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I'm working on this problem, and would like some help:

Let $X=[0,1)\times [0,1)\subset \mathbb R^2$, and for fixed $\alpha\in\mathbb R$, let $E\subset X$ be the range of the map: for $t\geq 0$, $$t\to (t\pmod 1, \alpha t \pmod 1)$$ I want to show that $E$ is Borel measurable, and find $m^2(E)$, the product measure of the set $E$.

I think the idea is to consider each sections of the set both in $x$ and in $y$. Then I have a theorem that I can use to calculate the measure. But I'm having a hard time understanding what the set E should be like.

What I have in mind is that, if $E$ were to be a countable union of line segments, then I can take sections of $E$ with fixed $x,y$. Since the Lebesgue measure is $\sigma$-finite on $[0,1)$, then $$E_x=\{y\in [0,1):(x,y)\in E\}$$ $E_x$ would then be a countable set of single points for a fixed $x$, which means that it is Borel and thus measurable since it is a countable union of singletons. The same logic would apply to $E^y$ as well, the sections with fixed $y\in [0,1)$. If that's the case, I then have that $$m^2(E)=\int_{[0,1)}m(E_x)dm=\int_{[0,1)}m(E^y)dm$$ Now, since $E_x,E^y$ are countable union of points, they have Lebesgue measure 0, and $m^2(E)=0$.

Now, what I have issue with is that I'm not sure how to explicitly show that $E$ is a countable union of line segments, and I'm also not sure if my argument is sound assuming that $E$ is indeed a countable union of line segments with slope $\alpha$.

I would love to get some feedback on this, as I have been struggling with this problem for quite a while.

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    $\begingroup$ $E$ is a countable union of line segments of slope $\alpha$ $\endgroup$ – Berci Nov 6 '18 at 22:10
  • $\begingroup$ HINT: the map have only countable many points of discontinuities, depending on the value of $\alpha$. And between these points of discontinuities the image of the map is a product of intervals. $\endgroup$ – Masacroso Nov 6 '18 at 22:16
  • $\begingroup$ @Masacroso Hmm.. I'm having a hard time visualizing why that is so. Could you possibly write out an answer to help me out? $\endgroup$ – Sank Nov 6 '18 at 22:23
  • $\begingroup$ note that $t\mapsto t\pmod 1$ is discontinuous just when $t\in\Bbb N_{\ge 1}$, and the map $t\mapsto\alpha t\pmod 1$ is discontinuous whenever $\alpha t\in\Bbb Z\setminus\{0\}$. Hence you map to $\Bbb R^2$ is discontinuous just when $t\in\Bbb N$ or when $\alpha t\in\Bbb Z\setminus\{0\}$. $\endgroup$ – Masacroso Nov 6 '18 at 22:31
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    $\begingroup$ Well every straight line has Lebesgue measure 0 in $\mathbb R^2$, which seems to align with your argument. $\endgroup$ – Mog Nov 7 '18 at 6:43

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