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Prove that under Lebesgue Measure on $\mathbb{R}^{2}$ every straight line has a measure of zero.

My try:

Let A = $\{ (x,f(x)), x \in \mathbb{R} \}$ be the set of all points lying on the straight line $y = f(x)$.

Show that $\lambda(A) = 0$.

Let $\{ (a_{i}, b_{i})\}$ be a sequence covering $\mathbb{R}$ with $ b_{i} - a_{i} = 1$ $ \forall i \in \mathbb{N}$.

Define $K_{i} = \{ (x,f(x) - \frac{\epsilon}{2^{i}})\times (x,f(x) + \frac{\epsilon}{2^{i}}), x \in (a_{i}, b_{i}) \} $

Then K = $ \cup_i K_{i}$ covers A and

$\lambda(K) \leq \sum_{i}(b_{i} - a_{i}) \frac{2\epsilon}{2^{i}} = 2\epsilon \rightarrow 0, \epsilon \rightarrow 0$.

Since $K$ covers $A$,

$\lambda(A) \leq \lambda(K)$

Is this ok or I have I missed something or done something wrong?

El_Loco
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    Your definition of $K_i$ seems off to me. Do you want $K_i$ to be $K_i={ ,(x,y) \mid x\in(a_i, b_i), f(x)-\epsilon/2\le y\le f(x)+\epsilon/2,}$? (So, the "diagonal strip" of height $2\epsilon$, parallel to, and centered on, the line.) – David Mitra Feb 21 '13 at 16:06
  • Almost, $f(x) - \epsilon/2^{i} \leq y \leq f(x) + \epsilon/2^{i}$, so you get a rectangle covering a line segment for each $i$ – El_Loco Feb 21 '13 at 16:20
  • Each segment $K_{i}$ is a parallelogram with height $b_{i} - a_{i} = 1$ and width $\frac{2\epsilon}{2^{i}}$. I am interested in the measure of the union K of $K_{i}$ since it includes A. – El_Loco Feb 21 '13 at 16:41
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    Yes. Perhaps a nitpick; but I think it should be stated in your argument why $\lambda(K_i)=(b_i-a_i)\cdot{2\epsilon\over 2^i}$. – David Mitra Feb 21 '13 at 16:43
  • Ok, then I understand. :D Thanks for your comments! – El_Loco Feb 21 '13 at 16:45

2 Answers2

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Great work. Here is an alternative that could make things slightly (just slightly) easier to write down.

First observe that there exists an affine translation of $\mathbb{R}^2$ which preserves measure (a rotation + a translation), and takes your straight line to $$ L=\{(x,0)\;;\;x\in\mathbb{R}\}. $$

Now your line has the same measure as $L$.

Now by Fubini $$ \lambda(L)=\int_{\mathbb{R}^2}1_Ldxdy=\int_{x\in \mathbb{R}} \left(\int_{y\in\mathbb{R}}1_L(x,y)dy \right)dx=\int_{x\in\mathbb{R}}\left(\int_{y\in\{0\}}1dy \right)dx=0 $$ since $\lambda(\{0\})=0$.

Julien
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Excellent work. The only thing left to take care of now are vertical lines.

Cameron Buie
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    Thx for the advice, it is always something you forget. :D – El_Loco Feb 21 '13 at 15:51
  • Is it not possible to skip the whole proof by saying that in $\Bbb{R}^2$ the Lebesgue measure computes the area of rectangles and that the area of a line is $0$? – sound wave Dec 03 '18 at 17:43
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    @sound wave: Not really. That's a nice heuristic argument, but lines are not rectangles. It's certainly true that Lebesgue measure gives areas, when the set involved has an area, but consider $[0,1]\times(\Bbb R\setminus\Bbb Q).$ This is merely a union of non-adjacent line segments, so it "should" have area $0,$ but it has infinite measure. – Cameron Buie Dec 03 '18 at 18:06
  • You can see a line as a rectangle of zero height and infinite basis, so its area is zero. The set you wrote is "seen" by the Lebesgue measure as a rectangle of height one and infinite basis, so its area is infinite, as you said. – sound wave Dec 03 '18 at 18:19
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    $0\cdot\infty$ isn't well-defined, so while you could certainly see a rectangle of $0$ height and infinite base, that doesn't allow you to find its "area." – Cameron Buie Dec 03 '18 at 18:22
  • We don't need to compute zero times infinity, consider the sequence of rectangles with basis $n$ and height $1/n^2$, as $n$ goes to infinity the area tends to $0$. – sound wave Dec 03 '18 at 19:18
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    @sound wave: By contrast, consider the sequence of rectangles with base $n$ and height $\frac1n.$ The area remains at $1$ for all $n,$ so tends to $1$ as $n\to\infty.$ Does that mean that the area of a line is $1$? Of course not. If you want to treat a line as a rectangle, then to calculate area, you multiply its base by its height. If you can't do that, then you aren't viewing it as a rectangle at all, but as a pathological rectangle-like object. Thus, the rectangle heuristic breaks down. – Cameron Buie Dec 03 '18 at 23:28
  • Good point. There should be a way tho to extend the concept of an interval with radius $0$ in dimension $1$: in $\Bbb{R}$ the Lebesgue measure computes the length of intervals, so the length of a singleton is $0$. – sound wave Dec 03 '18 at 23:57
  • Let us continue this discussion in chat. – Cameron Buie Dec 04 '18 at 00:23
  • Oops! I forgot to tag @sound wave about the chat. – Cameron Buie Dec 04 '18 at 11:16