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Find a nowhere dense closed subset of $mathbbR$ with the following property
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Let $m$ be the Lebesgue measure on $mathbbR$. Prove that for every $epsilon>0$, there exists a nowhere dense closed subset $F$ of $mathbbR$ such that for every interval $(a,b)$ of $mathbbR$, $m(F cap (a,b)) > b-a-epsilon$.
To be honest I'm not really sure where to start with this. The statement of the problem reminds me of Lusin's theorem, but I don't know how to apply that here. I've also thought that I might need some version of the Cantor set.
measure-theory lebesgue-measure
asked Jul 18 at 21:29
emma
1,5801319
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up vote
1
down vote
favorite
Let $m$ be the Lebesgue measure on $mathbbR$. Prove that for every $epsilon>0$, there exists a nowhere dense closed subset $F$ of $mathbbR$ such that for every interval $(a,b)$ of $mathbbR$, $m(F cap (a,b)) > b-a-epsilon$.
To be honest I'm not really sure where to start with this. The statement of the problem reminds me of Lusin's theorem, but I don't know how to apply that here. I've also thought that I might need some version of the Cantor set.
measure-theory lebesgue-measure
asked Jul 18 at 21:29
emma
1,5801319
Do you know how to construct a Cantor set of positive measure?
– Eric Wofsey
Jul 18 at 21:31
@EricWofsey Yes - by doing something like taking out the middle fourths, right? I think I could do this problem if it was for a set interval, but I'm not sure how to provide an F that would work for any interval.
– emma
Jul 18 at 21:33
Middle fourths doesn't work: you have to make the proportions you are removing in each step get smaller and smaller fast enough.
– Eric Wofsey
Jul 18 at 21:36
@EricWofsey Sorry I misspoke - I meant I could for example remove the middle $(1/4)^n$-s for the $n$th step, yes?
– emma
Jul 18 at 22:11
Right, that would work. And you can make the measure as close to the whole measure of the interval by adjusting the numbers.
– Eric Wofsey
Jul 18 at 22:16
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $m$ be the Lebesgue measure on $mathbbR$. Prove that for every $epsilon>0$, there exists a nowhere dense closed subset $F$ of $mathbbR$ such that for every interval $(a,b)$ of $mathbbR$, $m(F cap (a,b)) > b-a-epsilon$.
To be honest I'm not really sure where to start with this. The statement of the problem reminds me of Lusin's theorem, but I don't know how to apply that here. I've also thought that I might need some version of the Cantor set.
measure-theory lebesgue-measure
asked Jul 18 at 21:29
emma
1,5801319
Let $m$ be the Lebesgue measure on $mathbbR$. Prove that for every $epsilon>0$, there exists a nowhere dense closed subset $F$ of $mathbbR$ such that for every interval $(a,b)$ of $mathbbR$, $m(F cap (a,b)) > b-a-epsilon$.
To be honest I'm not really sure where to start with this. The statement of the problem reminds me of Lusin's theorem, but I don't know how to apply that here. I've also thought that I might need some version of the Cantor set.
measure-theory lebesgue-measure
asked Jul 18 at 21:29
emma
1,5801319
asked Jul 18 at 21:29
emma
1,5801319
asked Jul 18 at 21:29
emma
1,5801319
asked Jul 18 at 21:29
emma
1,5801319
1,5801319
Do you know how to construct a Cantor set of positive measure?
– Eric Wofsey
Jul 18 at 21:31
@EricWofsey Yes - by doing something like taking out the middle fourths, right? I think I could do this problem if it was for a set interval, but I'm not sure how to provide an F that would work for any interval.
– emma
Jul 18 at 21:33
Middle fourths doesn't work: you have to make the proportions you are removing in each step get smaller and smaller fast enough.
– Eric Wofsey
Jul 18 at 21:36
@EricWofsey Sorry I misspoke - I meant I could for example remove the middle $(1/4)^n$-s for the $n$th step, yes?
– emma
Jul 18 at 22:11
Right, that would work. And you can make the measure as close to the whole measure of the interval by adjusting the numbers.
– Eric Wofsey
Jul 18 at 22:16
add a comment |Â
Do you know how to construct a Cantor set of positive measure?
– Eric Wofsey
Jul 18 at 21:31
@EricWofsey Yes - by doing something like taking out the middle fourths, right? I think I could do this problem if it was for a set interval, but I'm not sure how to provide an F that would work for any interval.
– emma
Jul 18 at 21:33
Middle fourths doesn't work: you have to make the proportions you are removing in each step get smaller and smaller fast enough.
– Eric Wofsey
Jul 18 at 21:36
@EricWofsey Sorry I misspoke - I meant I could for example remove the middle $(1/4)^n$-s for the $n$th step, yes?
– emma
Jul 18 at 22:11
Right, that would work. And you can make the measure as close to the whole measure of the interval by adjusting the numbers.
– Eric Wofsey
Jul 18 at 22:16
Do you know how to construct a Cantor set of positive measure?
– Eric Wofsey
Jul 18 at 21:31
Do you know how to construct a Cantor set of positive measure?
– Eric Wofsey
Jul 18 at 21:31
@EricWofsey Yes - by doing something like taking out the middle fourths, right? I think I could do this problem if it was for a set interval, but I'm not sure how to provide an F that would work for any interval.
– emma
Jul 18 at 21:33
@EricWofsey Yes - by doing something like taking out the middle fourths, right? I think I could do this problem if it was for a set interval, but I'm not sure how to provide an F that would work for any interval.
– emma
Jul 18 at 21:33
Middle fourths doesn't work: you have to make the proportions you are removing in each step get smaller and smaller fast enough.
– Eric Wofsey
Jul 18 at 21:36
Middle fourths doesn't work: you have to make the proportions you are removing in each step get smaller and smaller fast enough.
– Eric Wofsey
Jul 18 at 21:36
@EricWofsey Sorry I misspoke - I meant I could for example remove the middle $(1/4)^n$-s for the $n$th step, yes?
– emma
Jul 18 at 22:11
@EricWofsey Sorry I misspoke - I meant I could for example remove the middle $(1/4)^n$-s for the $n$th step, yes?
– emma
Jul 18 at 22:11
Right, that would work. And you can make the measure as close to the whole measure of the interval by adjusting the numbers.
– Eric Wofsey
Jul 18 at 22:16
Right, that would work. And you can make the measure as close to the whole measure of the interval by adjusting the numbers.
– Eric Wofsey
Jul 18 at 22:16
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Hint: It would be enough to get a nowhere dense closed set $F$ such that $m(mathbbRsetminus F)<epsilon$. As a starting point, think about how to do this if you were on $[0,1]$ instead of $mathbbR$: how could you construct such an $Fsubset[0,1]$ with $m([0,1]setminus F)<epsilon$? Then think about how you can combine such sets to work on all of $mathbbR$.
Further hint for the last step:
If you choose a closed nowhere dense subset $F_nsubset[n,n+1]$ for each $ninmathbbZ$, then $F=bigcup_ninmathbbZF_n$ will still be closed and nowhere dense.
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
Thank you for putting this as a hint - I'm practicing for an exam so it's really helpful! Okay, so I can find $F subset [0,1]$ with $m([0,1]-F)< epsilon$, and I understand that I could replicate this with any interval, or union together several such sets. But - say I do this for each $F_n$ as you suggested. If $m([n,n+1]-F_n) < epsilon$, wouldn't $m(mathbbR -F) = sum_n in mathbbZ m([n,n+1]-F_n)$, which could be much larger than $epsilon$? How would I make the measure of the difference small enough?
– emma
Jul 18 at 22:38
Don't make $m([n,n+1]-F_n)$ equal to $epsilon$ for each $n$--make it vary with $n$.
– Eric Wofsey
Jul 18 at 22:54
Okay, I see what you're getting at. Thanks for your help!
– emma
Jul 18 at 23:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: It would be enough to get a nowhere dense closed set $F$ such that $m(mathbbRsetminus F)<epsilon$. As a starting point, think about how to do this if you were on $[0,1]$ instead of $mathbbR$: how could you construct such an $Fsubset[0,1]$ with $m([0,1]setminus F)<epsilon$? Then think about how you can combine such sets to work on all of $mathbbR$.
Further hint for the last step:
If you choose a closed nowhere dense subset $F_nsubset[n,n+1]$ for each $ninmathbbZ$, then $F=bigcup_ninmathbbZF_n$ will still be closed and nowhere dense.
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
Thank you for putting this as a hint - I'm practicing for an exam so it's really helpful! Okay, so I can find $F subset [0,1]$ with $m([0,1]-F)< epsilon$, and I understand that I could replicate this with any interval, or union together several such sets. But - say I do this for each $F_n$ as you suggested. If $m([n,n+1]-F_n) < epsilon$, wouldn't $m(mathbbR -F) = sum_n in mathbbZ m([n,n+1]-F_n)$, which could be much larger than $epsilon$? How would I make the measure of the difference small enough?
– emma
Jul 18 at 22:38
Don't make $m([n,n+1]-F_n)$ equal to $epsilon$ for each $n$--make it vary with $n$.
– Eric Wofsey
Jul 18 at 22:54
Okay, I see what you're getting at. Thanks for your help!
– emma
Jul 18 at 23:09
add a comment |Â
up vote
1
down vote
accepted
Hint: It would be enough to get a nowhere dense closed set $F$ such that $m(mathbbRsetminus F)<epsilon$. As a starting point, think about how to do this if you were on $[0,1]$ instead of $mathbbR$: how could you construct such an $Fsubset[0,1]$ with $m([0,1]setminus F)<epsilon$? Then think about how you can combine such sets to work on all of $mathbbR$.
Further hint for the last step:
If you choose a closed nowhere dense subset $F_nsubset[n,n+1]$ for each $ninmathbbZ$, then $F=bigcup_ninmathbbZF_n$ will still be closed and nowhere dense.
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
Thank you for putting this as a hint - I'm practicing for an exam so it's really helpful! Okay, so I can find $F subset [0,1]$ with $m([0,1]-F)< epsilon$, and I understand that I could replicate this with any interval, or union together several such sets. But - say I do this for each $F_n$ as you suggested. If $m([n,n+1]-F_n) < epsilon$, wouldn't $m(mathbbR -F) = sum_n in mathbbZ m([n,n+1]-F_n)$, which could be much larger than $epsilon$? How would I make the measure of the difference small enough?
– emma
Jul 18 at 22:38
Don't make $m([n,n+1]-F_n)$ equal to $epsilon$ for each $n$--make it vary with $n$.
– Eric Wofsey
Jul 18 at 22:54
Okay, I see what you're getting at. Thanks for your help!
– emma
Jul 18 at 23:09
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: It would be enough to get a nowhere dense closed set $F$ such that $m(mathbbRsetminus F)<epsilon$. As a starting point, think about how to do this if you were on $[0,1]$ instead of $mathbbR$: how could you construct such an $Fsubset[0,1]$ with $m([0,1]setminus F)<epsilon$? Then think about how you can combine such sets to work on all of $mathbbR$.
Further hint for the last step:
If you choose a closed nowhere dense subset $F_nsubset[n,n+1]$ for each $ninmathbbZ$, then $F=bigcup_ninmathbbZF_n$ will still be closed and nowhere dense.
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
Hint: It would be enough to get a nowhere dense closed set $F$ such that $m(mathbbRsetminus F)<epsilon$. As a starting point, think about how to do this if you were on $[0,1]$ instead of $mathbbR$: how could you construct such an $Fsubset[0,1]$ with $m([0,1]setminus F)<epsilon$? Then think about how you can combine such sets to work on all of $mathbbR$.
Further hint for the last step:
If you choose a closed nowhere dense subset $F_nsubset[n,n+1]$ for each $ninmathbbZ$, then $F=bigcup_ninmathbbZF_n$ will still be closed and nowhere dense.
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
answered Jul 18 at 21:38
Eric Wofsey
162k12189300
162k12189300
Thank you for putting this as a hint - I'm practicing for an exam so it's really helpful! Okay, so I can find $F subset [0,1]$ with $m([0,1]-F)< epsilon$, and I understand that I could replicate this with any interval, or union together several such sets. But - say I do this for each $F_n$ as you suggested. If $m([n,n+1]-F_n) < epsilon$, wouldn't $m(mathbbR -F) = sum_n in mathbbZ m([n,n+1]-F_n)$, which could be much larger than $epsilon$? How would I make the measure of the difference small enough?
– emma
Jul 18 at 22:38
Don't make $m([n,n+1]-F_n)$ equal to $epsilon$ for each $n$--make it vary with $n$.
– Eric Wofsey
Jul 18 at 22:54
Okay, I see what you're getting at. Thanks for your help!
– emma
Jul 18 at 23:09
add a comment |Â
Thank you for putting this as a hint - I'm practicing for an exam so it's really helpful! Okay, so I can find $F subset [0,1]$ with $m([0,1]-F)< epsilon$, and I understand that I could replicate this with any interval, or union together several such sets. But - say I do this for each $F_n$ as you suggested. If $m([n,n+1]-F_n) < epsilon$, wouldn't $m(mathbbR -F) = sum_n in mathbbZ m([n,n+1]-F_n)$, which could be much larger than $epsilon$? How would I make the measure of the difference small enough?
– emma
Jul 18 at 22:38
Don't make $m([n,n+1]-F_n)$ equal to $epsilon$ for each $n$--make it vary with $n$.
– Eric Wofsey
Jul 18 at 22:54
Okay, I see what you're getting at. Thanks for your help!
– emma
Jul 18 at 23:09
Thank you for putting this as a hint - I'm practicing for an exam so it's really helpful! Okay, so I can find $F subset [0,1]$ with $m([0,1]-F)< epsilon$, and I understand that I could replicate this with any interval, or union together several such sets. But - say I do this for each $F_n$ as you suggested. If $m([n,n+1]-F_n) < epsilon$, wouldn't $m(mathbbR -F) = sum_n in mathbbZ m([n,n+1]-F_n)$, which could be much larger than $epsilon$? How would I make the measure of the difference small enough?
– emma
Jul 18 at 22:38
Thank you for putting this as a hint - I'm practicing for an exam so it's really helpful! Okay, so I can find $F subset [0,1]$ with $m([0,1]-F)< epsilon$, and I understand that I could replicate this with any interval, or union together several such sets. But - say I do this for each $F_n$ as you suggested. If $m([n,n+1]-F_n) < epsilon$, wouldn't $m(mathbbR -F) = sum_n in mathbbZ m([n,n+1]-F_n)$, which could be much larger than $epsilon$? How would I make the measure of the difference small enough?
– emma
Jul 18 at 22:38
Don't make $m([n,n+1]-F_n)$ equal to $epsilon$ for each $n$--make it vary with $n$.
– Eric Wofsey
Jul 18 at 22:54
Don't make $m([n,n+1]-F_n)$ equal to $epsilon$ for each $n$--make it vary with $n$.
– Eric Wofsey
Jul 18 at 22:54
Okay, I see what you're getting at. Thanks for your help!
– emma
Jul 18 at 23:09
Okay, I see what you're getting at. Thanks for your help!
– emma
Jul 18 at 23:09
add a comment |Â
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Do you know how to construct a Cantor set of positive measure?
– Eric Wofsey
Jul 18 at 21:31
@EricWofsey Yes - by doing something like taking out the middle fourths, right? I think I could do this problem if it was for a set interval, but I'm not sure how to provide an F that would work for any interval.
– emma
Jul 18 at 21:33
Middle fourths doesn't work: you have to make the proportions you are removing in each step get smaller and smaller fast enough.
– Eric Wofsey
Jul 18 at 21:36
@EricWofsey Sorry I misspoke - I meant I could for example remove the middle $(1/4)^n$-s for the $n$th step, yes?
– emma
Jul 18 at 22:11
Right, that would work. And you can make the measure as close to the whole measure of the interval by adjusting the numbers.
– Eric Wofsey
Jul 18 at 22:16