$\mathbb{R}^n\times\{0\}$ has measure zero in $\mathbb{R}^{n+1}$
You can do something similar with cubes in $\mathbb{R}^n$ :
- Show that $[-k,k]^n \times \{0\}$ has measure zero for each $k \in \mathbb{N}$
Proof : For $\epsilon > 0$, choose $\delta > 0$ so that $$ 2\delta \prod_{i=1}^n (2k+2\epsilon) < \epsilon $$ Then note that $$ U = (-k-\epsilon, k+\epsilon)^n \times (-\delta, \delta) $$ contains $[-k,k]^n\times \{0\}$, and has measure $< \epsilon$
- Note that $$ m(\mathbb{R}^n \times \{0\}) = \lim_{k\to \infty} m([-k,k]^n\times \{0\}) $$
It is enough to do this in $\Bbb{R}^2$. We want to show the $x$-axis has measure zero. To do this we write the $x$-axis as a countable disjoint union of intervals $(k,k+1]$. Since a countable disjoint union of measure zero sets has measure zero it is enough to show one of these has measure zero as a subset of $\Bbb{R}^2$.
The crux now is this: If $\mu((0,1])$ didn't have measure zero, we could fill the unit square with as many of these as we like contradicting the unit square having finite measure. So the measure of $(k,k+1]$ is zero and we are done.