Mixing
Proper Measurable subgroups of $mathbb R$
$begingroup$
If $(mathbb{R},+)$ is a group and $H$ is a proper subgroup of $mathbb{R}$ then prove that $H$ is of measure zero.
real-analysis
$endgroup$
add a comment |
$begingroup$
If $(mathbb{R},+)$ is a group and $H$ is a proper subgroup of $mathbb{R}$ then prove that $H$ is of measure zero.
real-analysis
$endgroup$
add a comment |
$begingroup$
If $(mathbb{R},+)$ is a group and $H$ is a proper subgroup of $mathbb{R}$ then prove that $H$ is of measure zero.
real-analysis
$endgroup$
If $(mathbb{R},+)$ is a group and $H$ is a proper subgroup of $mathbb{R}$ then prove that $H$ is of measure zero.
real-analysis
real-analysis
edited Jan 7 '13 at 5:38
asked Aug 4 '10 at 6:30
anonymous
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As was noted by Jason DeVito if H is measurable then measure of H is 0.
From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $mathbb{R}$ is a vector space over $mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $mathbb{R}$, generated by others. Then H is a subgroup of ($mathbb{R}$, +).
Lemma: H is not measurable.
Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, hin H}$, where $qin Q$, has measure 0 (because it is just a shift of H). But $mathbb{R}$ is equal to union of countable many sets with measure 0: $mathbb{R} = cup_{qinmathbb{Q}}(H+qe)$. Therefore $m(mathbb{R})=0$. We have come to a contradiction.
Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.
Lemma: If H is measurable proper subgroup of $mathbb{R}$, then m(H)=0.
If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = Hcap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=delta>0$. Let's take integer N, such that $delta N > z+1$ (we will see later, why).
Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x<1, such that $xnotin H$. Suppose y=x/N!. Then for $n=1,dots,N$ number ny obeys the following properties:
1. 1>ny>0.
2. ny is not in H.
Then sets $H_0, H_0+y, dots, H_0+(N-1)y$ are disjoint subsets of $[0, 1+z)$. Therefore,
$displaystyle 1+z = mBig( [0, 1+z) Big) geq mBigg(bigcup_{n=0}^{N-1} (H_0 + n y)Bigg) = N delta.$ Here we have a contradiction with definition of N.
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
$endgroup$
1
$begingroup$
-1: Too much detail for an answer to a homeworklike question.
$endgroup$
– Charles Stewart
Aug 9 '10 at 8:37
$begingroup$
I don't get why $Mbig([0,1+z)big) leq mleft(bigcuplimits_{n=0}^{N-1}(H_0 + ny)right) = N delta$ is true.
$endgroup$
– M.G
Feb 22 '16 at 17:21
$begingroup$
There was a mistake in my proof: this inequality should be $geq$, not $leq$ (fixed above now). Clarified now.
$endgroup$
– Fiktor
Jan 28 at 8:20
add a comment |
$begingroup$
Your title talks about "proper measurable subgroups of R", but the body of your post doesn't require that H be measurable. The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0. I'm not sure if every subgroup must be measurable or not....
Here is a rough outline of the proof:
Lemma 1: If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0.
Proof: See http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/
For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).
Lemma 2: If H is a subgroup of G, then H-H=H.
Lemma 3: If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
As was noted by Jason DeVito if H is measurable then measure of H is 0.
From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $mathbb{R}$ is a vector space over $mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $mathbb{R}$, generated by others. Then H is a subgroup of ($mathbb{R}$, +).
Lemma: H is not measurable.
Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, hin H}$, where $qin Q$, has measure 0 (because it is just a shift of H). But $mathbb{R}$ is equal to union of countable many sets with measure 0: $mathbb{R} = cup_{qinmathbb{Q}}(H+qe)$. Therefore $m(mathbb{R})=0$. We have come to a contradiction.
Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.
Lemma: If H is measurable proper subgroup of $mathbb{R}$, then m(H)=0.
If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = Hcap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=delta>0$. Let's take integer N, such that $delta N > z+1$ (we will see later, why).
Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x<1, such that $xnotin H$. Suppose y=x/N!. Then for $n=1,dots,N$ number ny obeys the following properties:
1. 1>ny>0.
2. ny is not in H.
Then sets $H_0, H_0+y, dots, H_0+(N-1)y$ are disjoint subsets of $[0, 1+z)$. Therefore,
$displaystyle 1+z = mBig( [0, 1+z) Big) geq mBigg(bigcup_{n=0}^{N-1} (H_0 + n y)Bigg) = N delta.$ Here we have a contradiction with definition of N.
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
$endgroup$
1
$begingroup$
-1: Too much detail for an answer to a homeworklike question.
$endgroup$
– Charles Stewart
Aug 9 '10 at 8:37
$begingroup$
I don't get why $Mbig([0,1+z)big) leq mleft(bigcuplimits_{n=0}^{N-1}(H_0 + ny)right) = N delta$ is true.
$endgroup$
– M.G
Feb 22 '16 at 17:21
$begingroup$
There was a mistake in my proof: this inequality should be $geq$, not $leq$ (fixed above now). Clarified now.
$endgroup$
– Fiktor
Jan 28 at 8:20
add a comment |
$begingroup$
As was noted by Jason DeVito if H is measurable then measure of H is 0.
From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $mathbb{R}$ is a vector space over $mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $mathbb{R}$, generated by others. Then H is a subgroup of ($mathbb{R}$, +).
Lemma: H is not measurable.
Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, hin H}$, where $qin Q$, has measure 0 (because it is just a shift of H). But $mathbb{R}$ is equal to union of countable many sets with measure 0: $mathbb{R} = cup_{qinmathbb{Q}}(H+qe)$. Therefore $m(mathbb{R})=0$. We have come to a contradiction.
Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.
Lemma: If H is measurable proper subgroup of $mathbb{R}$, then m(H)=0.
If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = Hcap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=delta>0$. Let's take integer N, such that $delta N > z+1$ (we will see later, why).
Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x<1, such that $xnotin H$. Suppose y=x/N!. Then for $n=1,dots,N$ number ny obeys the following properties:
1. 1>ny>0.
2. ny is not in H.
Then sets $H_0, H_0+y, dots, H_0+(N-1)y$ are disjoint subsets of $[0, 1+z)$. Therefore,
$displaystyle 1+z = mBig( [0, 1+z) Big) geq mBigg(bigcup_{n=0}^{N-1} (H_0 + n y)Bigg) = N delta.$ Here we have a contradiction with definition of N.
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
$endgroup$
1
$begingroup$
-1: Too much detail for an answer to a homeworklike question.
$endgroup$
– Charles Stewart
Aug 9 '10 at 8:37
$begingroup$
I don't get why $Mbig([0,1+z)big) leq mleft(bigcuplimits_{n=0}^{N-1}(H_0 + ny)right) = N delta$ is true.
$endgroup$
– M.G
Feb 22 '16 at 17:21
$begingroup$
There was a mistake in my proof: this inequality should be $geq$, not $leq$ (fixed above now). Clarified now.
$endgroup$
– Fiktor
Jan 28 at 8:20
add a comment |
$begingroup$
As was noted by Jason DeVito if H is measurable then measure of H is 0.
From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $mathbb{R}$ is a vector space over $mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $mathbb{R}$, generated by others. Then H is a subgroup of ($mathbb{R}$, +).
Lemma: H is not measurable.
Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, hin H}$, where $qin Q$, has measure 0 (because it is just a shift of H). But $mathbb{R}$ is equal to union of countable many sets with measure 0: $mathbb{R} = cup_{qinmathbb{Q}}(H+qe)$. Therefore $m(mathbb{R})=0$. We have come to a contradiction.
Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.
Lemma: If H is measurable proper subgroup of $mathbb{R}$, then m(H)=0.
If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = Hcap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=delta>0$. Let's take integer N, such that $delta N > z+1$ (we will see later, why).
Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x<1, such that $xnotin H$. Suppose y=x/N!. Then for $n=1,dots,N$ number ny obeys the following properties:
1. 1>ny>0.
2. ny is not in H.
Then sets $H_0, H_0+y, dots, H_0+(N-1)y$ are disjoint subsets of $[0, 1+z)$. Therefore,
$displaystyle 1+z = mBig( [0, 1+z) Big) geq mBigg(bigcup_{n=0}^{N-1} (H_0 + n y)Bigg) = N delta.$ Here we have a contradiction with definition of N.
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
$endgroup$
As was noted by Jason DeVito if H is measurable then measure of H is 0.
From the other hand, if we suppose, that the axiom of choice holds, there is a possibility, that H is not measurable. The proof is quite simple. $mathbb{R}$ is a vector space over $mathbb{Q}$. Therefore there is some basis. Suppose e is one of it's elements and H is a subspace in $mathbb{R}$, generated by others. Then H is a subgroup of ($mathbb{R}$, +).
Lemma: H is not measurable.
Suppose H is measurable. Then as was noted above, it's measure m(H)=0. Then every set of the form $H+q e = {h+qe, hin H}$, where $qin Q$, has measure 0 (because it is just a shift of H). But $mathbb{R}$ is equal to union of countable many sets with measure 0: $mathbb{R} = cup_{qinmathbb{Q}}(H+qe)$. Therefore $m(mathbb{R})=0$. We have come to a contradiction.
Also there is simple proof of the fact, that if H is measurable, then H is of measure 0.
Lemma: If H is measurable proper subgroup of $mathbb{R}$, then m(H)=0.
If H={0} then proposition of the lemma is obvious. Otherwise we can find positive element z in H. Suppose, $H_0 = Hcap [0,z)$. If $m(H_0)=0$ then $m(H)=0$. Otherwise $m(H_0)=delta>0$. Let's take integer N, such that $delta N > z+1$ (we will see later, why).
Note that if x is not in H, then x/n (for every positive integer n) and -x are also not in H. Therefore, using the fact that H is proper, we can find positive x<1, such that $xnotin H$. Suppose y=x/N!. Then for $n=1,dots,N$ number ny obeys the following properties:
1. 1>ny>0.
2. ny is not in H.
Then sets $H_0, H_0+y, dots, H_0+(N-1)y$ are disjoint subsets of $[0, 1+z)$. Therefore,
$displaystyle 1+z = mBig( [0, 1+z) Big) geq mBigg(bigcup_{n=0}^{N-1} (H_0 + n y)Bigg) = N delta.$ Here we have a contradiction with definition of N.
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
edited Jan 28 at 8:15
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
answered Aug 4 '10 at 11:46
FiktorFiktor
1,5281218
1,5281218
1
$begingroup$
-1: Too much detail for an answer to a homeworklike question.
$endgroup$
– Charles Stewart
Aug 9 '10 at 8:37
$begingroup$
I don't get why $Mbig([0,1+z)big) leq mleft(bigcuplimits_{n=0}^{N-1}(H_0 + ny)right) = N delta$ is true.
$endgroup$
– M.G
Feb 22 '16 at 17:21
$begingroup$
There was a mistake in my proof: this inequality should be $geq$, not $leq$ (fixed above now). Clarified now.
$endgroup$
– Fiktor
Jan 28 at 8:20
add a comment |
1
$begingroup$
-1: Too much detail for an answer to a homeworklike question.
$endgroup$
– Charles Stewart
Aug 9 '10 at 8:37
$begingroup$
I don't get why $Mbig([0,1+z)big) leq mleft(bigcuplimits_{n=0}^{N-1}(H_0 + ny)right) = N delta$ is true.
$endgroup$
– M.G
Feb 22 '16 at 17:21
$begingroup$
There was a mistake in my proof: this inequality should be $geq$, not $leq$ (fixed above now). Clarified now.
$endgroup$
– Fiktor
Jan 28 at 8:20
1
1
$begingroup$
-1: Too much detail for an answer to a homeworklike question.
$endgroup$
– Charles Stewart
Aug 9 '10 at 8:37
$begingroup$
-1: Too much detail for an answer to a homeworklike question.
$endgroup$
– Charles Stewart
Aug 9 '10 at 8:37
$begingroup$
I don't get why $Mbig([0,1+z)big) leq mleft(bigcuplimits_{n=0}^{N-1}(H_0 + ny)right) = N delta$ is true.
$endgroup$
– M.G
Feb 22 '16 at 17:21
$begingroup$
I don't get why $Mbig([0,1+z)big) leq mleft(bigcuplimits_{n=0}^{N-1}(H_0 + ny)right) = N delta$ is true.
$endgroup$
– M.G
Feb 22 '16 at 17:21
$begingroup$
There was a mistake in my proof: this inequality should be $geq$, not $leq$ (fixed above now). Clarified now.
$endgroup$
– Fiktor
Jan 28 at 8:20
$begingroup$
There was a mistake in my proof: this inequality should be $geq$, not $leq$ (fixed above now). Clarified now.
$endgroup$
– Fiktor
Jan 28 at 8:20
add a comment |
$begingroup$
Your title talks about "proper measurable subgroups of R", but the body of your post doesn't require that H be measurable. The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0. I'm not sure if every subgroup must be measurable or not....
Here is a rough outline of the proof:
Lemma 1: If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0.
Proof: See http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/
For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).
Lemma 2: If H is a subgroup of G, then H-H=H.
Lemma 3: If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
$endgroup$
add a comment |
$begingroup$
Your title talks about "proper measurable subgroups of R", but the body of your post doesn't require that H be measurable. The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0. I'm not sure if every subgroup must be measurable or not....
Here is a rough outline of the proof:
Lemma 1: If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0.
Proof: See http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/
For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).
Lemma 2: If H is a subgroup of G, then H-H=H.
Lemma 3: If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
$endgroup$
add a comment |
$begingroup$
Your title talks about "proper measurable subgroups of R", but the body of your post doesn't require that H be measurable. The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0. I'm not sure if every subgroup must be measurable or not....
Here is a rough outline of the proof:
Lemma 1: If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0.
Proof: See http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/
For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).
Lemma 2: If H is a subgroup of G, then H-H=H.
Lemma 3: If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
$endgroup$
Your title talks about "proper measurable subgroups of R", but the body of your post doesn't require that H be measurable. The following outline shows that if H is a proper subgroup of R and is measurable, then it must be measure 0. I'm not sure if every subgroup must be measurable or not....
Here is a rough outline of the proof:
Lemma 1: If H is a subset of R and has positive measure, then H-H = {a-b| a,b, in H} contains an interval around 0.
Proof: See http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/
For Lemma 2 and 3, I'll leave the proofs to you (but I can add details if you need them).
Lemma 2: If H is a subgroup of G, then H-H=H.
Lemma 3: If H is a subgroup of the real numbers R and contains an interval around 0, then H = R.
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
answered Aug 4 '10 at 6:57
Jason DeVitoJason DeVito
31.1k475136
31.1k475136
add a comment |
add a comment |
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