Given a Borel set $B$ prove: for every $\epsilon$, $\exists$ compact and closed sets and a continuous $\phi$...

I am trying to solve the following problem:

Given a Borel set $B$ of finite Lebesgue measure show that for every $\epsilon>0$, there exist: a compact set $C_1 \subseteq B$, a closed set $C_2 \subseteq \mathbb{R}\setminus B$ and a continuous function $\phi: \mathbb{R}\rightarrow [0,1]$ s.t.: $$ \chi_{C_1}\leq \phi \leq \chi_{\mathbb{R}\setminus C_2}$$ $$ || \chi_B - \phi||_1 < \epsilon$$

I have only the vague intuition of using the monotone convergence theorem somehow - e.g. perhaps setting $B\cap[n,n+1]$ as sets on which corresponding simple functions $f_n$ (such as the indicator function) can be defined so that they are wedged between $\chi_{C_1}$ and $\chi_{\mathbb{R}\setminus C_2}$ (thus satisfying the first condition) and so that taking the limit on their integral would satisfy the second condition.

I am not sure how to go about accomplishing this, especially since the example I suggested for $B\cap[n,n+1]$ is not necessarily closed and therefore not necessarily compact. In addition, I am not sure how to define the required closed set $C_2$.

Any guidance on the steps of applying MCT (assuming I am at least right on this) to this problem would be much appreciated.

EDIT: Another thought I had was perhaps using somehow the regularity of the Lebesgue measure to define the compact and closed sets (?), i.e. there are a compact set $K$ and an open set $U$ such that:

$$\mu(B)\leq \mu(K)+\epsilon$$ and $$\mu(B)\geq \mu(U)-\epsilon$$

Maybe these could be used to define the indicator function I mentioned above?

The problem is stated a bit weirdly. It is more commonly stated with $U = \mathbb{R} \setminus C_2$. To solve your problem, you can use regularity of Lebesgue measure to get $K$ and $U$, and then use Urysohn's lemma to get $\phi \in C_c(\mathbb{R})$ (or even in $C_c^{\infty}(\mathbb{R})$) with $K \prec \phi \prec U$.