Mixing
Lebesgue outer measure regularity
$begingroup$
In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.
Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.
measure-theory lebesgue-measure outer-measure
edited Jan 10 at 16:22
amWhy
1
asked Jan 31 '17 at 15:10
ArindamArindam
338213
$endgroup$
add a comment |
$begingroup$
In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.
Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.
measure-theory lebesgue-measure outer-measure
edited Jan 10 at 16:22
amWhy
1
asked Jan 31 '17 at 15:10
ArindamArindam
338213
$endgroup$
add a comment |
$begingroup$
In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.
Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.
measure-theory lebesgue-measure outer-measure
edited Jan 10 at 16:22
amWhy
1
asked Jan 31 '17 at 15:10
ArindamArindam
338213
$endgroup$
In $mathbb{R}$, the outer regularity of Lebesgue measure gives $m^*(A)=inf{m(A)mid E text{ is open and } A subset E}$.
Can we replace $E$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.
measure-theory lebesgue-measure outer-measure
measure-theory lebesgue-measure outer-measure
edited Jan 10 at 16:22
amWhy
1
asked Jan 31 '17 at 15:10
ArindamArindam
338213
edited Jan 10 at 16:22
amWhy
1
asked Jan 31 '17 at 15:10
ArindamArindam
338213
edited Jan 10 at 16:22
amWhy
1
edited Jan 10 at 16:22
amWhy
1
edited Jan 10 at 16:22
amWhy
1
1
asked Jan 31 '17 at 15:10
ArindamArindam
338213
asked Jan 31 '17 at 15:10
ArindamArindam
338213
asked Jan 31 '17 at 15:10
ArindamArindam
338213
338213
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
$$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
This follows because the left infimum has a greater range.
By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.
Therefore, we have
$$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$
edited Jan 10 at 16:06
amWhy
1
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
$$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
This follows because the left infimum has a greater range.
By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.
Therefore, we have
$$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$
edited Jan 10 at 16:06
amWhy
1
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
$endgroup$
add a comment |
$begingroup$
You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
$$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
This follows because the left infimum has a greater range.
By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.
Therefore, we have
$$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$
edited Jan 10 at 16:06
amWhy
1
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
$endgroup$
add a comment |
$begingroup$
You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
$$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
This follows because the left infimum has a greater range.
By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.
Therefore, we have
$$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$
edited Jan 10 at 16:06
amWhy
1
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
$endgroup$
You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have
$$inf{m(E)mid E text{ is measurable},; A subseteq E} leq inf{m(E)mid E text{ is open}, ;Asubseteq E} = m^*(A).$$
This follows because the left infimum has a greater range.
By monotonicity, $m^{*}(A) leq m^{*}(E) = m(E)$ holds for all measurable $E$.
Therefore, we have
$$m^*(A) leq inf{m(E)mid E text{ is measurable}, A subseteq E}$$
edited Jan 10 at 16:06
amWhy
1
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
edited Jan 10 at 16:06
amWhy
1
edited Jan 10 at 16:06
amWhy
1
edited Jan 10 at 16:06
amWhy
1
1
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
answered Jan 31 '17 at 15:14
Henricus V.Henricus V.
15.1k22048
15.1k22048
add a comment |
add a comment |
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