Hyperplane has Lebesgue measure zero

By | April 26, 2015

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

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


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

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

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


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