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First I'll have to mention this is not homework or something like that. It's training for an exam and I couldn't find resources on how it is done. If you have any URL showing how to solve these kinds of Lebesgue exercises, please tell me.

I have the following to find:

  1. $\mu(([0,10]\setminus \mathbb{Q}) \cup ([0,20]\cap \mathbb{Q})) $ where $\mu$ is the Lebesgue measure on $\mathbb{R}$.

  2. $\lambda(\{(x,x)\in \mathbb{R}^2 \; : \; x\in \mathbb{R} \})$ where $\lambda$ is the Lebesgue measure on $\mathbb{R}^2$.

I know that for an interval, let's say $[a, b]$, the Lebesgue measure would be $b-a$, right?

But for these I do not know how, plus it's $\mathbb{R}^2$ in the second case, not $\mathbb{R}$.

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Tyraxor
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  • I think that number one is equal to 10. The reason is that the rationals have lebesgue measure 0 and the irrationals are the complement of the rationals and so the measure of the set $[0,10]\backslash \mathbb{Q}$ is the measure of $[0,10]$ minus the measure of $\mathbb{Q}$. This gives $10-0=0$. There is no contribution from $[0,20]\cap \mathbb{Q}$ because intersecting with a countable set forms a countable set and thus has measure 0. – Jürgen Sukumaran Jun 14 '16 at 19:12
  • Agree with Tony S.F. one problem one, for problem two, the Lebesgue measure is zero: rotate by 45 degrees and you are looking at the Lebesgue measure of a coordinate axis, which is of product form $\lambda({0} \times \mathbb R) = \lambda({0}) \times \lambda(\mathbb R)$, and $\lambda({0}) = 0$. – Joris Bierkens Jun 14 '16 at 19:18
  • Why don't you try writing the set out in the first part? That will make it abundantly clear what the Lebesgue measure is (note that the Lebesgue measure of the rationals is zero). – Cameron Williams Jun 14 '16 at 19:27
  • For b the answer is 0, right? I found a method in my course that applies for R^2. You draw the actual segments and "sum" them. At b) it would be segments (x, x) which is a straight line so that Lebesgue measure is 0 – Tyraxor Jun 14 '16 at 20:18

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Hint for (2): Use Fubini's theorem.

Robert Israel
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