Outer Measure of the complement of a Vitali Set in [0,1] equal to 1
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
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
262
add a comment |Â
up vote
5
down vote
favorite
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
262
1
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 |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
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
262
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
262
edited Sep 30 '12 at 23:08
user940
asked Sep 30 '12 at 22:16
joe
262
asked Sep 30 '12 at 22:16
joe
262
asked Sep 30 '12 at 22:16
joe
262
262
1
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 |Â
1
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
1
1
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 |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
add a comment |Â
up vote
0
down vote
Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f205076%2fouter-measure-of-the-complement-of-a-vitali-set-in-0-1-equal-to-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
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