Outer Measure of the complement of a Vitali Set in [0,1] equal to 1
I am trying to prove the first part of exercise 33, ch. 1 in Stein and Shakarchi (Real Analysis). I am running into some difficulties following the hint though. Here is the problem (note, $N$ is a Vitali set constructed in $[0,1]$):
Show that the set $[0,1]-N$ has outer measure $m_*(N^c)=1$. [Hint: argue by contradiction, and pick a measurable set such that $N^c subset U subset [0,1]$ and $m_*(U) le 1-epsilon$.
I know that both $N$ and its complement are not measurable, so neither are countable. I know that measurable subsets of non-measurable sets have measure 0. I am not sure how to proceed given the proof though. A point to note: the book does not work with the inner measure at all, and even though I have used the inner measure in a previous course, I do not think I am allowed to for this proof.
real-analysis measure-theory
asked Sep 30 '12 at 22:16
joe
312
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I am trying to prove the first part of exercise 33, ch. 1 in Stein and Shakarchi (Real Analysis). I am running into some difficulties following the hint though. Here is the problem (note, $N$ is a Vitali set constructed in $[0,1]$):
Show that the set $[0,1]-N$ has outer measure $m_*(N^c)=1$. [Hint: argue by contradiction, and pick a measurable set such that $N^c subset U subset [0,1]$ and $m_*(U) le 1-epsilon$.
I know that both $N$ and its complement are not measurable, so neither are countable. I know that measurable subsets of non-measurable sets have measure 0. I am not sure how to proceed given the proof though. A point to note: the book does not work with the inner measure at all, and even though I have used the inner measure in a previous course, I do not think I am allowed to for this proof.
real-analysis measure-theory
asked Sep 30 '12 at 22:16
joe
312
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
2
You wrote something false. Take a Vitali set in $[0,frac12]$, union the interval $[frac12,1]$. This set is certainly non-measurable, but it has a measurable subset of measure $frac12$. It is true, however, that the Vitali set has inner measure zero.
– Asaf Karagila♦
Sep 30 '12 at 22:30
One way to see that Vitali set have inner measure zero is Lemma 2 in here
– leo
Oct 1 '12 at 22:56
add a comment |
I am trying to prove the first part of exercise 33, ch. 1 in Stein and Shakarchi (Real Analysis). I am running into some difficulties following the hint though. Here is the problem (note, $N$ is a Vitali set constructed in $[0,1]$):
Show that the set $[0,1]-N$ has outer measure $m_*(N^c)=1$. [Hint: argue by contradiction, and pick a measurable set such that $N^c subset U subset [0,1]$ and $m_*(U) le 1-epsilon$.
I know that both $N$ and its complement are not measurable, so neither are countable. I know that measurable subsets of non-measurable sets have measure 0. I am not sure how to proceed given the proof though. A point to note: the book does not work with the inner measure at all, and even though I have used the inner measure in a previous course, I do not think I am allowed to for this proof.
real-analysis measure-theory
asked Sep 30 '12 at 22:16
joe
312
I am trying to prove the first part of exercise 33, ch. 1 in Stein and Shakarchi (Real Analysis). I am running into some difficulties following the hint though. Here is the problem (note, $N$ is a Vitali set constructed in $[0,1]$):
Show that the set $[0,1]-N$ has outer measure $m_*(N^c)=1$. [Hint: argue by contradiction, and pick a measurable set such that $N^c subset U subset [0,1]$ and $m_*(U) le 1-epsilon$.
I know that both $N$ and its complement are not measurable, so neither are countable. I know that measurable subsets of non-measurable sets have measure 0. I am not sure how to proceed given the proof though. A point to note: the book does not work with the inner measure at all, and even though I have used the inner measure in a previous course, I do not think I am allowed to for this proof.
real-analysis measure-theory
real-analysis measure-theory
asked Sep 30 '12 at 22:16
joe
312
asked Sep 30 '12 at 22:16
joe
312
edited Sep 30 '12 at 23:08
user940
asked Sep 30 '12 at 22:16
joe
312
asked Sep 30 '12 at 22:16
joe
312
asked Sep 30 '12 at 22:16
joe
312
312
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
2
You wrote something false. Take a Vitali set in $[0,frac12]$, union the interval $[frac12,1]$. This set is certainly non-measurable, but it has a measurable subset of measure $frac12$. It is true, however, that the Vitali set has inner measure zero.
– Asaf Karagila♦
Sep 30 '12 at 22:30
One way to see that Vitali set have inner measure zero is Lemma 2 in here
– leo
Oct 1 '12 at 22:56
add a comment |
2
You wrote something false. Take a Vitali set in $[0,frac12]$, union the interval $[frac12,1]$. This set is certainly non-measurable, but it has a measurable subset of measure $frac12$. It is true, however, that the Vitali set has inner measure zero.
– Asaf Karagila♦
Sep 30 '12 at 22:30
One way to see that Vitali set have inner measure zero is Lemma 2 in here
– leo
Oct 1 '12 at 22:56
2
2
You wrote something false. Take a Vitali set in $[0,frac12]$, union the interval $[frac12,1]$. This set is certainly non-measurable, but it has a measurable subset of measure $frac12$. It is true, however, that the Vitali set has inner measure zero.
– Asaf Karagila♦
Sep 30 '12 at 22:30
You wrote something false. Take a Vitali set in $[0,frac12]$, union the interval $[frac12,1]$. This set is certainly non-measurable, but it has a measurable subset of measure $frac12$. It is true, however, that the Vitali set has inner measure zero.
– Asaf Karagila♦
Sep 30 '12 at 22:30
One way to see that Vitali set have inner measure zero is Lemma 2 in here
– leo
Oct 1 '12 at 22:56
One way to see that Vitali set have inner measure zero is Lemma 2 in here
– leo
Oct 1 '12 at 22:56
add a comment |
1 Answer
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Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
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1 Answer
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Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
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Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
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Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
answered Sep 30 '12 at 23:11
user940
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2
You wrote something false. Take a Vitali set in $[0,frac12]$, union the interval $[frac12,1]$. This set is certainly non-measurable, but it has a measurable subset of measure $frac12$. It is true, however, that the Vitali set has inner measure zero.
– Asaf Karagila♦
Sep 30 '12 at 22:30
One way to see that Vitali set have inner measure zero is Lemma 2 in here
– leo
Oct 1 '12 at 22:56