# Show that the natural copy of $\mathbb{R^{n-1}}$ inside $\mathbb{R^{n}}$ - namely, $\{(x_1, x_2,…, x_{n-1},0)\}$ - has measure zero

Question P.202 (Differential Topology - Guillemin, Pollack) : Show that the natural copy of $\mathbb{R^{n-1}}$ inside $\mathbb{R^{n}}$ - namely, $\{(x_1, x_2,..., x_{n-1},0)\}$ - has measure zero. [Hint : Show that every compact subset of $\mathbb{R^{n-1}}$ sits inside a single rectangle in $\mathbb{R^{n}}$ with volume $< \epsilon$ ]

Here, I believe it is necessary to use the differentiable map $i: \mathbb{R^{n-1}} \to \mathbb{R^{n}} ; (x_1, ... , x_{n-1}) \to (x_1, ... , x_{n-1}, 0)$ with the idea of using the Lipschitz condition on a compact set C of $\mathbb{R^{n-1}}$ . Obviously, I can see how to interpret the problem. However, I do not find the necessary tools for their applications. Someone would it be able to guide me a bit about how to start the problem?

• You can work pretty much directly: cover the rectangle $[-M,M]^{n-1}$ by the rectangle $[-M,M]^{n-1} \times [-\delta_M,\delta_M]$, for $\delta_M$ sufficiently small that $(2M)^{n-1} (2\delta_M)<\varepsilon$ for every $M$. So I guess $\delta_M < \frac{\varepsilon}{2^n M^{n-1}}$ is enough. – Ian Jul 14 '15 at 15:38
• I had the idea, but I thought I was mistaken. Yes, indeed, you are right, thank you! – user230283 Jul 14 '15 at 15:53
• TeX question: Did you really mean to use $\Bbb {R^n}$ and not $\Bbb R^n$? – PVAL-inactive Jul 14 '15 at 16:01
• Sorry, minor error, I meant the rectangle $[-M,M]^{n-1} \times \{ 0 \}$. – Ian Jul 14 '15 at 16:27

Hint: Since clearly ${\mathbb R}^{n-1} \times \{0\}$ is a countable union of unit-side-length hypercubes whose last coordinate is zero, it suffices to show that $[0,1]^{n-1} \times \{0\}$ has measure zero in ${\mathbb R}^n$. This will make your proof a bit easier than considering arbitrarily large hypercubes.
Let's do this in $\mathbb R^{2}$ to get an idea of how this would go:
Let $\epsilon$ be given and consider sets of the form $\frac{1}{2}[-n,n]\times \frac{1}{2}\left [ \frac{-\epsilon}{n2^{n}},\frac{\epsilon}{n2^{n}} \right ]$. Then
$\mathbb R\times \left \{ 0 \right \}\subseteq \bigcup _{n\geq 1} \frac{1}{2}[-n,n]\times \frac{1}{2}\left [ \frac{-\epsilon}{n2^{n}},\frac{\epsilon}{n2^{n}} \right ]$ and so
$\mu ^{*}\left ( \mathbb R\times \left \{ 0 \right \} \right )\leq \sum^{\infty } _{n=1}n\left ( \frac{\epsilon }{n2^{n}} \right )=\epsilon$.
This extends easily to the case $n>2$.