# Lebesgue measure on RxR

• Aug 16th 2011, 12:21 PM
anselmoalko
Lebesgue measure on RxR
Hi,

I have the following prblem:

Let $m_2$ be the Lebesgue measure on $\mathbb{R} ^2$.

(i) Given $c\in\mathbb{R}$ find the measure of the set $\{(x,y)\in\mathbb{R}^2; x+y=c\}$.
My solution: I simply drew the set and jcomputed the length using the Pythagorean Thm, so $m_2(\{(x,y)\in\mathbb{R}^2; x+y=c\})=\sqrt 2 |c|$.

(ii) Find $m_2(A)$, where $A$ consists of all pairs $(x,y)$ in $[0,\pi/2]\times[0,\pi/2]$ such that $\cos (x)\ge 1/2$ and $\sin(y)$ is irrational.
My attempt: I tried to do the same thing here, but as you can see the set is a bit more complicated. This is what I've got so far: since $\cos (x)\ge 1/2$ iff. $x\le\pi/3$ we have $0\le x\le \pi/3$.

What next?
• Aug 16th 2011, 12:47 PM
girdav
Re: Lebesgue measure on RxR
(i) You computed the measure of the triangle whose vertexes are $(0,0)$, $(0,c)$, $(c,0)$, but it's not what it's asked. You can write $\left\{(x,y)\in\mathbb R^2, x+y=c\right\} =\bigcup_{n\in\mathbb Z}\left\{(x,y)\in\mathbb R^2, x+y=c, n\leq x\leq n+1\right\}$ and show that for all $n$, $m_2(\left\{(x,y)\in\mathbb R^2, x+y=c, n\leq x\leq n+1\right\})=0$. To do that, you can cover the line for a fixed $k\in\mathbb N$ by $k$ squares with measure $\frac 1{k^2}$.
(ii) Write $A = \left[0,\frac{\pi}3\right]\times B$ where $B =\left\{y\in\mathbb R, 0\leq y\leq \frac{\pi}2, \sin y \notin \mathbb Q\right\}$. We can compute the measure of the complement of $B$ in $\left[0,\frac{\pi}2\right]$ writing $\left[0,\frac{\pi}2\right]\setminus B = \bigcup_{r\in \mathbb Q}\left\{x\in \left[0,\frac{\pi}2\right],\sin x=r\right\}$.
• Aug 17th 2011, 05:35 AM
anselmoalko
Re: Lebesgue measure on RxR
(i) Oh, I see. And then one lets $k \rightarrow \infty$ and we obtain $m_2(\{(x,y)\in\mathbb{R}^2; x+y=c, n\le x\le n+1\})=0$ and since the original set is a disjoint union of sets of measure zero we get at it is also of measure zero.

(ii) Here I fully understand what you mean but I'm not sure of my result.

So what I get is $m_2([0,\pi/2]\setminus B) = \sum_{r\in\mathbb{Q}}m(\{y\in\mathbb{R}; 0\le y\le \pi/2, \sin(y)=r\})=0$. And then $m([0,\pi/2])=m([0,\pi/2]\setminus B\cup B) = m([0,\pi/2]\setminus B)+m(B)=m(B)$. So $m_2([0,\pi/3]\times B)=m([0,\pi/3])m(B)=(\pi/3)(\pi/2)=\pi^2/6$.

Is this correct? What I'm not sure about is the step $m_2([0,\pi/3]\times B)=m([0,\pi/3])m(B)$, from what I've read the set needs to be a "rectangle" for that to be allowed.
• Aug 17th 2011, 06:24 AM
girdav
Re: Lebesgue measure on RxR
(i) It's ok.
(ii) We can write, since the measures of the following sets are finite: $m_2\left(\left[0,\frac{\pi}3\right]\times B\right) =m_2\left(\left[0,\frac{\pi}3\right]\times \left[0,\frac{\pi}2\right]\right)- m_2\left(\left[0,\frac{\pi}3\right]\times \left(\left[0,\frac{\pi}2\right]\setminus B\right)\right)$, and conclude since $\left[0,\frac{\pi}2\right]\setminus B$ is countable.