow$), I don't even know how to start I just have that $B \subset A \subset C$ and I have a null set but I need some help proving this please, I'll appreciate your time. Thanks!
The other direction can proved as:
$B,C$ are measurable for they are $G_δ$ and $F_σ$ sets. Since $B⊂A⊂C$ and $C-B$ is a null set, we have $$ \lambda(B)\leqslant\lambda(A)\leqslant\lambda(C)\quad\text{and }\quad\lambda(C)-\lambda(B)=0 $$ So $A$ is measurable and $\lambda(A)=\lambda(C)$.