$\mathbb{R}^n\times\{0\}$ has measure zero in $\mathbb{R}^{n+1}$

You can do something similar with cubes in $\mathbb{R}^n$ :

1. 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 $< \epsilon$

1. Note that $$ m(\mathbb{R}^n \times \{0\}) = \lim_{k\to \infty} m([-k,k]^n\times \{0\}) $$