For $A \subset \mathbb{R}$ a bounded Borel set. Show that for all $\epsilon > 0$ there exists a set $U$ which is a finite union of intervals such that $\lambda_1(A \triangle U)\leq \epsilon$ With $\triangle$ being the symmetric difference of $A$ and $U$
I'm currently in a measure theory class and I have this question and I just don't know where to start. I have a feeling I may need to use the definition of an outer measure, as well as the fact that the complement of an interval in the semi-algebra of intervals in $\mathbb{R}$ is a finite union of intervals, but I'm not sure. I'm really struggling in this module in knowing how to approach questions so if anyone had any advice I'd really appreciate that.
Edit: $\lambda_1$ is the Lebesgue measure on $\mathbb{R}$ So for an interval $(a,b]$ we have that $\lambda_1((a,b])=b-a$
Hints: (1) Use outer measure to approximate $A$, to within measure $\varepsilon/2$, by an open superset. (2) Recall that an open subset of $\mathbb R$ is the union of countably many disjoint intervals. The measure of the open set is the sum of the measures of those intervals. (3) Approximate that infinite sum, to within measure $\varepsilon/2$, by a finite partial sum.