Let $A \subseteq \mathbb R^{d}$, and $\bar{B^{d}}:=\{B \cup \bar{N} \subseteq \mathbb R^{d}: B \in B^{d},\exists N \in B^{d}, \lambda^{d}(N) = 0, \bar{N} \subseteq N\}$. Show that:
$\forall \epsilon > 0$, $\exists U \subseteq \mathbb R^{d}$ open set, and $C \subseteq \mathbb R^{d}$ closed set with $C \subseteq A \subseteq U$ and $\lambda^{d}(U-C) < \epsilon \iff A \in \bar{B^{d}}$
I have already been able to prove the rather trivial direction "$\Leftarrow$", but in terms of the next direction "$\Rightarrow$" I am having a difficult time utilizing both the open and the closed sets at the same time, as my previous proofs only revolved around either the open or the closed sets.
My failed attempt so far to use both: Let $\epsilon > 0$ and $C_{n} \subseteq A \subseteq U_{n}$, such that $C_{n}$ is closed, $U_{n}$ is open $\forall n \in \mathbb N$ and $\lambda^{d}(U_{n}-C_{n}) < \epsilon$
Since $A \subseteq U_{n}$, and $C_{n} \subseteq A, \forall n \in \mathbb N$, $\lambda^{d}(U_{n}-A) \leq \lambda^{d}(U_{n}-C_{n})< \epsilon$. Any ideas as to how I could use this to prove that $A \in \bar{B^{d}}$?
Perhaps if I am able to prove that $U_{n}, C_{n} \in \bar{B^{d}}, \forall n \in \mathbb N \Rightarrow A \in \bar{B^{d}}$?
First suppose that $\lambda^*(A)<\infty.$ Let $(U_n)_n$ and $(C_n)_n$ have the properties ascribed to them in the exercise. Observe that $\bigcup_nC_n\subseteq A\subseteq \bigcap_nU_n$ and this implies that $A=\bigcup_nC_n\cup (A\setminus \bigcup_nC_n).$ But, $\bigcup_nC_n$ is a Borel set, $A\setminus \bigcup_nC_n\subseteq \bigcap_nU_n\setminus\bigcup_n C_n$ and $\lambda(\bigcap_nU_n\setminus\bigcup_n C_n)=0,\ $ which proves the finite case. Now the general case follows because $\lambda$ is $\sigma-$finite.