Mixing
Non-Measurable set construction on a unit circle
$begingroup$
I am wondering about part 3 of the following question. I was hoping to get another hint. Thank you.
Identify the interval $[0,1]$ with the unit circle $S^1={e^{itheta}:thetain[0,2pi)}subsetmathbb C$.
We will construct a set in $S^1$ that is not Borel-measurable. To this end, for $z=e^{ialpha},w=e^{ibeta}in S^1$, we say $zsim w$ if $alpha-betainmathbb{Q}$. This is clearly an equivalence relation.
Using the axiom of choice we can construct a set $Lambda$ whose elements are one from each equivalence class. For $alphain[0,2pi)capmathbb{Q}$ let $Lambda_alpha=e^{ialpha}Lambda$ be the rotation of $Lambda$ by the angle $alpha$.
1) Prove that if $alpha,betain[0,2pi)capmathbb{Q}$ are distinct, then $Lambda_alphacapLambda_beta=varnothing$.
2) Prove that $S^1$ is the union over $alphain[0,2pi)capmathbb{Q}$ of $Lambda_alpha$.
3) Let $m$ be the Lebesgue measure on $S^1$ (equipped with the Borel $sigma$-algebra). Prove that $m(Lambda)$ cannot exist. (Hint: use the thing you just proved and the fact that $m$ is shift-invariant.)
Some quick notes on the problem:
A relation of equivalence breaks the entire circle into disjoint equivalence classes. In this problem, we have countable many equivalence classes (ie. one from each rational number). In other words, to get an equivalence class, we pick one element from each class. To get an equivalence class, I pick an element, say $gamma$ from $Lambda$ and then we have ${gamma+alpha: alpha in mathbb{Q}}$ is the set of all elements equivalent to $gamma$.
real-analysis measure-theory
edited Jan 18 at 21:41
J.F
33812
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
$endgroup$
add a comment |
$begingroup$
I am wondering about part 3 of the following question. I was hoping to get another hint. Thank you.
Identify the interval $[0,1]$ with the unit circle $S^1={e^{itheta}:thetain[0,2pi)}subsetmathbb C$.
We will construct a set in $S^1$ that is not Borel-measurable. To this end, for $z=e^{ialpha},w=e^{ibeta}in S^1$, we say $zsim w$ if $alpha-betainmathbb{Q}$. This is clearly an equivalence relation.
Using the axiom of choice we can construct a set $Lambda$ whose elements are one from each equivalence class. For $alphain[0,2pi)capmathbb{Q}$ let $Lambda_alpha=e^{ialpha}Lambda$ be the rotation of $Lambda$ by the angle $alpha$.
1) Prove that if $alpha,betain[0,2pi)capmathbb{Q}$ are distinct, then $Lambda_alphacapLambda_beta=varnothing$.
2) Prove that $S^1$ is the union over $alphain[0,2pi)capmathbb{Q}$ of $Lambda_alpha$.
3) Let $m$ be the Lebesgue measure on $S^1$ (equipped with the Borel $sigma$-algebra). Prove that $m(Lambda)$ cannot exist. (Hint: use the thing you just proved and the fact that $m$ is shift-invariant.)
Some quick notes on the problem:
A relation of equivalence breaks the entire circle into disjoint equivalence classes. In this problem, we have countable many equivalence classes (ie. one from each rational number). In other words, to get an equivalence class, we pick one element from each class. To get an equivalence class, I pick an element, say $gamma$ from $Lambda$ and then we have ${gamma+alpha: alpha in mathbb{Q}}$ is the set of all elements equivalent to $gamma$.
real-analysis measure-theory
edited Jan 18 at 21:41
J.F
33812
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
$endgroup$
1
$begingroup$
Hint: if it exists, $m(Lambda)$ is either zero or some positive number. Now sum over all the distinct $Lambda_alpha$
$endgroup$
– Henry
Jan 18 at 18:34
$begingroup$
Thank you for the hint. Is it because if I sum over all the measures of distinct $Lambda_{alpha}$ then I would get 0 since they are made up of points and the measure of a bunch of disjoint points is 0? But If I were to shift it in the right way I could get nonzero values... Sorry I am having a tough time with this problem.
$endgroup$
– MathIsHard
Jan 18 at 18:59
add a comment |
$begingroup$
I am wondering about part 3 of the following question. I was hoping to get another hint. Thank you.
Identify the interval $[0,1]$ with the unit circle $S^1={e^{itheta}:thetain[0,2pi)}subsetmathbb C$.
We will construct a set in $S^1$ that is not Borel-measurable. To this end, for $z=e^{ialpha},w=e^{ibeta}in S^1$, we say $zsim w$ if $alpha-betainmathbb{Q}$. This is clearly an equivalence relation.
Using the axiom of choice we can construct a set $Lambda$ whose elements are one from each equivalence class. For $alphain[0,2pi)capmathbb{Q}$ let $Lambda_alpha=e^{ialpha}Lambda$ be the rotation of $Lambda$ by the angle $alpha$.
1) Prove that if $alpha,betain[0,2pi)capmathbb{Q}$ are distinct, then $Lambda_alphacapLambda_beta=varnothing$.
2) Prove that $S^1$ is the union over $alphain[0,2pi)capmathbb{Q}$ of $Lambda_alpha$.
3) Let $m$ be the Lebesgue measure on $S^1$ (equipped with the Borel $sigma$-algebra). Prove that $m(Lambda)$ cannot exist. (Hint: use the thing you just proved and the fact that $m$ is shift-invariant.)
Some quick notes on the problem:
A relation of equivalence breaks the entire circle into disjoint equivalence classes. In this problem, we have countable many equivalence classes (ie. one from each rational number). In other words, to get an equivalence class, we pick one element from each class. To get an equivalence class, I pick an element, say $gamma$ from $Lambda$ and then we have ${gamma+alpha: alpha in mathbb{Q}}$ is the set of all elements equivalent to $gamma$.
real-analysis measure-theory
edited Jan 18 at 21:41
J.F
33812
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
$endgroup$
I am wondering about part 3 of the following question. I was hoping to get another hint. Thank you.
Identify the interval $[0,1]$ with the unit circle $S^1={e^{itheta}:thetain[0,2pi)}subsetmathbb C$.
We will construct a set in $S^1$ that is not Borel-measurable. To this end, for $z=e^{ialpha},w=e^{ibeta}in S^1$, we say $zsim w$ if $alpha-betainmathbb{Q}$. This is clearly an equivalence relation.
Using the axiom of choice we can construct a set $Lambda$ whose elements are one from each equivalence class. For $alphain[0,2pi)capmathbb{Q}$ let $Lambda_alpha=e^{ialpha}Lambda$ be the rotation of $Lambda$ by the angle $alpha$.
1) Prove that if $alpha,betain[0,2pi)capmathbb{Q}$ are distinct, then $Lambda_alphacapLambda_beta=varnothing$.
2) Prove that $S^1$ is the union over $alphain[0,2pi)capmathbb{Q}$ of $Lambda_alpha$.
3) Let $m$ be the Lebesgue measure on $S^1$ (equipped with the Borel $sigma$-algebra). Prove that $m(Lambda)$ cannot exist. (Hint: use the thing you just proved and the fact that $m$ is shift-invariant.)
Some quick notes on the problem:
A relation of equivalence breaks the entire circle into disjoint equivalence classes. In this problem, we have countable many equivalence classes (ie. one from each rational number). In other words, to get an equivalence class, we pick one element from each class. To get an equivalence class, I pick an element, say $gamma$ from $Lambda$ and then we have ${gamma+alpha: alpha in mathbb{Q}}$ is the set of all elements equivalent to $gamma$.
real-analysis measure-theory
real-analysis measure-theory
edited Jan 18 at 21:41
J.F
33812
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
edited Jan 18 at 21:41
J.F
33812
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
edited Jan 18 at 21:41
J.F
33812
edited Jan 18 at 21:41
J.F
33812
edited Jan 18 at 21:41
J.F
33812
33812
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
asked Jan 18 at 18:26
MathIsHardMathIsHard
1,278516
1,278516
1
$begingroup$
Hint: if it exists, $m(Lambda)$ is either zero or some positive number. Now sum over all the distinct $Lambda_alpha$
$endgroup$
– Henry
Jan 18 at 18:34
$begingroup$
Thank you for the hint. Is it because if I sum over all the measures of distinct $Lambda_{alpha}$ then I would get 0 since they are made up of points and the measure of a bunch of disjoint points is 0? But If I were to shift it in the right way I could get nonzero values... Sorry I am having a tough time with this problem.
$endgroup$
– MathIsHard
Jan 18 at 18:59
add a comment |
1
$begingroup$
Hint: if it exists, $m(Lambda)$ is either zero or some positive number. Now sum over all the distinct $Lambda_alpha$
$endgroup$
– Henry
Jan 18 at 18:34
$begingroup$
Thank you for the hint. Is it because if I sum over all the measures of distinct $Lambda_{alpha}$ then I would get 0 since they are made up of points and the measure of a bunch of disjoint points is 0? But If I were to shift it in the right way I could get nonzero values... Sorry I am having a tough time with this problem.
$endgroup$
– MathIsHard
Jan 18 at 18:59
1
1
$begingroup$
Hint: if it exists, $m(Lambda)$ is either zero or some positive number. Now sum over all the distinct $Lambda_alpha$
$endgroup$
– Henry
Jan 18 at 18:34
$begingroup$
Hint: if it exists, $m(Lambda)$ is either zero or some positive number. Now sum over all the distinct $Lambda_alpha$
$endgroup$
– Henry
Jan 18 at 18:34
$begingroup$
Thank you for the hint. Is it because if I sum over all the measures of distinct $Lambda_{alpha}$ then I would get 0 since they are made up of points and the measure of a bunch of disjoint points is 0? But If I were to shift it in the right way I could get nonzero values... Sorry I am having a tough time with this problem.
$endgroup$
– MathIsHard
Jan 18 at 18:59
$begingroup$
Thank you for the hint. Is it because if I sum over all the measures of distinct $Lambda_{alpha}$ then I would get 0 since they are made up of points and the measure of a bunch of disjoint points is 0? But If I were to shift it in the right way I could get nonzero values... Sorry I am having a tough time with this problem.
$endgroup$
– MathIsHard
Jan 18 at 18:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As Henry hinted you, you should assume by contradiction that $m(Lambda)=c > 0$.
Then you get $$infty > m(S^1) = m(bigcup_alpha Lambda_alpha) = sum_{alpha} m(Lambda_alpha) = sum_{alpha} m(Lambda) = sum_{alpha} c = infty$$.
I would have loved commenting this as a comment instead but I don't have enough reputation yet.
answered Jan 18 at 19:21
J.FJ.F
33812
$endgroup$
1
$begingroup$
Thank you so much. I really appreciate it.
$endgroup$
– MathIsHard
Jan 18 at 19:24
add a comment |
$begingroup$
You wrote:
In this problem, we have countable many equivalence classes (ie. one
from each rational number)
That's incorrect. Each equivalence class has countably many members, and therefore there are uncountably many equivalence classes. The rational numbers are all in the same class, and then there's a class with $sqrt{2}-1$, and a class with $pi-3$, and so on. A set of representatives, such as $Lambda$, is therefore uncountable. If it had been countable, it would have been measurable, of measure $0$.
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
$endgroup$
1
$begingroup$
Oh thank you. I appreciate the clarification on that.
$endgroup$
– MathIsHard
Jan 18 at 19:35
add a comment |
Your Answer
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2 Answers
2
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2 Answers
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$begingroup$
As Henry hinted you, you should assume by contradiction that $m(Lambda)=c > 0$.
Then you get $$infty > m(S^1) = m(bigcup_alpha Lambda_alpha) = sum_{alpha} m(Lambda_alpha) = sum_{alpha} m(Lambda) = sum_{alpha} c = infty$$.
I would have loved commenting this as a comment instead but I don't have enough reputation yet.
answered Jan 18 at 19:21
J.FJ.F
33812
$endgroup$
1
$begingroup$
Thank you so much. I really appreciate it.
$endgroup$
– MathIsHard
Jan 18 at 19:24
add a comment |
$begingroup$
As Henry hinted you, you should assume by contradiction that $m(Lambda)=c > 0$.
Then you get $$infty > m(S^1) = m(bigcup_alpha Lambda_alpha) = sum_{alpha} m(Lambda_alpha) = sum_{alpha} m(Lambda) = sum_{alpha} c = infty$$.
I would have loved commenting this as a comment instead but I don't have enough reputation yet.
answered Jan 18 at 19:21
J.FJ.F
33812
$endgroup$
1
$begingroup$
Thank you so much. I really appreciate it.
$endgroup$
– MathIsHard
Jan 18 at 19:24
add a comment |
$begingroup$
As Henry hinted you, you should assume by contradiction that $m(Lambda)=c > 0$.
Then you get $$infty > m(S^1) = m(bigcup_alpha Lambda_alpha) = sum_{alpha} m(Lambda_alpha) = sum_{alpha} m(Lambda) = sum_{alpha} c = infty$$.
I would have loved commenting this as a comment instead but I don't have enough reputation yet.
answered Jan 18 at 19:21
J.FJ.F
33812
$endgroup$
As Henry hinted you, you should assume by contradiction that $m(Lambda)=c > 0$.
Then you get $$infty > m(S^1) = m(bigcup_alpha Lambda_alpha) = sum_{alpha} m(Lambda_alpha) = sum_{alpha} m(Lambda) = sum_{alpha} c = infty$$.
I would have loved commenting this as a comment instead but I don't have enough reputation yet.
answered Jan 18 at 19:21
J.FJ.F
33812
answered Jan 18 at 19:21
J.FJ.F
33812
answered Jan 18 at 19:21
J.FJ.F
33812
answered Jan 18 at 19:21
J.FJ.F
33812
33812
1
$begingroup$
Thank you so much. I really appreciate it.
$endgroup$
– MathIsHard
Jan 18 at 19:24
add a comment |
1
$begingroup$
Thank you so much. I really appreciate it.
$endgroup$
– MathIsHard
Jan 18 at 19:24
1
1
$begingroup$
Thank you so much. I really appreciate it.
$endgroup$
– MathIsHard
Jan 18 at 19:24
$begingroup$
Thank you so much. I really appreciate it.
$endgroup$
– MathIsHard
Jan 18 at 19:24
add a comment |
$begingroup$
You wrote:
In this problem, we have countable many equivalence classes (ie. one
from each rational number)
That's incorrect. Each equivalence class has countably many members, and therefore there are uncountably many equivalence classes. The rational numbers are all in the same class, and then there's a class with $sqrt{2}-1$, and a class with $pi-3$, and so on. A set of representatives, such as $Lambda$, is therefore uncountable. If it had been countable, it would have been measurable, of measure $0$.
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
$endgroup$
1
$begingroup$
Oh thank you. I appreciate the clarification on that.
$endgroup$
– MathIsHard
Jan 18 at 19:35
add a comment |
$begingroup$
You wrote:
In this problem, we have countable many equivalence classes (ie. one
from each rational number)
That's incorrect. Each equivalence class has countably many members, and therefore there are uncountably many equivalence classes. The rational numbers are all in the same class, and then there's a class with $sqrt{2}-1$, and a class with $pi-3$, and so on. A set of representatives, such as $Lambda$, is therefore uncountable. If it had been countable, it would have been measurable, of measure $0$.
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
$endgroup$
1
$begingroup$
Oh thank you. I appreciate the clarification on that.
$endgroup$
– MathIsHard
Jan 18 at 19:35
add a comment |
$begingroup$
You wrote:
In this problem, we have countable many equivalence classes (ie. one
from each rational number)
That's incorrect. Each equivalence class has countably many members, and therefore there are uncountably many equivalence classes. The rational numbers are all in the same class, and then there's a class with $sqrt{2}-1$, and a class with $pi-3$, and so on. A set of representatives, such as $Lambda$, is therefore uncountable. If it had been countable, it would have been measurable, of measure $0$.
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
$endgroup$
You wrote:
In this problem, we have countable many equivalence classes (ie. one
from each rational number)
That's incorrect. Each equivalence class has countably many members, and therefore there are uncountably many equivalence classes. The rational numbers are all in the same class, and then there's a class with $sqrt{2}-1$, and a class with $pi-3$, and so on. A set of representatives, such as $Lambda$, is therefore uncountable. If it had been countable, it would have been measurable, of measure $0$.
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
answered Jan 18 at 19:27
Alon AmitAlon Amit
10.6k3768
10.6k3768
1
$begingroup$
Oh thank you. I appreciate the clarification on that.
$endgroup$
– MathIsHard
Jan 18 at 19:35
add a comment |
1
$begingroup$
Oh thank you. I appreciate the clarification on that.
$endgroup$
– MathIsHard
Jan 18 at 19:35
1
1
$begingroup$
Oh thank you. I appreciate the clarification on that.
$endgroup$
– MathIsHard
Jan 18 at 19:35
$begingroup$
Oh thank you. I appreciate the clarification on that.
$endgroup$
– MathIsHard
Jan 18 at 19:35
add a comment |
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$begingroup$
Hint: if it exists, $m(Lambda)$ is either zero or some positive number. Now sum over all the distinct $Lambda_alpha$
$endgroup$
– Henry
Jan 18 at 18:34
$begingroup$
Thank you for the hint. Is it because if I sum over all the measures of distinct $Lambda_{alpha}$ then I would get 0 since they are made up of points and the measure of a bunch of disjoint points is 0? But If I were to shift it in the right way I could get nonzero values... Sorry I am having a tough time with this problem.
$endgroup$
– MathIsHard
Jan 18 at 18:59