Open sets in $\mathbb{R}^d$ written as union of partially open cubes We know that an open set in $\mathbb{R}^d$ can be written as union of disjoint closed cubes (here for example).
In Zygmund's book it is left to reader the fact that the same open set can be written as a countable union of disjoint partly open cubes. I don't know how to proceed in this case. I believe one should follow the idea in 1-dimensional case, taking intervals like
$$ [a_1,b_1) \cup [a_2,b_2) \cup \ldots$$
How does the proof linked above change? Thanks in advance!!
 A: I think you mean open sets in $\Bbb R^n$ are unions of "almost" disjoint cubes
(interiors are disjoint).
The idea in the linked post works here too.
For a start consider the set of cubes $[a_1,a_1+1)\times\cdots\times[a_n,a_n+1)$
with $a_i\in\Bbb Z$. Start off by taking all those contained in your open set $U$
and let their union be $V_0$.
Now consider cubes $[a_1/2,a_1/2+1/2)\times\cdots\times[a_n/2,a_n/2+1/2)$
again with $a_i\in\Bbb Z$ and take those contained in $U-V_0$. Let $V_1$
be the union of $V_0$ and these cubes.
After the $k$-th stage you have a subset $V_{k-1}$ of $U$ which is
a disjoint union of cubes and $U-V_{k-1}$ contains no cube
$[a_1/2^{k-1},a_1/2^{k-1}+1/2^{k-1})\times\cdots\times[a_n/2^{k-1},a_n/2^{k-1}+1/2^{k-1})$ with integer $a_i$.
Then $V_k$ will be the union of $V_{k-1}$ with the
$[a_1/2^k,a_1/2^k+1/2^k)\times\cdots\times[a_n/2^k,a_n/2^k+1/2^k)$ within
$U-V_k$.
Finally, $\bigcup_k V_k$ is a disjoint union of "half-open" cubes
and one shows that $U$ being open entails that $U=V_k$.
