Lebesgue Measurable via open set
$begingroup$
I am a noob at measure theory so please forgive me for being naive, I just need help, here is what I have albeit this is a very very standard result:
Let $m$ denote Lebesgue measure on $mathbb{R}$ and $E$ a measurable subset with FINITE measure.
Show that,
$forall$ $epsilon > 0$, $exists$ open set $U$ containing $E$ s.t.
$m(U backslash E)$ < $epsilon$ (**)
What is the standard way to go about it because my intuition is that since $E$ is already measurable, then (**) becomes
$m(U backslash E)$ $leq$ $epsilon$
then we assume further that $E$ has FINITE measure,
so $E$ is somewhat "small"er than if it has infinite measure, but how does this guarantee that the open set containing $E$ has a "smaller difference" to $E$ which in turn gives us $<$ and not $leq$.
OR is there a subtlety on the fact that $m$ is Lebesgue measure and my problem only states $E$ is measurable, but when it states finite measure it says $m(E) < infty$ which assumes $E$ is Lebesgue Measurable? so This thought can't be correct.
I am using Stein Shakarchi mainly although I have papa Rudin for reference.
Any helps is greatly appreciated, if theres ANYTHING I said incorrectly or am doing wrong, PLEASE LET ME KNOW!!
real-analysis measure-theory lebesgue-measure
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
$endgroup$
|
show 4 more comments
$begingroup$
I am a noob at measure theory so please forgive me for being naive, I just need help, here is what I have albeit this is a very very standard result:
Let $m$ denote Lebesgue measure on $mathbb{R}$ and $E$ a measurable subset with FINITE measure.
Show that,
$forall$ $epsilon > 0$, $exists$ open set $U$ containing $E$ s.t.
$m(U backslash E)$ < $epsilon$ (**)
What is the standard way to go about it because my intuition is that since $E$ is already measurable, then (**) becomes
$m(U backslash E)$ $leq$ $epsilon$
then we assume further that $E$ has FINITE measure,
so $E$ is somewhat "small"er than if it has infinite measure, but how does this guarantee that the open set containing $E$ has a "smaller difference" to $E$ which in turn gives us $<$ and not $leq$.
OR is there a subtlety on the fact that $m$ is Lebesgue measure and my problem only states $E$ is measurable, but when it states finite measure it says $m(E) < infty$ which assumes $E$ is Lebesgue Measurable? so This thought can't be correct.
I am using Stein Shakarchi mainly although I have papa Rudin for reference.
Any helps is greatly appreciated, if theres ANYTHING I said incorrectly or am doing wrong, PLEASE LET ME KNOW!!
real-analysis measure-theory lebesgue-measure
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
$endgroup$
1
$begingroup$
Don't apologize for asking questions! I appreciate well-researched "noob" questions more than PSQs.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:20
1
$begingroup$
$<$ and $leq$ in this situation are exactly equivalent, since $epsilon$ is arbitrary and kept strictly positive. In any case, this result is called outer regularity, and exactly how you would like to prove it will depend somewhat on what you already know/assume about Lebesgue measurable sets. In particular, if you've defined the Lebesgue measure as the restriction of the Lebesgue outer measure to the Lebesgue measurable sets, then this result follows immediately from the definition of the Lebesgue outer measure.
$endgroup$
– Ian
Dec 18 '18 at 20:24
1
$begingroup$
(Cont.) By contrast, if, say, you define the Lebesgue measure by defining it on intervals and then applying an extension theorem to obtain the Lebesgue measure on the Borel $sigma$-algebra and then extend it by completion, then there is more work to be done. (By the way, the result is also true when $E$ has infinite measure.)
$endgroup$
– Ian
Dec 18 '18 at 20:26
$begingroup$
OMG you're right I cannot believe I missed that, the two inequalities are the same since we change by $epsilon$!! ok, so if the way I wrote up there IS how we defined a Lebesgue Measurable set Then like you say the finite case doesn't require much proof, its more definitions shoving around? that makes sense. And you're correct the result holds in general for any measurable set you can always find an open set small enough so my inequality holds. Another question I had was could you go about proving it using exterior measure and union of almost disjoint cubes?
$endgroup$
– Hossien Sahebjame
Dec 18 '18 at 20:32
1
$begingroup$
For any measurable $E,$ and $epsilon >0:$ For $nin Bbb N$ let $E_n=Ecap (-n,n)$ and let $U_n$ be open with $E_nsubset U_n$ and $m(U_n$ $E_n)<2^{-n}epsilon$. Let $U=cup_{nin Bbb N}U_n.$ Then $Usupset E$ and $U$ $E=$ $=(cup_{nin Bbb N}U_n) setminus E$ $=cup_{nin Bbb N}(U_n$ $E)subset$ $ cup_{nin Bbb N}(U_n$ $E_n).$ Therefore $m(U$ $E)leq$ $leq sum_{nin Bbb N}m(U_n setminus E_n)<epsilon.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:58
|
show 4 more comments
$begingroup$
I am a noob at measure theory so please forgive me for being naive, I just need help, here is what I have albeit this is a very very standard result:
Let $m$ denote Lebesgue measure on $mathbb{R}$ and $E$ a measurable subset with FINITE measure.
Show that,
$forall$ $epsilon > 0$, $exists$ open set $U$ containing $E$ s.t.
$m(U backslash E)$ < $epsilon$ (**)
What is the standard way to go about it because my intuition is that since $E$ is already measurable, then (**) becomes
$m(U backslash E)$ $leq$ $epsilon$
then we assume further that $E$ has FINITE measure,
so $E$ is somewhat "small"er than if it has infinite measure, but how does this guarantee that the open set containing $E$ has a "smaller difference" to $E$ which in turn gives us $<$ and not $leq$.
OR is there a subtlety on the fact that $m$ is Lebesgue measure and my problem only states $E$ is measurable, but when it states finite measure it says $m(E) < infty$ which assumes $E$ is Lebesgue Measurable? so This thought can't be correct.
I am using Stein Shakarchi mainly although I have papa Rudin for reference.
Any helps is greatly appreciated, if theres ANYTHING I said incorrectly or am doing wrong, PLEASE LET ME KNOW!!
real-analysis measure-theory lebesgue-measure
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
$endgroup$
I am a noob at measure theory so please forgive me for being naive, I just need help, here is what I have albeit this is a very very standard result:
Let $m$ denote Lebesgue measure on $mathbb{R}$ and $E$ a measurable subset with FINITE measure.
Show that,
$forall$ $epsilon > 0$, $exists$ open set $U$ containing $E$ s.t.
$m(U backslash E)$ < $epsilon$ (**)
What is the standard way to go about it because my intuition is that since $E$ is already measurable, then (**) becomes
$m(U backslash E)$ $leq$ $epsilon$
then we assume further that $E$ has FINITE measure,
so $E$ is somewhat "small"er than if it has infinite measure, but how does this guarantee that the open set containing $E$ has a "smaller difference" to $E$ which in turn gives us $<$ and not $leq$.
OR is there a subtlety on the fact that $m$ is Lebesgue measure and my problem only states $E$ is measurable, but when it states finite measure it says $m(E) < infty$ which assumes $E$ is Lebesgue Measurable? so This thought can't be correct.
I am using Stein Shakarchi mainly although I have papa Rudin for reference.
Any helps is greatly appreciated, if theres ANYTHING I said incorrectly or am doing wrong, PLEASE LET ME KNOW!!
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
edited Dec 18 '18 at 20:21
Hossien Sahebjame
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
asked Dec 18 '18 at 20:19
Hossien SahebjameHossien Sahebjame
1049
1049
1
$begingroup$
Don't apologize for asking questions! I appreciate well-researched "noob" questions more than PSQs.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:20
1
$begingroup$
$<$ and $leq$ in this situation are exactly equivalent, since $epsilon$ is arbitrary and kept strictly positive. In any case, this result is called outer regularity, and exactly how you would like to prove it will depend somewhat on what you already know/assume about Lebesgue measurable sets. In particular, if you've defined the Lebesgue measure as the restriction of the Lebesgue outer measure to the Lebesgue measurable sets, then this result follows immediately from the definition of the Lebesgue outer measure.
$endgroup$
– Ian
Dec 18 '18 at 20:24
1
$begingroup$
(Cont.) By contrast, if, say, you define the Lebesgue measure by defining it on intervals and then applying an extension theorem to obtain the Lebesgue measure on the Borel $sigma$-algebra and then extend it by completion, then there is more work to be done. (By the way, the result is also true when $E$ has infinite measure.)
$endgroup$
– Ian
Dec 18 '18 at 20:26
$begingroup$
OMG you're right I cannot believe I missed that, the two inequalities are the same since we change by $epsilon$!! ok, so if the way I wrote up there IS how we defined a Lebesgue Measurable set Then like you say the finite case doesn't require much proof, its more definitions shoving around? that makes sense. And you're correct the result holds in general for any measurable set you can always find an open set small enough so my inequality holds. Another question I had was could you go about proving it using exterior measure and union of almost disjoint cubes?
$endgroup$
– Hossien Sahebjame
Dec 18 '18 at 20:32
1
$begingroup$
For any measurable $E,$ and $epsilon >0:$ For $nin Bbb N$ let $E_n=Ecap (-n,n)$ and let $U_n$ be open with $E_nsubset U_n$ and $m(U_n$ $E_n)<2^{-n}epsilon$. Let $U=cup_{nin Bbb N}U_n.$ Then $Usupset E$ and $U$ $E=$ $=(cup_{nin Bbb N}U_n) setminus E$ $=cup_{nin Bbb N}(U_n$ $E)subset$ $ cup_{nin Bbb N}(U_n$ $E_n).$ Therefore $m(U$ $E)leq$ $leq sum_{nin Bbb N}m(U_n setminus E_n)<epsilon.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:58
|
show 4 more comments
1
$begingroup$
Don't apologize for asking questions! I appreciate well-researched "noob" questions more than PSQs.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:20
1
$begingroup$
$<$ and $leq$ in this situation are exactly equivalent, since $epsilon$ is arbitrary and kept strictly positive. In any case, this result is called outer regularity, and exactly how you would like to prove it will depend somewhat on what you already know/assume about Lebesgue measurable sets. In particular, if you've defined the Lebesgue measure as the restriction of the Lebesgue outer measure to the Lebesgue measurable sets, then this result follows immediately from the definition of the Lebesgue outer measure.
$endgroup$
– Ian
Dec 18 '18 at 20:24
1
$begingroup$
(Cont.) By contrast, if, say, you define the Lebesgue measure by defining it on intervals and then applying an extension theorem to obtain the Lebesgue measure on the Borel $sigma$-algebra and then extend it by completion, then there is more work to be done. (By the way, the result is also true when $E$ has infinite measure.)
$endgroup$
– Ian
Dec 18 '18 at 20:26
$begingroup$
OMG you're right I cannot believe I missed that, the two inequalities are the same since we change by $epsilon$!! ok, so if the way I wrote up there IS how we defined a Lebesgue Measurable set Then like you say the finite case doesn't require much proof, its more definitions shoving around? that makes sense. And you're correct the result holds in general for any measurable set you can always find an open set small enough so my inequality holds. Another question I had was could you go about proving it using exterior measure and union of almost disjoint cubes?
$endgroup$
– Hossien Sahebjame
Dec 18 '18 at 20:32
1
$begingroup$
For any measurable $E,$ and $epsilon >0:$ For $nin Bbb N$ let $E_n=Ecap (-n,n)$ and let $U_n$ be open with $E_nsubset U_n$ and $m(U_n$ $E_n)<2^{-n}epsilon$. Let $U=cup_{nin Bbb N}U_n.$ Then $Usupset E$ and $U$ $E=$ $=(cup_{nin Bbb N}U_n) setminus E$ $=cup_{nin Bbb N}(U_n$ $E)subset$ $ cup_{nin Bbb N}(U_n$ $E_n).$ Therefore $m(U$ $E)leq$ $leq sum_{nin Bbb N}m(U_n setminus E_n)<epsilon.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:58
1
1
$begingroup$
Don't apologize for asking questions! I appreciate well-researched "noob" questions more than PSQs.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:20
$begingroup$
Don't apologize for asking questions! I appreciate well-researched "noob" questions more than PSQs.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:20
1
1
$begingroup$
$<$ and $leq$ in this situation are exactly equivalent, since $epsilon$ is arbitrary and kept strictly positive. In any case, this result is called outer regularity, and exactly how you would like to prove it will depend somewhat on what you already know/assume about Lebesgue measurable sets. In particular, if you've defined the Lebesgue measure as the restriction of the Lebesgue outer measure to the Lebesgue measurable sets, then this result follows immediately from the definition of the Lebesgue outer measure.
$endgroup$
– Ian
Dec 18 '18 at 20:24
$begingroup$
$<$ and $leq$ in this situation are exactly equivalent, since $epsilon$ is arbitrary and kept strictly positive. In any case, this result is called outer regularity, and exactly how you would like to prove it will depend somewhat on what you already know/assume about Lebesgue measurable sets. In particular, if you've defined the Lebesgue measure as the restriction of the Lebesgue outer measure to the Lebesgue measurable sets, then this result follows immediately from the definition of the Lebesgue outer measure.
$endgroup$
– Ian
Dec 18 '18 at 20:24
1
1
$begingroup$
(Cont.) By contrast, if, say, you define the Lebesgue measure by defining it on intervals and then applying an extension theorem to obtain the Lebesgue measure on the Borel $sigma$-algebra and then extend it by completion, then there is more work to be done. (By the way, the result is also true when $E$ has infinite measure.)
$endgroup$
– Ian
Dec 18 '18 at 20:26
$begingroup$
(Cont.) By contrast, if, say, you define the Lebesgue measure by defining it on intervals and then applying an extension theorem to obtain the Lebesgue measure on the Borel $sigma$-algebra and then extend it by completion, then there is more work to be done. (By the way, the result is also true when $E$ has infinite measure.)
$endgroup$
– Ian
Dec 18 '18 at 20:26
$begingroup$
OMG you're right I cannot believe I missed that, the two inequalities are the same since we change by $epsilon$!! ok, so if the way I wrote up there IS how we defined a Lebesgue Measurable set Then like you say the finite case doesn't require much proof, its more definitions shoving around? that makes sense. And you're correct the result holds in general for any measurable set you can always find an open set small enough so my inequality holds. Another question I had was could you go about proving it using exterior measure and union of almost disjoint cubes?
$endgroup$
– Hossien Sahebjame
Dec 18 '18 at 20:32
$begingroup$
OMG you're right I cannot believe I missed that, the two inequalities are the same since we change by $epsilon$!! ok, so if the way I wrote up there IS how we defined a Lebesgue Measurable set Then like you say the finite case doesn't require much proof, its more definitions shoving around? that makes sense. And you're correct the result holds in general for any measurable set you can always find an open set small enough so my inequality holds. Another question I had was could you go about proving it using exterior measure and union of almost disjoint cubes?
$endgroup$
– Hossien Sahebjame
Dec 18 '18 at 20:32
1
1
$begingroup$
For any measurable $E,$ and $epsilon >0:$ For $nin Bbb N$ let $E_n=Ecap (-n,n)$ and let $U_n$ be open with $E_nsubset U_n$ and $m(U_n$ $E_n)<2^{-n}epsilon$. Let $U=cup_{nin Bbb N}U_n.$ Then $Usupset E$ and $U$ $E=$ $=(cup_{nin Bbb N}U_n) setminus E$ $=cup_{nin Bbb N}(U_n$ $E)subset$ $ cup_{nin Bbb N}(U_n$ $E_n).$ Therefore $m(U$ $E)leq$ $leq sum_{nin Bbb N}m(U_n setminus E_n)<epsilon.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:58
$begingroup$
For any measurable $E,$ and $epsilon >0:$ For $nin Bbb N$ let $E_n=Ecap (-n,n)$ and let $U_n$ be open with $E_nsubset U_n$ and $m(U_n$ $E_n)<2^{-n}epsilon$. Let $U=cup_{nin Bbb N}U_n.$ Then $Usupset E$ and $U$ $E=$ $=(cup_{nin Bbb N}U_n) setminus E$ $=cup_{nin Bbb N}(U_n$ $E)subset$ $ cup_{nin Bbb N}(U_n$ $E_n).$ Therefore $m(U$ $E)leq$ $leq sum_{nin Bbb N}m(U_n setminus E_n)<epsilon.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:58
|
show 4 more comments
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1
$begingroup$
Don't apologize for asking questions! I appreciate well-researched "noob" questions more than PSQs.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:20
1
$begingroup$
$<$ and $leq$ in this situation are exactly equivalent, since $epsilon$ is arbitrary and kept strictly positive. In any case, this result is called outer regularity, and exactly how you would like to prove it will depend somewhat on what you already know/assume about Lebesgue measurable sets. In particular, if you've defined the Lebesgue measure as the restriction of the Lebesgue outer measure to the Lebesgue measurable sets, then this result follows immediately from the definition of the Lebesgue outer measure.
$endgroup$
– Ian
Dec 18 '18 at 20:24
1
$begingroup$
(Cont.) By contrast, if, say, you define the Lebesgue measure by defining it on intervals and then applying an extension theorem to obtain the Lebesgue measure on the Borel $sigma$-algebra and then extend it by completion, then there is more work to be done. (By the way, the result is also true when $E$ has infinite measure.)
$endgroup$
– Ian
Dec 18 '18 at 20:26
$begingroup$
OMG you're right I cannot believe I missed that, the two inequalities are the same since we change by $epsilon$!! ok, so if the way I wrote up there IS how we defined a Lebesgue Measurable set Then like you say the finite case doesn't require much proof, its more definitions shoving around? that makes sense. And you're correct the result holds in general for any measurable set you can always find an open set small enough so my inequality holds. Another question I had was could you go about proving it using exterior measure and union of almost disjoint cubes?
$endgroup$
– Hossien Sahebjame
Dec 18 '18 at 20:32
1
$begingroup$
For any measurable $E,$ and $epsilon >0:$ For $nin Bbb N$ let $E_n=Ecap (-n,n)$ and let $U_n$ be open with $E_nsubset U_n$ and $m(U_n$ $E_n)<2^{-n}epsilon$. Let $U=cup_{nin Bbb N}U_n.$ Then $Usupset E$ and $U$ $E=$ $=(cup_{nin Bbb N}U_n) setminus E$ $=cup_{nin Bbb N}(U_n$ $E)subset$ $ cup_{nin Bbb N}(U_n$ $E_n).$ Therefore $m(U$ $E)leq$ $leq sum_{nin Bbb N}m(U_n setminus E_n)<epsilon.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:58