Hyperplane has Lebesgue measure zero

Theorem. Let $\mathcal{P}$ be a hyperplane of $\mathbb{R}^{d}$ . Then $\mathcal{P}$ has Lebesgue measure zero.

Proof. To show it, we first show

$\mathcal{P}_{0}=\left\{ \left(y_{1},\dots,y_{d}\right)\in\mathbb{R}^{d}:y_{1}=0\right\}$

has measure zero.

Let $\varepsilon>0$ be given. Then set

$Q_{n}=\left[-\frac{\varepsilon}{\left(2n\right)^{d-1}2^{n+1}},\frac{\varepsilon}{\left(2n\right)^{d-1}2^{n+1}}\right]\times\left[-n,n\right]\times\cdots\times\left[-n,n\right].$

By archimedian property, we get $\mathcal{P}_{0}\subset\bigcup_{n=1}^{\infty}Q_{n}$ .
Note that

$m\left(Q_{n}\right)=\left(2n\right)^{d-1}\times\frac{\varepsilon}{\left(2n\right)^{d-1}2^{n}}=\frac{\varepsilon}{2^{n}}.$

$m\left(\mathcal{P}_{0}\right)\le\sum_{n=1}^{\infty}m\left(Q_{n}\right)=\sum_{n=1}^{\infty}\frac{\varepsilon}{2^{n}}=\varepsilon.$

This implies $m\left(\mathcal{P}_{0}\right)=0$ .

So we prove the our theorem. Since $\mathcal{P}$ is a hyperplane of $\mathbb{R}^{d}$ , there is $\mathbf{n}\in\mathbb{R}^{d}\setminus\left\{ 0\right\}$ and $\gamma\in\mathbb{R}^{d}$ such that

$\mathcal{P}=\left\{ \mathbf{x}\in\mathbb{R}^{d}:\mathbf{x}\cdot\mathbf{n}=\gamma\right\} .$

Define

$\mathcal{P}^{\prime}=\left\{ x\in\mathbb{R}^{d}:\mathbf{x}\cdot\mathbf{n}=0\right\}$

Note that by translation invariant of Lebesgue measure, we have

$m\left(\mathcal{P}\right)=m\left(\mathcal{P}^{\prime}\right).$

Set $\mathbf{e}_{1}=\frac{\mathbf{n}}{\left|\mathbf{n}\right|}$ and extend to the orthonormal basis for $\mathbb{R}^{d}$ . From this, there is a orthogonal transform $A:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d}$ such that $A\left(\mathcal{P}_{0}\right)=\mathcal{P}^{\prime}$ . So

$m\left(\mathcal{P}\right)=m\left(\mathcal{P}^{\prime}\right)=m\left(A\left(\mathcal{P}_{0}\right)\right)=\left|\det A\right|m\left(\mathcal{P}_{0}\right)=m\left(\mathcal{P}_{0}\right)=0.$

$\qedbox$

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