Let $\mu$ be a finite positive regular Borel measure on $\mathbb{R}^{d}$ and let $S$ be the family of finite unions of squares of the form $\{a_{1}2^{n} \leq x_{1} \leq (a_{1} + 1)2^{n}, \ldots, a_{d}2^{n} \leq x_{d} \leq (a_{d} + 1)2^{n}\}$ where $a_{i}$ and $n$ are integers (that is the dyadic cubes). Show that the set of linear combinations of indicator functions of elements of $S$ is dense in $L^{1}(d\mu)$.
Here's the work I've done so far. Fix $f \in L^{1}$ and $\varepsilon > 0$. By density of simple functions, it suffices to approximation $\chi_{E}$ where $E$ is a measurable set. By outer regularity of $\mu$, there exists an open set $U \supset E$ such that $\mu(U) \leq \mu(E) + \varepsilon$. Thus it remains to approximate $\chi_{U}$ as the linear combination of indicator functions of dyadic cubes.
Since $U \subset \mathbb{R}^{d}$ is open, $U$ is the countable union of almost disjoint cubes $D_{j}$ (see Stein Shakarchi Real Analysis, page 7 and its proof). I would like to say that since $U = \bigcup_{j = 1}^{\infty}D_{j}$, then $\mu(U) = \sum_{j = 1}^{\infty}\mu(D_{j})$, however I only have an almost disjoint union rather than a disjoint union. The expression is however true in the case of Lebesgue measure, I was wondering how to prove/get around this in the case of general $\mu$ as in the problem?
If this expression is true, then choose $N$ so that $\sum_{j = N + 1}^{\infty}\mu(D_{j}) < \varepsilon$ and consider $\sum_{i = 1}^{N}\chi_{D_{i}}$ and we are done.
Remark: It suffices to deal with bounded open sets (take intersections with a family of balls of radius $R\to\infty$).
If you made your cubes half-closed, like $a_{1}2^{n} \leq x_{1} < (a_{1} + 1)2^{n}$, you would not have this problem. Indeed, given an open set, you would pick maximal (by inclusion) dyadic cubes contained in it, and that gives a representation of the open set as a countable disjoint union of dyadic cubes. The conclusion follows.
Working with closed cubes, you have to suffer a little. Here is one way. For each fixed $n$, one can color all dyadic cubes in $\mathbb R^d$ using $3^d$ colors, so that any two cubes of the same color are disjoint. (This follows from the fact that each cube has $3^d-1$ neighbors.) Let $U_0=U$, and build $U_1,\dots,$ inductively. Given $U_k$, choose $n$ so that the union of all dyadic cubes of size $2^n$ contained in $U_k$ has at least half of the measure of $U_k$ (referring to $\mu$ measure). Then pick a color such that the cubes of size $2^n$ with that color have total measure at least $3^{-d} \frac12 \mu(U_k)$. Remove these cubes from $U_k$. What remains is $U_{k+1}$; repeat.
By construction, $\mu(U_k)\to 0$ as $k\to\infty$. Hence, the $L^1$ distance between $\chi_U$ and the (disjoint!) union of cubes removed from $U$ after $k$ steps tends to zero.