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

A: 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.
A: Choose a small $\epsilon_0 > 0$, then the $\{0\} \times R^{n-1}$ hyperplane can be covered by countably many open cubes $\{C_k\}$ in $R^{n-1}$ with side length $1+2\epsilon_0$(Just extend each unit grid for a little). For any $\epsilon > 0$ let $h_k=\frac{\epsilon}{(1+2\epsilon_0)^{n-1}2^{k+1}}$and the countable many open rectangles in $R^n$: $\{D_k:= (-h_k,h_k) \times C_k\}$ covers the hyperplane with total volume less than $\epsilon$.
