In Chapter 1 of the book Measure & Integral by Wheeden & Zygmund (2nd Edition), Q22 goes as follows:
For an integer $k=1,2,\dots,n$ and a real number $\alpha$, consider the hyperplane $H=\{x=(x_1,x_2,\dots,x_n) : x_k = \alpha\}$. Show that for every $\varepsilon>0$, there is a collection $\{Q_j\}_{j = 1}^\infty$ of cubes in $\mathbb{R}^n$ with edges parallel to the coordinate axes such that $H \subseteq \bigcup Q_j$ and $\sum v(Q_j) < \varepsilon$.
I understand that the above question is equivalent to showing that the measure of $H$ is $0$. However, all the proofs I've found online uses rectangles (rather than cubes in $\mathbb{R}^n$) to prove this fact.
I've also tried using harmonic series to form the sides of the cubes but I ultimately still fail to show that the total volume of these cubes (after covering the entire $H$) will be finite.
I am starting to get doubtful if this is indeed provable with cubes in $\mathbb{R}^n$. Is there a typo in this question and that it should be "cubes in $\mathbb{R}^{n-1}$" or something to that effect? If not, how do we go about proving this with cubes?
A set has measure zero with respect to rectangles if and only if it has measure zero with respect to cubes (doesn't matter if they're open or closed... and there are actually a whole bunch of other shapes possible too, but that gets more tricky). I'll let you fiddle around with $\epsilon$'s here to justify the equivalence in general.
In the case of a hyperplane, it's pretty easy to give a proof. Let $E=[0,1]^{n-1}\times\{0\}\subset\Bbb{R}^n$ be the unit $(n-1)$-dimensional cube. Then, $H$ is a countable union of such cubes, so it is sufficient to show $E$ has measure zero with respect to cubes. Now, for each $p\in\Bbb{N}$, consider the cover of $E$ by $n$-cubes of length $\frac{1}{p}$. Such cubes have volume $\frac{1}{p^n}$, and we can cover $E$ by $p^{n-1}$ many such cubes, so their total volume is $\frac{1}{p}$, which vanishes as $p\to \infty$. So, given $\epsilon>0$, just choose $k$ large enough such that $\frac{1}{p}<\epsilon$.
In your post, the hyperplane $H$ is obtained by setting $x_k=\alpha$, but clearly, we can assume $\alpha=0$ and $k=n$, because the argument remains the same.