Need help badly: n-dimensional Lebesgue measure of a hyperplane is zero.

Question

Let $\alpha \in \mathbb{R}$, $a \neq 0$, and $\mu \in \mathbb{R}^{n}$. Let $H$ be the hyperplane in $R^{n}$ given by $h = \{ x \in \mathbb{R}^{n} : \langle x-\mu , a \rangle = 0 \}$.

Show that $m_{n}(H)=0$, where $m_{n}$ is $n-$dimensional Lebesgue measure, and deduce that $\int_{H}f(x)dx = \int_{\infty}^{\infty}\cdots \int_{-\infty}^{\infty} f(x_{1},\cdots, x_{n})1_{H}(x_{1},\cdots,x_{n})dx_{1},\cdots, dx_{n} = 0$ for any Borel function $f$ on $\mathbb{R}^{n}$.

We need to find a linear basis for $\mathbb{R}^{n}$, then remove one of the vectors that doesn't define the hyperplane, so that we get that it is sufficient to have at most $n-1$ vectors, then redefine the muktivariable Lebesgue measure according to the rotation and translation, but I am really at a loss as to how to do this. Please help!

Answer 1

Since the Lebesgue measure is invariant to rigid motions, it is enough to prove that the measure of the hyperplane $\{x_n =0 \}$ is zero. Using the fact that the Lebesgue measure is countably additive (and subadditive), it is enough to prove the result for a $n-1$ dimensional box of sizes $B=r_1\times r_2 \times ... \times r_{n-1} \times 0$.

This box can be included in a $n$ dimensional box of size $r_1 \times r_2 \times ... \times r_{n-1} \times h$ (imagine that you consider the portions "above" and "below" the $n-1$ dimensional set from heights $-h/2$ to $h/2$, or more precisely, if $B=\{[0,r_1]\times ... \times [0,r_{n-1}] \times \{0\}\}$ you can consider $B=\{[0,r_1]\times ... \times [0,r_{n-1}] \times [-h/2,h/2]\}$; the inclusion $B\subset B_h$ is obvious now)

It is known (from the definition) that the Lebesgue measure of the box is $r_1...r_{n-1}h$ so as $h$ goes to zero we have $|B_h| \to 0$, so $|B|\leq |B_h|$ which implies that $|B|=0$.

Once you have proven the result for a $n-1$ dimensional box, you divide your hyperplane into small, not necessary disjoint rectangles which all have measure zero, so the measure of your hyperplane is zero.

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This seems a bit like reasoning in a circle... If we already know the measure of a $n$ dimensional box, then for a $n-1$ dimensional box, the measure needs to be zero since one of the edges of the box has length zero.

Answer 2

Let us consider the case $n=2$ for simplicity (the general case is a straightforward generalization). Using that Lebesgue measure is invariant under rotations and translations we may assume that the hyperplane is equal to $\mathbb R$. Note that $\mathbb R=\bigcup_{n\in\mathbb Z}[n,n+1]$, so it suffices to show that the interval $[n,n+1]$ has two-dimensional measure zero. This is clear since it is contained in a rectangle of lenght 1 and height arbitrarily small.

For the second part: First show the result for simple functions. Fix $\phi=a_1\chi_{A_1}+...+a_n\chi_{A_n}$ then $\int_H \phi=a_1\mu(H\cap A_1)+...+a_n\mu(H\cap A_n)=0$ since $\mu(H)=0$, for the general result just approximate an arbitrary Borel function using simple functions.