Outer Measure of the complement of a Vitali Set in [0,1] equal to 1
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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
joejoe
312
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add a comment |
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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
joejoe
312
$endgroup$
2
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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.
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– Asaf Karagila♦
Sep 30 '12 at 22:30
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One way to see that Vitali set have inner measure zero is Lemma 2 in here
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– leo
Oct 1 '12 at 22:56
add a comment |
$begingroup$
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
joejoe
312
$endgroup$
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
joejoe
312
asked Sep 30 '12 at 22:16
joejoe
312
edited Sep 30 '12 at 23:08
user940
asked Sep 30 '12 at 22:16
joejoe
312
asked Sep 30 '12 at 22:16
joejoe
312
asked Sep 30 '12 at 22:16
joejoe
312
312
2
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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.
$endgroup$
– Asaf Karagila♦
Sep 30 '12 at 22:30
$begingroup$
One way to see that Vitali set have inner measure zero is Lemma 2 in here
$endgroup$
– leo
Oct 1 '12 at 22:56
add a comment |
2
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
Sep 30 '12 at 22:30
$begingroup$
One way to see that Vitali set have inner measure zero is Lemma 2 in here
$endgroup$
– leo
Oct 1 '12 at 22:56
2
2
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
Sep 30 '12 at 22:30
$begingroup$
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.
$endgroup$
– Asaf Karagila♦
Sep 30 '12 at 22:30
$begingroup$
One way to see that Vitali set have inner measure zero is Lemma 2 in here
$endgroup$
– leo
Oct 1 '12 at 22:56
$begingroup$
One way to see that Vitali set have inner measure zero is Lemma 2 in here
$endgroup$
– 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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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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add a comment |
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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, $m_*(I)=m_*(U)+m_*(U^c).$
answered Sep 30 '12 at 23:11
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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.
$endgroup$
– Asaf Karagila♦
Sep 30 '12 at 22:30
$begingroup$
One way to see that Vitali set have inner measure zero is Lemma 2 in here
$endgroup$
– leo
Oct 1 '12 at 22:56