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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?

Mika H.
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  • Are you allowed to use the fact that the Lebesgue measure on $\mathbb{R}^n$ is a product measure? – Michael Albanese Oct 13 '13 at 04:03
  • @MichaelAlbanese For the purpose of this, let's say no :) – Mika H. Oct 13 '13 at 04:04
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    $\Bbb R^n$ is a countable union of $n$-dimensional hypercubes, so it suffices to show these have measure zero in $\Bbb R^{n+1}$. One can bound this $n$-cube's measure by that of $(n+1)$-cubes containing it with length $\epsilon$ into the $(n+1)$th dimension, and take $\epsilon\to0$. – anon Oct 13 '13 at 04:05

3 Answers3

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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 $$ m(\mathbb{R}^n \times \{0\}) = \lim_{k\to \infty} m([-k,k]^n\times \{0\}) $$
Prahlad Vaidyanathan
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2

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.

0

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 $\epsilon$.

H.Deng
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