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Math Help - Lebesgue measure on RxR

  1. #1
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    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?
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  2. #2
    Super Member girdav's Avatar
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    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\}.
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  3. #3
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    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.
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  4. #4
    Super Member girdav's Avatar
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    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.
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