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Lebesgue measure of a straight line in $mathbb{R}^2$ is zero
$begingroup$
I'm trying to prove that the Lebesgue measure of a straight line in $mathbb{R}^2$ is zero. I've found certain posts here on Stackexchange concerning this, but I have a slightly different solution which I would like to discuss.
Let $L subset mathbb{R}^2$ be a straight line, and let $(p_n, q_n)_{n=1}^infty$ be an enumeration of the countable set $L cap mathbb{Q}^2$. Fix $epsilon > 0$ and define
$$ A_n = left{ (x,y)in mathbb{R}^2 ,mid, -frac{sqrt{epsilon}}{2sqrt{2^n}} leq x-p_n < frac{sqrt{epsilon}}{2sqrt{2^n}}, ;
-frac{sqrt{epsilon}}{2sqrt{2^n}} leq y-q_n < frac{sqrt{epsilon}}{2sqrt{2^n}} right}. $$
Since $L cap mathbb{Q}^2$ is dense in $L$, it follows that
$$ L subset bigcup_{n=1}^infty A_n, $$
and so
$$ lambda(L) leq sum_{n=1}^infty lambda(A_n) = sum_{n=1}^infty text{vol}(A_n) = sum_{n=1}^infty epsilon/2^n = epsilon. $$
But $epsilon$ was arbitrary, so $lambda(L) = 0$.
Is this argument fine?
measure-theory lebesgue-measure
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
$endgroup$
|
show 4 more comments
$begingroup$
I'm trying to prove that the Lebesgue measure of a straight line in $mathbb{R}^2$ is zero. I've found certain posts here on Stackexchange concerning this, but I have a slightly different solution which I would like to discuss.
Let $L subset mathbb{R}^2$ be a straight line, and let $(p_n, q_n)_{n=1}^infty$ be an enumeration of the countable set $L cap mathbb{Q}^2$. Fix $epsilon > 0$ and define
$$ A_n = left{ (x,y)in mathbb{R}^2 ,mid, -frac{sqrt{epsilon}}{2sqrt{2^n}} leq x-p_n < frac{sqrt{epsilon}}{2sqrt{2^n}}, ;
-frac{sqrt{epsilon}}{2sqrt{2^n}} leq y-q_n < frac{sqrt{epsilon}}{2sqrt{2^n}} right}. $$
Since $L cap mathbb{Q}^2$ is dense in $L$, it follows that
$$ L subset bigcup_{n=1}^infty A_n, $$
and so
$$ lambda(L) leq sum_{n=1}^infty lambda(A_n) = sum_{n=1}^infty text{vol}(A_n) = sum_{n=1}^infty epsilon/2^n = epsilon. $$
But $epsilon$ was arbitrary, so $lambda(L) = 0$.
Is this argument fine?
measure-theory lebesgue-measure
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
$endgroup$
1
$begingroup$
I don't see why $Lsubseteqbigcup A_n$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 18:36
1
$begingroup$
Why is $Lcapmathbb Q^2$ non-empty?
$endgroup$
– kimchi lover
Feb 2 at 18:36
2
$begingroup$
Let $L$ be all $(x,y)$ such that $y=pi+sqrt 2 x$, for example. If $x$ is rational, $y$ is irrational.
$endgroup$
– kimchi lover
Feb 2 at 18:46
1
$begingroup$
However the implication $Lcapmathbb{Q}^2$ is dense in $L$ $Rightarrow$ $Lsubset cup A_n$ is a bit unclear to me
$endgroup$
– Nick
Feb 2 at 18:53
1
$begingroup$
@MisterRiemann sorry I made a mistake, only diffeomorphisms preserve measure zero sets.
$endgroup$
– Nick
Feb 2 at 19:00
|
show 4 more comments
$begingroup$
I'm trying to prove that the Lebesgue measure of a straight line in $mathbb{R}^2$ is zero. I've found certain posts here on Stackexchange concerning this, but I have a slightly different solution which I would like to discuss.
Let $L subset mathbb{R}^2$ be a straight line, and let $(p_n, q_n)_{n=1}^infty$ be an enumeration of the countable set $L cap mathbb{Q}^2$. Fix $epsilon > 0$ and define
$$ A_n = left{ (x,y)in mathbb{R}^2 ,mid, -frac{sqrt{epsilon}}{2sqrt{2^n}} leq x-p_n < frac{sqrt{epsilon}}{2sqrt{2^n}}, ;
-frac{sqrt{epsilon}}{2sqrt{2^n}} leq y-q_n < frac{sqrt{epsilon}}{2sqrt{2^n}} right}. $$
Since $L cap mathbb{Q}^2$ is dense in $L$, it follows that
$$ L subset bigcup_{n=1}^infty A_n, $$
and so
$$ lambda(L) leq sum_{n=1}^infty lambda(A_n) = sum_{n=1}^infty text{vol}(A_n) = sum_{n=1}^infty epsilon/2^n = epsilon. $$
But $epsilon$ was arbitrary, so $lambda(L) = 0$.
Is this argument fine?
measure-theory lebesgue-measure
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
$endgroup$
I'm trying to prove that the Lebesgue measure of a straight line in $mathbb{R}^2$ is zero. I've found certain posts here on Stackexchange concerning this, but I have a slightly different solution which I would like to discuss.
Let $L subset mathbb{R}^2$ be a straight line, and let $(p_n, q_n)_{n=1}^infty$ be an enumeration of the countable set $L cap mathbb{Q}^2$. Fix $epsilon > 0$ and define
$$ A_n = left{ (x,y)in mathbb{R}^2 ,mid, -frac{sqrt{epsilon}}{2sqrt{2^n}} leq x-p_n < frac{sqrt{epsilon}}{2sqrt{2^n}}, ;
-frac{sqrt{epsilon}}{2sqrt{2^n}} leq y-q_n < frac{sqrt{epsilon}}{2sqrt{2^n}} right}. $$
Since $L cap mathbb{Q}^2$ is dense in $L$, it follows that
$$ L subset bigcup_{n=1}^infty A_n, $$
and so
$$ lambda(L) leq sum_{n=1}^infty lambda(A_n) = sum_{n=1}^infty text{vol}(A_n) = sum_{n=1}^infty epsilon/2^n = epsilon. $$
But $epsilon$ was arbitrary, so $lambda(L) = 0$.
Is this argument fine?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
asked Feb 2 at 18:34
MisterRiemannMisterRiemann
5,8951625
5,8951625
1
$begingroup$
I don't see why $Lsubseteqbigcup A_n$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 18:36
1
$begingroup$
Why is $Lcapmathbb Q^2$ non-empty?
$endgroup$
– kimchi lover
Feb 2 at 18:36
2
$begingroup$
Let $L$ be all $(x,y)$ such that $y=pi+sqrt 2 x$, for example. If $x$ is rational, $y$ is irrational.
$endgroup$
– kimchi lover
Feb 2 at 18:46
1
$begingroup$
However the implication $Lcapmathbb{Q}^2$ is dense in $L$ $Rightarrow$ $Lsubset cup A_n$ is a bit unclear to me
$endgroup$
– Nick
Feb 2 at 18:53
1
$begingroup$
@MisterRiemann sorry I made a mistake, only diffeomorphisms preserve measure zero sets.
$endgroup$
– Nick
Feb 2 at 19:00
|
show 4 more comments
1
$begingroup$
I don't see why $Lsubseteqbigcup A_n$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 18:36
1
$begingroup$
Why is $Lcapmathbb Q^2$ non-empty?
$endgroup$
– kimchi lover
Feb 2 at 18:36
2
$begingroup$
Let $L$ be all $(x,y)$ such that $y=pi+sqrt 2 x$, for example. If $x$ is rational, $y$ is irrational.
$endgroup$
– kimchi lover
Feb 2 at 18:46
1
$begingroup$
However the implication $Lcapmathbb{Q}^2$ is dense in $L$ $Rightarrow$ $Lsubset cup A_n$ is a bit unclear to me
$endgroup$
– Nick
Feb 2 at 18:53
1
$begingroup$
@MisterRiemann sorry I made a mistake, only diffeomorphisms preserve measure zero sets.
$endgroup$
– Nick
Feb 2 at 19:00
1
1
$begingroup$
I don't see why $Lsubseteqbigcup A_n$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 18:36
$begingroup$
I don't see why $Lsubseteqbigcup A_n$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 18:36
1
1
$begingroup$
Why is $Lcapmathbb Q^2$ non-empty?
$endgroup$
– kimchi lover
Feb 2 at 18:36
$begingroup$
Why is $Lcapmathbb Q^2$ non-empty?
$endgroup$
– kimchi lover
Feb 2 at 18:36
2
2
$begingroup$
Let $L$ be all $(x,y)$ such that $y=pi+sqrt 2 x$, for example. If $x$ is rational, $y$ is irrational.
$endgroup$
– kimchi lover
Feb 2 at 18:46
$begingroup$
Let $L$ be all $(x,y)$ such that $y=pi+sqrt 2 x$, for example. If $x$ is rational, $y$ is irrational.
$endgroup$
– kimchi lover
Feb 2 at 18:46
1
1
$begingroup$
However the implication $Lcapmathbb{Q}^2$ is dense in $L$ $Rightarrow$ $Lsubset cup A_n$ is a bit unclear to me
$endgroup$
– Nick
Feb 2 at 18:53
$begingroup$
However the implication $Lcapmathbb{Q}^2$ is dense in $L$ $Rightarrow$ $Lsubset cup A_n$ is a bit unclear to me
$endgroup$
– Nick
Feb 2 at 18:53
1
1
$begingroup$
@MisterRiemann sorry I made a mistake, only diffeomorphisms preserve measure zero sets.
$endgroup$
– Nick
Feb 2 at 19:00
$begingroup$
@MisterRiemann sorry I made a mistake, only diffeomorphisms preserve measure zero sets.
$endgroup$
– Nick
Feb 2 at 19:00
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The outline and overall plan of your argument are fine: cover $L$ with countably many sets of small total area, and so on.
But your execution is too complicated and (as the comments indicate) has technical flaws.
My recommendation would be: come up with a proof that the $x$-axis has measure $0$, and then shift and rotate your construction to take care of arbitrary $L$.
Exactly how you do this might depend on what kinds of sets you know how to measure, and what you already know about 2 dimensional measure. You might for instance cover $L$ with disks, the $n$th disk with radius $epsilon/n$ so the sum of their diameters is big enough to cover all of $L$, but yet with finite small total area.
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
Thanks for a thorough answer
$endgroup$
– MisterRiemann
Feb 2 at 19:02
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The outline and overall plan of your argument are fine: cover $L$ with countably many sets of small total area, and so on.
But your execution is too complicated and (as the comments indicate) has technical flaws.
My recommendation would be: come up with a proof that the $x$-axis has measure $0$, and then shift and rotate your construction to take care of arbitrary $L$.
Exactly how you do this might depend on what kinds of sets you know how to measure, and what you already know about 2 dimensional measure. You might for instance cover $L$ with disks, the $n$th disk with radius $epsilon/n$ so the sum of their diameters is big enough to cover all of $L$, but yet with finite small total area.
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
Thanks for a thorough answer
$endgroup$
– MisterRiemann
Feb 2 at 19:02
add a comment |
$begingroup$
The outline and overall plan of your argument are fine: cover $L$ with countably many sets of small total area, and so on.
But your execution is too complicated and (as the comments indicate) has technical flaws.
My recommendation would be: come up with a proof that the $x$-axis has measure $0$, and then shift and rotate your construction to take care of arbitrary $L$.
Exactly how you do this might depend on what kinds of sets you know how to measure, and what you already know about 2 dimensional measure. You might for instance cover $L$ with disks, the $n$th disk with radius $epsilon/n$ so the sum of their diameters is big enough to cover all of $L$, but yet with finite small total area.
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
Thanks for a thorough answer
$endgroup$
– MisterRiemann
Feb 2 at 19:02
add a comment |
$begingroup$
The outline and overall plan of your argument are fine: cover $L$ with countably many sets of small total area, and so on.
But your execution is too complicated and (as the comments indicate) has technical flaws.
My recommendation would be: come up with a proof that the $x$-axis has measure $0$, and then shift and rotate your construction to take care of arbitrary $L$.
Exactly how you do this might depend on what kinds of sets you know how to measure, and what you already know about 2 dimensional measure. You might for instance cover $L$ with disks, the $n$th disk with radius $epsilon/n$ so the sum of their diameters is big enough to cover all of $L$, but yet with finite small total area.
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
$endgroup$
The outline and overall plan of your argument are fine: cover $L$ with countably many sets of small total area, and so on.
But your execution is too complicated and (as the comments indicate) has technical flaws.
My recommendation would be: come up with a proof that the $x$-axis has measure $0$, and then shift and rotate your construction to take care of arbitrary $L$.
Exactly how you do this might depend on what kinds of sets you know how to measure, and what you already know about 2 dimensional measure. You might for instance cover $L$ with disks, the $n$th disk with radius $epsilon/n$ so the sum of their diameters is big enough to cover all of $L$, but yet with finite small total area.
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
edited Feb 2 at 20:23
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 18:53
kimchi loverkimchi lover
11.8k31229
11.8k31229
$begingroup$
Thanks for a thorough answer
$endgroup$
– MisterRiemann
Feb 2 at 19:02
add a comment |
$begingroup$
Thanks for a thorough answer
$endgroup$
– MisterRiemann
Feb 2 at 19:02
$begingroup$
Thanks for a thorough answer
$endgroup$
– MisterRiemann
Feb 2 at 19:02
$begingroup$
Thanks for a thorough answer
$endgroup$
– MisterRiemann
Feb 2 at 19:02
add a comment |
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1
$begingroup$
I don't see why $Lsubseteqbigcup A_n$.
$endgroup$
– Lord Shark the Unknown
Feb 2 at 18:36
1
$begingroup$
Why is $Lcapmathbb Q^2$ non-empty?
$endgroup$
– kimchi lover
Feb 2 at 18:36
2
$begingroup$
Let $L$ be all $(x,y)$ such that $y=pi+sqrt 2 x$, for example. If $x$ is rational, $y$ is irrational.
$endgroup$
– kimchi lover
Feb 2 at 18:46
1
$begingroup$
However the implication $Lcapmathbb{Q}^2$ is dense in $L$ $Rightarrow$ $Lsubset cup A_n$ is a bit unclear to me
$endgroup$
– Nick
Feb 2 at 18:53
1
$begingroup$
@MisterRiemann sorry I made a mistake, only diffeomorphisms preserve measure zero sets.
$endgroup$
– Nick
Feb 2 at 19:00