This feels wrong to me, but I seem to be able to perfectly approximate every bounded Borel set using a countable union of disjoint open intervals. By perfectly, I mean that the Lebesgue measure of the symmetric difference between the Borel set and my approximation is zero.
Assume a Borel set $B \in \mathcal{B}(\mathbb{R})$ is bounded, and so $B \subseteq (a_0, b_0)$ for some $a_0, b_0 \in \mathbb{R}$. For each $x \in B$, we consider the maximal interval $(a_x, b_x) \ni x$ such that $\lambda((a_x, b_x) \triangle B) = 0$, where $\lambda$ is the Lebesgue measure and $X \triangle Y := (X\backslash Y) \cup (Y \backslash X)$ denotes the symmetric difference. More precisely, \begin{align*} a_x &:= \inf \{ a \in \mathbb{R}: a \leq x, \lambda((a, x) \triangle B) = 0 \}, \\ b_x &:= \sup \{ b \in \mathbb{R}: b \geq x, \lambda((x, b) \triangle B) = 0 \}.\end{align*} We end up with disjoint intervals, because if $x \in I$ and $x \in J$ for some intervals $I$ and $J$, then we must have $I \cup J \in (a_x, b_x)$ by maximality of the interval $(a_x, b_x)$. Moreover, every open interval contains a rational number, and iF yOu BeLiEvE iN tHe AxIoM oF cHoIcE, we can choose a rational representative of every interval, and so this is a countable union.
Denote this countable union by $S := \cup_{n \in \mathbb{N}}I_n$. Then $\lambda(S \triangle B) = \sum_{n=1}^{\infty} \lambda (I_n \triangle B) = \sum_{n=1}^{\infty} 0 = 0$ since the $I_n \triangle B$ are disjoint and a countable sum of zeros is zero.
The above feels wrong, but I can't see what's wrong with my reasoning. Where did I make a mistake?
EDIT
@user14111 pointed out a typo above that's been fixed now.
ANOTHER EDIT
Thanks to @user14111 in his first comment below, I think I see my mistake now. Essentially, I implicitly sort of assumed that "only dense bits of a set can generate positive measure", if that makes sense. In my argument above, for each $x \in B$, we could have the case $a_x = x = b_x$, but I didn't mention it because then trivially we have $(a_x, b_x) = \emptyset$ and we could just ignore it, but we can't just ignore it, because for example the fat Cantor set is nowhere dense (meaning all of the $(a_x, b_x)$ would be $\emptyset$) but it nevertheless has positive measure.