Vitali Covering
Clash Royale CLAN TAG#URR8PPP
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Let $E$ be a set of finite outer measure and $F$ a collection of closed, bounded intervals that cover $E$ in the sense of Vitali. Show that there is a countable disjoint collection $I_k$$_k=1^infty$ of intervals in $F$ for which
$m$*$[E$~$bigcup_k=1^inftyI_k] = 0$.
This is the same as the Vitali Covering Lemma except for each $epsilon > 0$, $m$*$[E$~$bigcup_k=1^inftyI_k] < epsilon$. In that proof, my book (Royden, 4th) uses
Theorem 11: Let $E$ be any set of real numbers. Then, each of the following four assertions is equivalent to the measurability of E.
i) For each $epsilon > 0$, there is an open set $O$ containing $E$ for which $m$*$(O$~$E)<epsilon$.
ii) There is a $G_delta$ set $G$ containing $E$ for which $m$*$(G$~$E)=0$.
iii), iv) etc.
Would this proof boil down to i) -> ii)?
measure-theory
asked Dec 9 '12 at 18:31
Jake Casey
398216
add a comment |Â
up vote
4
down vote
favorite
Let $E$ be a set of finite outer measure and $F$ a collection of closed, bounded intervals that cover $E$ in the sense of Vitali. Show that there is a countable disjoint collection $I_k$$_k=1^infty$ of intervals in $F$ for which
$m$*$[E$~$bigcup_k=1^inftyI_k] = 0$.
This is the same as the Vitali Covering Lemma except for each $epsilon > 0$, $m$*$[E$~$bigcup_k=1^inftyI_k] < epsilon$. In that proof, my book (Royden, 4th) uses
Theorem 11: Let $E$ be any set of real numbers. Then, each of the following four assertions is equivalent to the measurability of E.
i) For each $epsilon > 0$, there is an open set $O$ containing $E$ for which $m$*$(O$~$E)<epsilon$.
ii) There is a $G_delta$ set $G$ containing $E$ for which $m$*$(G$~$E)=0$.
iii), iv) etc.
Would this proof boil down to i) -> ii)?
measure-theory
asked Dec 9 '12 at 18:31
Jake Casey
398216
1
Note that in the Vitali covering theorem $E$ need not be measurable. Also, Royden's "Vitali Covering Lemma" yields a finite collection of intervals.
– T. Eskin
Dec 9 '12 at 20:43
From my book: The Vitali Covering Lemma - Let E be a set of finite outer measure and F a collection of closed, bounded intervals that covers E in the sense of Vitali. Then for each epsilon>0, there is a finite disjoint subcollection, Ik's, of F for which etc. How are they different?
– Jake Casey
Dec 10 '12 at 20:21
In the Vitali Covering Lemma, the collection of intervals obtained is finite, while in this exercise, it is countably infinite. Note also that we only know $mu^*(E)<infty$, not necessarily that $E$ is measurable.
– T. Eskin
Dec 10 '12 at 22:16
Ok, so what would be a good way to approach this problem?
– Jake Casey
Dec 10 '12 at 22:23
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $E$ be a set of finite outer measure and $F$ a collection of closed, bounded intervals that cover $E$ in the sense of Vitali. Show that there is a countable disjoint collection $I_k$$_k=1^infty$ of intervals in $F$ for which
$m$*$[E$~$bigcup_k=1^inftyI_k] = 0$.
This is the same as the Vitali Covering Lemma except for each $epsilon > 0$, $m$*$[E$~$bigcup_k=1^inftyI_k] < epsilon$. In that proof, my book (Royden, 4th) uses
Theorem 11: Let $E$ be any set of real numbers. Then, each of the following four assertions is equivalent to the measurability of E.
i) For each $epsilon > 0$, there is an open set $O$ containing $E$ for which $m$*$(O$~$E)<epsilon$.
ii) There is a $G_delta$ set $G$ containing $E$ for which $m$*$(G$~$E)=0$.
iii), iv) etc.
Would this proof boil down to i) -> ii)?
measure-theory
asked Dec 9 '12 at 18:31
Jake Casey
398216
Let $E$ be a set of finite outer measure and $F$ a collection of closed, bounded intervals that cover $E$ in the sense of Vitali. Show that there is a countable disjoint collection $I_k$$_k=1^infty$ of intervals in $F$ for which
$m$*$[E$~$bigcup_k=1^inftyI_k] = 0$.
This is the same as the Vitali Covering Lemma except for each $epsilon > 0$, $m$*$[E$~$bigcup_k=1^inftyI_k] < epsilon$. In that proof, my book (Royden, 4th) uses
Theorem 11: Let $E$ be any set of real numbers. Then, each of the following four assertions is equivalent to the measurability of E.
i) For each $epsilon > 0$, there is an open set $O$ containing $E$ for which $m$*$(O$~$E)<epsilon$.
ii) There is a $G_delta$ set $G$ containing $E$ for which $m$*$(G$~$E)=0$.
iii), iv) etc.
Would this proof boil down to i) -> ii)?
measure-theory
measure-theory
asked Dec 9 '12 at 18:31
Jake Casey
398216
asked Dec 9 '12 at 18:31
Jake Casey
398216
asked Dec 9 '12 at 18:31
Jake Casey
398216
asked Dec 9 '12 at 18:31
Jake Casey
398216
asked Dec 9 '12 at 18:31
Jake Casey
398216
398216
1
Note that in the Vitali covering theorem $E$ need not be measurable. Also, Royden's "Vitali Covering Lemma" yields a finite collection of intervals.
– T. Eskin
Dec 9 '12 at 20:43
From my book: The Vitali Covering Lemma - Let E be a set of finite outer measure and F a collection of closed, bounded intervals that covers E in the sense of Vitali. Then for each epsilon>0, there is a finite disjoint subcollection, Ik's, of F for which etc. How are they different?
– Jake Casey
Dec 10 '12 at 20:21
In the Vitali Covering Lemma, the collection of intervals obtained is finite, while in this exercise, it is countably infinite. Note also that we only know $mu^*(E)<infty$, not necessarily that $E$ is measurable.
– T. Eskin
Dec 10 '12 at 22:16
Ok, so what would be a good way to approach this problem?
– Jake Casey
Dec 10 '12 at 22:23
add a comment |Â
1
Note that in the Vitali covering theorem $E$ need not be measurable. Also, Royden's "Vitali Covering Lemma" yields a finite collection of intervals.
– T. Eskin
Dec 9 '12 at 20:43
From my book: The Vitali Covering Lemma - Let E be a set of finite outer measure and F a collection of closed, bounded intervals that covers E in the sense of Vitali. Then for each epsilon>0, there is a finite disjoint subcollection, Ik's, of F for which etc. How are they different?
– Jake Casey
Dec 10 '12 at 20:21
In the Vitali Covering Lemma, the collection of intervals obtained is finite, while in this exercise, it is countably infinite. Note also that we only know $mu^*(E)<infty$, not necessarily that $E$ is measurable.
– T. Eskin
Dec 10 '12 at 22:16
Ok, so what would be a good way to approach this problem?
– Jake Casey
Dec 10 '12 at 22:23
1
1
Note that in the Vitali covering theorem $E$ need not be measurable. Also, Royden's "Vitali Covering Lemma" yields a finite collection of intervals.
– T. Eskin
Dec 9 '12 at 20:43
Note that in the Vitali covering theorem $E$ need not be measurable. Also, Royden's "Vitali Covering Lemma" yields a finite collection of intervals.
– T. Eskin
Dec 9 '12 at 20:43
From my book: The Vitali Covering Lemma - Let E be a set of finite outer measure and F a collection of closed, bounded intervals that covers E in the sense of Vitali. Then for each epsilon>0, there is a finite disjoint subcollection, Ik's, of F for which etc. How are they different?
– Jake Casey
Dec 10 '12 at 20:21
From my book: The Vitali Covering Lemma - Let E be a set of finite outer measure and F a collection of closed, bounded intervals that covers E in the sense of Vitali. Then for each epsilon>0, there is a finite disjoint subcollection, Ik's, of F for which etc. How are they different?
– Jake Casey
Dec 10 '12 at 20:21
In the Vitali Covering Lemma, the collection of intervals obtained is finite, while in this exercise, it is countably infinite. Note also that we only know $mu^*(E)<infty$, not necessarily that $E$ is measurable.
– T. Eskin
Dec 10 '12 at 22:16
In the Vitali Covering Lemma, the collection of intervals obtained is finite, while in this exercise, it is countably infinite. Note also that we only know $mu^*(E)<infty$, not necessarily that $E$ is measurable.
– T. Eskin
Dec 10 '12 at 22:16
Ok, so what would be a good way to approach this problem?
– Jake Casey
Dec 10 '12 at 22:23
Ok, so what would be a good way to approach this problem?
– Jake Casey
Dec 10 '12 at 22:23
add a comment |Â
1 Answer
1
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Let E be a set of finite outer measure and :F a collection of
closed, bounded intervals that covers E in the sense of Vitali. Then for each E > 0, there is a
finite disjoint subcollection (Ik}k=1 of :F for which
m* [E"- U It] < E.
k=1
(2)
P
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
Welcome to MSE. Please format your answer using MathJax.
– user507623
Jan 1 at 10:00
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
Let E be a set of finite outer measure and :F a collection of
closed, bounded intervals that covers E in the sense of Vitali. Then for each E > 0, there is a
finite disjoint subcollection (Ik}k=1 of :F for which
m* [E"- U It] < E.
k=1
(2)
P
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
Welcome to MSE. Please format your answer using MathJax.
– user507623
Jan 1 at 10:00
add a comment |Â
up vote
0
down vote
Let E be a set of finite outer measure and :F a collection of
closed, bounded intervals that covers E in the sense of Vitali. Then for each E > 0, there is a
finite disjoint subcollection (Ik}k=1 of :F for which
m* [E"- U It] < E.
k=1
(2)
P
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
Welcome to MSE. Please format your answer using MathJax.
– user507623
Jan 1 at 10:00
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let E be a set of finite outer measure and :F a collection of
closed, bounded intervals that covers E in the sense of Vitali. Then for each E > 0, there is a
finite disjoint subcollection (Ik}k=1 of :F for which
m* [E"- U It] < E.
k=1
(2)
P
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
Let E be a set of finite outer measure and :F a collection of
closed, bounded intervals that covers E in the sense of Vitali. Then for each E > 0, there is a
finite disjoint subcollection (Ik}k=1 of :F for which
m* [E"- U It] < E.
k=1
(2)
P
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
answered Jan 1 at 9:41
Faruk Ibrahim Garko
1
1
Welcome to MSE. Please format your answer using MathJax.
– user507623
Jan 1 at 10:00
add a comment |Â
Welcome to MSE. Please format your answer using MathJax.
– user507623
Jan 1 at 10:00
Welcome to MSE. Please format your answer using MathJax.
– user507623
Jan 1 at 10:00
Welcome to MSE. Please format your answer using MathJax.
– user507623
Jan 1 at 10:00
add a comment |Â
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1
Note that in the Vitali covering theorem $E$ need not be measurable. Also, Royden's "Vitali Covering Lemma" yields a finite collection of intervals.
– T. Eskin
Dec 9 '12 at 20:43
From my book: The Vitali Covering Lemma - Let E be a set of finite outer measure and F a collection of closed, bounded intervals that covers E in the sense of Vitali. Then for each epsilon>0, there is a finite disjoint subcollection, Ik's, of F for which etc. How are they different?
– Jake Casey
Dec 10 '12 at 20:21
In the Vitali Covering Lemma, the collection of intervals obtained is finite, while in this exercise, it is countably infinite. Note also that we only know $mu^*(E)<infty$, not necessarily that $E$ is measurable.
– T. Eskin
Dec 10 '12 at 22:16
Ok, so what would be a good way to approach this problem?
– Jake Casey
Dec 10 '12 at 22:23