Mixing
Property regarding Lebesgue Outer Measure
$begingroup$
In real variables by Michael Spigel on page 34, the author says and proves that if $E$ is any given set contained in an open set $O$, then the Lebesgue outer measure of $E$ is less than Lebesgue outer measure of $O$ plus any arbitrary positive $varepsilon$.
However, on page 35 in proof of problem 2.4, it says that the Lebesgue outer measure of an open set $O_k$ is less than the Lebesgue outer measure of $E_k + frac{varepsilon}{2^k}$.
How can both be true?
measure-theory lebesgue-measure
edited Jan 10 at 16:17
amWhy
1
asked Mar 31 '16 at 22:49
saqibsaqib
383
$endgroup$
add a comment |
$begingroup$
In real variables by Michael Spigel on page 34, the author says and proves that if $E$ is any given set contained in an open set $O$, then the Lebesgue outer measure of $E$ is less than Lebesgue outer measure of $O$ plus any arbitrary positive $varepsilon$.
However, on page 35 in proof of problem 2.4, it says that the Lebesgue outer measure of an open set $O_k$ is less than the Lebesgue outer measure of $E_k + frac{varepsilon}{2^k}$.
How can both be true?
measure-theory lebesgue-measure
edited Jan 10 at 16:17
amWhy
1
asked Mar 31 '16 at 22:49
saqibsaqib
383
$endgroup$
1
$begingroup$
Please use mathjax.
$endgroup$
– Henricus V.
Mar 31 '16 at 22:52
1
$begingroup$
$varepsilon / 2^k$ is arbitrary as much as $varepsilon$ is. Aren't you familiar with arbitrary positive real numbers?
$endgroup$
– Crostul
Mar 31 '16 at 22:58
add a comment |
$begingroup$
In real variables by Michael Spigel on page 34, the author says and proves that if $E$ is any given set contained in an open set $O$, then the Lebesgue outer measure of $E$ is less than Lebesgue outer measure of $O$ plus any arbitrary positive $varepsilon$.
However, on page 35 in proof of problem 2.4, it says that the Lebesgue outer measure of an open set $O_k$ is less than the Lebesgue outer measure of $E_k + frac{varepsilon}{2^k}$.
How can both be true?
measure-theory lebesgue-measure
edited Jan 10 at 16:17
amWhy
1
asked Mar 31 '16 at 22:49
saqibsaqib
383
$endgroup$
In real variables by Michael Spigel on page 34, the author says and proves that if $E$ is any given set contained in an open set $O$, then the Lebesgue outer measure of $E$ is less than Lebesgue outer measure of $O$ plus any arbitrary positive $varepsilon$.
However, on page 35 in proof of problem 2.4, it says that the Lebesgue outer measure of an open set $O_k$ is less than the Lebesgue outer measure of $E_k + frac{varepsilon}{2^k}$.
How can both be true?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Jan 10 at 16:17
amWhy
1
asked Mar 31 '16 at 22:49
saqibsaqib
383
edited Jan 10 at 16:17
amWhy
1
asked Mar 31 '16 at 22:49
saqibsaqib
383
edited Jan 10 at 16:17
amWhy
1
edited Jan 10 at 16:17
amWhy
1
edited Jan 10 at 16:17
amWhy
1
1
asked Mar 31 '16 at 22:49
saqibsaqib
383
asked Mar 31 '16 at 22:49
saqibsaqib
383
asked Mar 31 '16 at 22:49
saqibsaqib
383
383
1
$begingroup$
Please use mathjax.
$endgroup$
– Henricus V.
Mar 31 '16 at 22:52
1
$begingroup$
$varepsilon / 2^k$ is arbitrary as much as $varepsilon$ is. Aren't you familiar with arbitrary positive real numbers?
$endgroup$
– Crostul
Mar 31 '16 at 22:58
add a comment |
1
$begingroup$
Please use mathjax.
$endgroup$
– Henricus V.
Mar 31 '16 at 22:52
1
$begingroup$
$varepsilon / 2^k$ is arbitrary as much as $varepsilon$ is. Aren't you familiar with arbitrary positive real numbers?
$endgroup$
– Crostul
Mar 31 '16 at 22:58
1
1
$begingroup$
Please use mathjax.
$endgroup$
– Henricus V.
Mar 31 '16 at 22:52
$begingroup$
Please use mathjax.
$endgroup$
– Henricus V.
Mar 31 '16 at 22:52
1
1
$begingroup$
$varepsilon / 2^k$ is arbitrary as much as $varepsilon$ is. Aren't you familiar with arbitrary positive real numbers?
$endgroup$
– Crostul
Mar 31 '16 at 22:58
$begingroup$
$varepsilon / 2^k$ is arbitrary as much as $varepsilon$ is. Aren't you familiar with arbitrary positive real numbers?
$endgroup$
– Crostul
Mar 31 '16 at 22:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As shown in Problem 2.3, there is an open set $O_k$ satisfying both inequalities, that is,
$$
m(E_k)le m(O_k)le m(E_k)+frac1{2^k},
$$
where $m$ is the Lebesgue outer measure.
But not all open sets $O_k$ satisfy both inequalities!
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
$endgroup$
$begingroup$
but isn't this contradictory? . How can both inequalities be satisfied if you have strict inequality
$endgroup$
– saqib
Mar 31 '16 at 23:15
add a comment |
$begingroup$
You're mixing up some different ideas. It is true that if $E subseteq O$, then $m^*(E) leq O$. More carefully, $m^*(E) = inf{ U mid E subseteq U , textrm{for open}, U}$. Consequently, you can always find some open $U$ so that $m(E_k) leq m(U) leq m^*(E_k)+frac{epsilon}{2^k}$.
It is true that if $E$ is measurable, then there exists some open $U$ so that $m^*(U-E)<epsilon$ etc.
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
$endgroup$
$begingroup$
measure of U is length of U and which is also Lebesgue outer measure of U. if both inequalities hold then isn't this contradictory?
$endgroup$
– saqib
Mar 31 '16 at 23:16
$begingroup$
the measure of $U$ is the length of disjoint open intervals that make up $U$
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:17
$begingroup$
They're both saying that the measure open covers come pretty close to the measure of $E$.
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:18
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As shown in Problem 2.3, there is an open set $O_k$ satisfying both inequalities, that is,
$$
m(E_k)le m(O_k)le m(E_k)+frac1{2^k},
$$
where $m$ is the Lebesgue outer measure.
But not all open sets $O_k$ satisfy both inequalities!
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
$endgroup$
$begingroup$
but isn't this contradictory? . How can both inequalities be satisfied if you have strict inequality
$endgroup$
– saqib
Mar 31 '16 at 23:15
add a comment |
$begingroup$
As shown in Problem 2.3, there is an open set $O_k$ satisfying both inequalities, that is,
$$
m(E_k)le m(O_k)le m(E_k)+frac1{2^k},
$$
where $m$ is the Lebesgue outer measure.
But not all open sets $O_k$ satisfy both inequalities!
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
$endgroup$
$begingroup$
but isn't this contradictory? . How can both inequalities be satisfied if you have strict inequality
$endgroup$
– saqib
Mar 31 '16 at 23:15
add a comment |
$begingroup$
As shown in Problem 2.3, there is an open set $O_k$ satisfying both inequalities, that is,
$$
m(E_k)le m(O_k)le m(E_k)+frac1{2^k},
$$
where $m$ is the Lebesgue outer measure.
But not all open sets $O_k$ satisfy both inequalities!
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
$endgroup$
As shown in Problem 2.3, there is an open set $O_k$ satisfying both inequalities, that is,
$$
m(E_k)le m(O_k)le m(E_k)+frac1{2^k},
$$
where $m$ is the Lebesgue outer measure.
But not all open sets $O_k$ satisfy both inequalities!
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
answered Mar 31 '16 at 22:59
John BJohn B
12.2k51840
12.2k51840
$begingroup$
but isn't this contradictory? . How can both inequalities be satisfied if you have strict inequality
$endgroup$
– saqib
Mar 31 '16 at 23:15
add a comment |
$begingroup$
but isn't this contradictory? . How can both inequalities be satisfied if you have strict inequality
$endgroup$
– saqib
Mar 31 '16 at 23:15
$begingroup$
but isn't this contradictory? . How can both inequalities be satisfied if you have strict inequality
$endgroup$
– saqib
Mar 31 '16 at 23:15
$begingroup$
but isn't this contradictory? . How can both inequalities be satisfied if you have strict inequality
$endgroup$
– saqib
Mar 31 '16 at 23:15
add a comment |
$begingroup$
You're mixing up some different ideas. It is true that if $E subseteq O$, then $m^*(E) leq O$. More carefully, $m^*(E) = inf{ U mid E subseteq U , textrm{for open}, U}$. Consequently, you can always find some open $U$ so that $m(E_k) leq m(U) leq m^*(E_k)+frac{epsilon}{2^k}$.
It is true that if $E$ is measurable, then there exists some open $U$ so that $m^*(U-E)<epsilon$ etc.
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
$endgroup$
$begingroup$
measure of U is length of U and which is also Lebesgue outer measure of U. if both inequalities hold then isn't this contradictory?
$endgroup$
– saqib
Mar 31 '16 at 23:16
$begingroup$
the measure of $U$ is the length of disjoint open intervals that make up $U$
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:17
$begingroup$
They're both saying that the measure open covers come pretty close to the measure of $E$.
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:18
add a comment |
$begingroup$
You're mixing up some different ideas. It is true that if $E subseteq O$, then $m^*(E) leq O$. More carefully, $m^*(E) = inf{ U mid E subseteq U , textrm{for open}, U}$. Consequently, you can always find some open $U$ so that $m(E_k) leq m(U) leq m^*(E_k)+frac{epsilon}{2^k}$.
It is true that if $E$ is measurable, then there exists some open $U$ so that $m^*(U-E)<epsilon$ etc.
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
$endgroup$
$begingroup$
measure of U is length of U and which is also Lebesgue outer measure of U. if both inequalities hold then isn't this contradictory?
$endgroup$
– saqib
Mar 31 '16 at 23:16
$begingroup$
the measure of $U$ is the length of disjoint open intervals that make up $U$
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:17
$begingroup$
They're both saying that the measure open covers come pretty close to the measure of $E$.
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:18
add a comment |
$begingroup$
You're mixing up some different ideas. It is true that if $E subseteq O$, then $m^*(E) leq O$. More carefully, $m^*(E) = inf{ U mid E subseteq U , textrm{for open}, U}$. Consequently, you can always find some open $U$ so that $m(E_k) leq m(U) leq m^*(E_k)+frac{epsilon}{2^k}$.
It is true that if $E$ is measurable, then there exists some open $U$ so that $m^*(U-E)<epsilon$ etc.
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
$endgroup$
You're mixing up some different ideas. It is true that if $E subseteq O$, then $m^*(E) leq O$. More carefully, $m^*(E) = inf{ U mid E subseteq U , textrm{for open}, U}$. Consequently, you can always find some open $U$ so that $m(E_k) leq m(U) leq m^*(E_k)+frac{epsilon}{2^k}$.
It is true that if $E$ is measurable, then there exists some open $U$ so that $m^*(U-E)<epsilon$ etc.
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
edited Apr 3 '16 at 18:27
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
answered Mar 31 '16 at 22:58
Andres MejiaAndres Mejia
16.1k21548
16.1k21548
$begingroup$
measure of U is length of U and which is also Lebesgue outer measure of U. if both inequalities hold then isn't this contradictory?
$endgroup$
– saqib
Mar 31 '16 at 23:16
$begingroup$
the measure of $U$ is the length of disjoint open intervals that make up $U$
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:17
$begingroup$
They're both saying that the measure open covers come pretty close to the measure of $E$.
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:18
add a comment |
$begingroup$
measure of U is length of U and which is also Lebesgue outer measure of U. if both inequalities hold then isn't this contradictory?
$endgroup$
– saqib
Mar 31 '16 at 23:16
$begingroup$
the measure of $U$ is the length of disjoint open intervals that make up $U$
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:17
$begingroup$
They're both saying that the measure open covers come pretty close to the measure of $E$.
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:18
$begingroup$
measure of U is length of U and which is also Lebesgue outer measure of U. if both inequalities hold then isn't this contradictory?
$endgroup$
– saqib
Mar 31 '16 at 23:16
$begingroup$
measure of U is length of U and which is also Lebesgue outer measure of U. if both inequalities hold then isn't this contradictory?
$endgroup$
– saqib
Mar 31 '16 at 23:16
$begingroup$
the measure of $U$ is the length of disjoint open intervals that make up $U$
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:17
$begingroup$
the measure of $U$ is the length of disjoint open intervals that make up $U$
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:17
$begingroup$
They're both saying that the measure open covers come pretty close to the measure of $E$.
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:18
$begingroup$
They're both saying that the measure open covers come pretty close to the measure of $E$.
$endgroup$
– Andres Mejia
Mar 31 '16 at 23:18
add a comment |
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1
$begingroup$
Please use mathjax.
$endgroup$
– Henricus V.
Mar 31 '16 at 22:52
1
$begingroup$
$varepsilon / 2^k$ is arbitrary as much as $varepsilon$ is. Aren't you familiar with arbitrary positive real numbers?
$endgroup$
– Crostul
Mar 31 '16 at 22:58