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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!

Community
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  • This sounds like a really convoluted proof direction, and I am very unclear as to both why your professor would insist it be done this way, and what he expects you to do. Frankly, I'd say it's easier to re-invent the wheel and just provide proof of rotation and translation invariance. This invariance is usually established very soon after defining the Lebesgue measure, so I am surprised this is result is not available to you. – zibadawa timmy Nov 12 '13 at 23:34

2 Answers2

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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.

azarel
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  • What about the second part? –  Nov 10 '13 at 20:31
  • @user100463 I'll add some details for the second part. – azarel Nov 10 '13 at 20:36
  • What are the $a_{i}$ and $A_{i}$? Are the A's Borel measurable sets? Do the a 's come from them, or are these the a's given in the definition of the hyperplane above? –  Nov 11 '13 at 04:00
  • Our professor said it is not acceptable to just show it for $n=2$ and say it follows for general $n$. I'm going to either edit or repost with more details as to how she wants us to tackle it after my class gets out. Basically, we have to show that Lebesgue Measure is invariant by using a basis for $\mathbb{R}^{n}$. –  Nov 11 '13 at 19:11
  • @user100463 When you see "the general case is a straightforward generalization", that is math speak for "experts already saw how to generalize this before I even mentioned the general case, and students should work out and provide the specific details explicitly on their own (possibly after much toil)". When I was in grad school, it was the joke of the graduate students that (to us) such a line meant "you will probably need several hours and several pages of fruitless effort to understand this". – zibadawa timmy Nov 12 '13 at 23:38
  • Yeah, it's sort of seeming that way, although I went to office hours tonight, and I think I understand what to do now. Can I collect my own bounty? LOL. –  Nov 13 '13 at 03:05
  • @user100463 I'm not sure how bounties work, but you can answer your own question and accept your own answer. You don't get reputation for accepting your own answer, so I would imagine you don't get your own bounty, either. – zibadawa timmy Nov 16 '13 at 05:02
  • @user100463: So after making people work to write a good answer for your question you want the bounty for yourself? – Beni Bogosel Nov 19 '13 at 14:22
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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.

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.

Beni Bogosel
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