Lebesgue measurable set with certain propertiesFinding a Lebesgue measurable set that contains a set and they...
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Lebesgue measurable set with certain properties
Finding a Lebesgue measurable set that contains a set and they have the same outer measureExistence of distinct points with rational difference in lebesgue measurable setAssume that $Nsubset mathbb{R}$ is not Lebesgue measurable. Is then the set $ Ntimes mathbb{R} $ also not Lebesgue measurable?Prove that Lebesgue measurable set is the union of a Borel measurable set and a set of Lebesgue measure zeroCharacterization of (Lebesgue) measurable setsAre these two set Lebesgue-measurable?Image of Lebesgue measurable set by $C^1$ function is measurableHow to prove every open set is Lebesgue measurable?Lebesgue Measurable Subset of $[0,1]times[0,1]$Subset of Lebesgue measurable subset of Vitali set is NOT Lebesgue measurable
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I am currently stuck on a past paper question. It asks whether there exists a Lebesgue measurable set $D⊃[0,1]$ such that $Dneq [0,1]$ and $λ^{*}(D) =1?$ The only Lebesgue measurable set I have encountered with $λ^{*}(D) =1?$ in my module so far is $[0,1]∩mathbb{Q}^{c}$ but $[0,1]notsubset [0,1]∩mathbb{Q}^{c}$. I have also tried to assume there does and doesn't but I can't seem to get anywhere. Any help would be appreciated.
measure-theory lebesgue-measure
asked Mar 15 at 18:02
RJSSRJSS
52
$endgroup$
add a comment |
$begingroup$
I am currently stuck on a past paper question. It asks whether there exists a Lebesgue measurable set $D⊃[0,1]$ such that $Dneq [0,1]$ and $λ^{*}(D) =1?$ The only Lebesgue measurable set I have encountered with $λ^{*}(D) =1?$ in my module so far is $[0,1]∩mathbb{Q}^{c}$ but $[0,1]notsubset [0,1]∩mathbb{Q}^{c}$. I have also tried to assume there does and doesn't but I can't seem to get anywhere. Any help would be appreciated.
measure-theory lebesgue-measure
asked Mar 15 at 18:02
RJSSRJSS
52
$endgroup$
1
$begingroup$
How about sets like $[0,1]cup{2}$?
$endgroup$
– drhab
Mar 15 at 18:09
$begingroup$
It seems to me that you are thinking on $Dsubset [0,1]$. If, as you have written $Dsupset [0,1]$, then it is enough to take $D=[0,1]cup{2}$, Then $lambda(D)=1.$
$endgroup$
– szw1710
Mar 15 at 18:09
add a comment |
$begingroup$
I am currently stuck on a past paper question. It asks whether there exists a Lebesgue measurable set $D⊃[0,1]$ such that $Dneq [0,1]$ and $λ^{*}(D) =1?$ The only Lebesgue measurable set I have encountered with $λ^{*}(D) =1?$ in my module so far is $[0,1]∩mathbb{Q}^{c}$ but $[0,1]notsubset [0,1]∩mathbb{Q}^{c}$. I have also tried to assume there does and doesn't but I can't seem to get anywhere. Any help would be appreciated.
measure-theory lebesgue-measure
asked Mar 15 at 18:02
RJSSRJSS
52
$endgroup$
I am currently stuck on a past paper question. It asks whether there exists a Lebesgue measurable set $D⊃[0,1]$ such that $Dneq [0,1]$ and $λ^{*}(D) =1?$ The only Lebesgue measurable set I have encountered with $λ^{*}(D) =1?$ in my module so far is $[0,1]∩mathbb{Q}^{c}$ but $[0,1]notsubset [0,1]∩mathbb{Q}^{c}$. I have also tried to assume there does and doesn't but I can't seem to get anywhere. Any help would be appreciated.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Mar 15 at 18:02
RJSSRJSS
52
asked Mar 15 at 18:02
RJSSRJSS
52
asked Mar 15 at 18:02
RJSSRJSS
52
asked Mar 15 at 18:02
RJSSRJSS
52
asked Mar 15 at 18:02
RJSSRJSS
52
52
1
$begingroup$
How about sets like $[0,1]cup{2}$?
$endgroup$
– drhab
Mar 15 at 18:09
$begingroup$
It seems to me that you are thinking on $Dsubset [0,1]$. If, as you have written $Dsupset [0,1]$, then it is enough to take $D=[0,1]cup{2}$, Then $lambda(D)=1.$
$endgroup$
– szw1710
Mar 15 at 18:09
add a comment |
1
$begingroup$
How about sets like $[0,1]cup{2}$?
$endgroup$
– drhab
Mar 15 at 18:09
$begingroup$
It seems to me that you are thinking on $Dsubset [0,1]$. If, as you have written $Dsupset [0,1]$, then it is enough to take $D=[0,1]cup{2}$, Then $lambda(D)=1.$
$endgroup$
– szw1710
Mar 15 at 18:09
1
1
$begingroup$
How about sets like $[0,1]cup{2}$?
$endgroup$
– drhab
Mar 15 at 18:09
$begingroup$
How about sets like $[0,1]cup{2}$?
$endgroup$
– drhab
Mar 15 at 18:09
$begingroup$
It seems to me that you are thinking on $Dsubset [0,1]$. If, as you have written $Dsupset [0,1]$, then it is enough to take $D=[0,1]cup{2}$, Then $lambda(D)=1.$
$endgroup$
– szw1710
Mar 15 at 18:09
$begingroup$
It seems to me that you are thinking on $Dsubset [0,1]$. If, as you have written $Dsupset [0,1]$, then it is enough to take $D=[0,1]cup{2}$, Then $lambda(D)=1.$
$endgroup$
– szw1710
Mar 15 at 18:09
add a comment |
1 Answer
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$begingroup$
Just add any null set to $[0,1]$ which has elements outside of $[0,1]$, the measure will not change from that. $[0,1]cup{2}$ will work, just as well as $[0,1]cupmathbb{Q}$.
answered Mar 15 at 18:09
MarkMark
10.4k1622
$endgroup$
$begingroup$
Ahh I see thank you very much!
$endgroup$
– RJSS
Mar 15 at 20:26
add a comment |
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$begingroup$
Just add any null set to $[0,1]$ which has elements outside of $[0,1]$, the measure will not change from that. $[0,1]cup{2}$ will work, just as well as $[0,1]cupmathbb{Q}$.
answered Mar 15 at 18:09
MarkMark
10.4k1622
$endgroup$
$begingroup$
Ahh I see thank you very much!
$endgroup$
– RJSS
Mar 15 at 20:26
add a comment |
$begingroup$
Just add any null set to $[0,1]$ which has elements outside of $[0,1]$, the measure will not change from that. $[0,1]cup{2}$ will work, just as well as $[0,1]cupmathbb{Q}$.
answered Mar 15 at 18:09
MarkMark
10.4k1622
$endgroup$
$begingroup$
Ahh I see thank you very much!
$endgroup$
– RJSS
Mar 15 at 20:26
add a comment |
$begingroup$
Just add any null set to $[0,1]$ which has elements outside of $[0,1]$, the measure will not change from that. $[0,1]cup{2}$ will work, just as well as $[0,1]cupmathbb{Q}$.
answered Mar 15 at 18:09
MarkMark
10.4k1622
$endgroup$
Just add any null set to $[0,1]$ which has elements outside of $[0,1]$, the measure will not change from that. $[0,1]cup{2}$ will work, just as well as $[0,1]cupmathbb{Q}$.
answered Mar 15 at 18:09
MarkMark
10.4k1622
answered Mar 15 at 18:09
MarkMark
10.4k1622
answered Mar 15 at 18:09
MarkMark
10.4k1622
answered Mar 15 at 18:09
MarkMark
10.4k1622
10.4k1622
$begingroup$
Ahh I see thank you very much!
$endgroup$
– RJSS
Mar 15 at 20:26
add a comment |
$begingroup$
Ahh I see thank you very much!
$endgroup$
– RJSS
Mar 15 at 20:26
$begingroup$
Ahh I see thank you very much!
$endgroup$
– RJSS
Mar 15 at 20:26
$begingroup$
Ahh I see thank you very much!
$endgroup$
– RJSS
Mar 15 at 20:26
add a comment |
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$begingroup$
How about sets like $[0,1]cup{2}$?
$endgroup$
– drhab
Mar 15 at 18:09
$begingroup$
It seems to me that you are thinking on $Dsubset [0,1]$. If, as you have written $Dsupset [0,1]$, then it is enough to take $D=[0,1]cup{2}$, Then $lambda(D)=1.$
$endgroup$
– szw1710
Mar 15 at 18:09