# Is this a Borel Measurable set in $[0,1)^2$?

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.

• $E$ is a countable union of line segments of slope $\alpha$ – Berci Nov 6 '18 at 22:10
• 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. – Masacroso Nov 6 '18 at 22:16
• @Masacroso Hmm.. I'm having a hard time visualizing why that is so. Could you possibly write out an answer to help me out? – Sank Nov 6 '18 at 22:23
• 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 $\alpha t\in\Bbb Z\setminus\{0\}$. – Masacroso Nov 6 '18 at 22:31
• Well every straight line has Lebesgue measure 0 in $\mathbb R^2$, which seems to align with your argument. – Mog Nov 7 '18 at 6:43