$\mathbb{R}^n\times\{0\}$ has measure zero in $\mathbb{R}^{n+1}$ I want to show that $\mathbb{R}^n\times\{0\}$ has measure zero in $\mathbb{R}^{n+1}$.
For example, take $n=1$. I want to show that the $x$-axis has measure zero in the plane. I cover it with the sets $[-1,1]\times[-\epsilon/8,\epsilon/8]$, $[-2,2]\times[-\epsilon/32,\epsilon/32]$, $\ldots$
The goal is to have the volumes be $\epsilon/2, \epsilon/4, \ldots$ so that they sum to $\epsilon$.
I think this method generalizes to $\mathbb{R}^n$, simply by choosing $[-1,1]^{n-1}\times[-\epsilon/2^{n+2},\epsilon/2^{n+2}],\ldots$. I don't think it is very elegant though. Is there a "better" way to do this?

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

