Example of a subset of $mathbb{R}^2$ which is not $sigma$-finite under Lebesgue measure?
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We know open and closed sets are $sigma$-finite. What would be the example of a subset of $mathbb{R}^2$ which is not $sigma$-finite under Lebesgue measure?
real-analysis analysis measure-theory lebesgue-measure
asked Jan 12 at 17:16
ershersh
573113
$endgroup$
add a comment |
$begingroup$
We know open and closed sets are $sigma$-finite. What would be the example of a subset of $mathbb{R}^2$ which is not $sigma$-finite under Lebesgue measure?
real-analysis analysis measure-theory lebesgue-measure
asked Jan 12 at 17:16
ershersh
573113
$endgroup$
$begingroup$
@Song If the subset is Lebesgue measurable, then it is $sigma$-finite.
$endgroup$
– zhw.
Jan 12 at 19:32
$begingroup$
@Song are you confusing $sigma$-compact with $sigma$ -finite?
$endgroup$
– zhw.
Jan 12 at 19:47
$begingroup$
@zhw For some reason, I've read $sigma$-finite as $sigma$-compact. Thanks for pointing out my error!
$endgroup$
– Song
Jan 13 at 20:17
add a comment |
$begingroup$
We know open and closed sets are $sigma$-finite. What would be the example of a subset of $mathbb{R}^2$ which is not $sigma$-finite under Lebesgue measure?
real-analysis analysis measure-theory lebesgue-measure
asked Jan 12 at 17:16
ershersh
573113
$endgroup$
We know open and closed sets are $sigma$-finite. What would be the example of a subset of $mathbb{R}^2$ which is not $sigma$-finite under Lebesgue measure?
real-analysis analysis measure-theory lebesgue-measure
real-analysis analysis measure-theory lebesgue-measure
asked Jan 12 at 17:16
ershersh
573113
asked Jan 12 at 17:16
ershersh
573113
asked Jan 12 at 17:16
ershersh
573113
asked Jan 12 at 17:16
ershersh
573113
asked Jan 12 at 17:16
ershersh
573113
573113
$begingroup$
@Song If the subset is Lebesgue measurable, then it is $sigma$-finite.
$endgroup$
– zhw.
Jan 12 at 19:32
$begingroup$
@Song are you confusing $sigma$-compact with $sigma$ -finite?
$endgroup$
– zhw.
Jan 12 at 19:47
$begingroup$
@zhw For some reason, I've read $sigma$-finite as $sigma$-compact. Thanks for pointing out my error!
$endgroup$
– Song
Jan 13 at 20:17
add a comment |
$begingroup$
@Song If the subset is Lebesgue measurable, then it is $sigma$-finite.
$endgroup$
– zhw.
Jan 12 at 19:32
$begingroup$
@Song are you confusing $sigma$-compact with $sigma$ -finite?
$endgroup$
– zhw.
Jan 12 at 19:47
$begingroup$
@zhw For some reason, I've read $sigma$-finite as $sigma$-compact. Thanks for pointing out my error!
$endgroup$
– Song
Jan 13 at 20:17
$begingroup$
@Song If the subset is Lebesgue measurable, then it is $sigma$-finite.
$endgroup$
– zhw.
Jan 12 at 19:32
$begingroup$
@Song If the subset is Lebesgue measurable, then it is $sigma$-finite.
$endgroup$
– zhw.
Jan 12 at 19:32
$begingroup$
@Song are you confusing $sigma$-compact with $sigma$ -finite?
$endgroup$
– zhw.
Jan 12 at 19:47
$begingroup$
@Song are you confusing $sigma$-compact with $sigma$ -finite?
$endgroup$
– zhw.
Jan 12 at 19:47
$begingroup$
@zhw For some reason, I've read $sigma$-finite as $sigma$-compact. Thanks for pointing out my error!
$endgroup$
– Song
Jan 13 at 20:17
$begingroup$
@zhw For some reason, I've read $sigma$-finite as $sigma$-compact. Thanks for pointing out my error!
$endgroup$
– Song
Jan 13 at 20:17
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
(Thanks to @zhw pointing out my mistake) If we allow $Esubsetmathbb{R}^2$ to be a non-measurable set, then obviously it cannot be a countable union of measurable sets of a finite measure. We can construct a non-measurable subset of $[0,1]^2$ in a similar way constructing a Vitali set. This can be an example of a non-$sigma$-finite set that you are looking for. However, we cannot find any examples among measurable sets. The reason is as follows. Note that every measurable set $E$ can be expressed as
$$
E=bigcup_{n=1}^infty E_n
$$ where $E_n =Ecap [-n,n]^2$ and that each $E_n$ has a finite measure $m(E_n)le m([-n,n]^2)= 4n^2$. This shows that there are no measurable, non-$sigma$-finite subset of $mathbb{R}^2$.
answered Jan 12 at 20:16
SongSong
18.6k21651
$endgroup$
$begingroup$
You deleted your comments that I corrected, didn't answer my question there, then provided an answer with zero acknowledgement. Not really a cool thing to do.
$endgroup$
– zhw.
Jan 13 at 20:04
$begingroup$
@zhw. I'm so sorry about that if it makes you feel bad. At first I did leave a comment appreciating your pointing out my mistake, but after a while I deleted it only because I thought you probably already read it. Again, I apologize for such a misbehavior.
$endgroup$
– Song
Jan 13 at 20:18
$begingroup$
OK, thanks for your comment here. In the future, at times, you might want to consider not deleting comments of yours even if they're mistaken. Often the resulting discussion can help readers.
$endgroup$
– zhw.
Jan 13 at 20:28
$begingroup$
@zhw Oh, I think it's a really good point. I'll keep it in my mind. Thank you!
$endgroup$
– Song
Jan 13 at 20:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
(Thanks to @zhw pointing out my mistake) If we allow $Esubsetmathbb{R}^2$ to be a non-measurable set, then obviously it cannot be a countable union of measurable sets of a finite measure. We can construct a non-measurable subset of $[0,1]^2$ in a similar way constructing a Vitali set. This can be an example of a non-$sigma$-finite set that you are looking for. However, we cannot find any examples among measurable sets. The reason is as follows. Note that every measurable set $E$ can be expressed as
$$
E=bigcup_{n=1}^infty E_n
$$ where $E_n =Ecap [-n,n]^2$ and that each $E_n$ has a finite measure $m(E_n)le m([-n,n]^2)= 4n^2$. This shows that there are no measurable, non-$sigma$-finite subset of $mathbb{R}^2$.
answered Jan 12 at 20:16
SongSong
18.6k21651
$endgroup$
$begingroup$
You deleted your comments that I corrected, didn't answer my question there, then provided an answer with zero acknowledgement. Not really a cool thing to do.
$endgroup$
– zhw.
Jan 13 at 20:04
$begingroup$
@zhw. I'm so sorry about that if it makes you feel bad. At first I did leave a comment appreciating your pointing out my mistake, but after a while I deleted it only because I thought you probably already read it. Again, I apologize for such a misbehavior.
$endgroup$
– Song
Jan 13 at 20:18
$begingroup$
OK, thanks for your comment here. In the future, at times, you might want to consider not deleting comments of yours even if they're mistaken. Often the resulting discussion can help readers.
$endgroup$
– zhw.
Jan 13 at 20:28
$begingroup$
@zhw Oh, I think it's a really good point. I'll keep it in my mind. Thank you!
$endgroup$
– Song
Jan 13 at 20:31
add a comment |
$begingroup$
(Thanks to @zhw pointing out my mistake) If we allow $Esubsetmathbb{R}^2$ to be a non-measurable set, then obviously it cannot be a countable union of measurable sets of a finite measure. We can construct a non-measurable subset of $[0,1]^2$ in a similar way constructing a Vitali set. This can be an example of a non-$sigma$-finite set that you are looking for. However, we cannot find any examples among measurable sets. The reason is as follows. Note that every measurable set $E$ can be expressed as
$$
E=bigcup_{n=1}^infty E_n
$$ where $E_n =Ecap [-n,n]^2$ and that each $E_n$ has a finite measure $m(E_n)le m([-n,n]^2)= 4n^2$. This shows that there are no measurable, non-$sigma$-finite subset of $mathbb{R}^2$.
answered Jan 12 at 20:16
SongSong
18.6k21651
$endgroup$
$begingroup$
You deleted your comments that I corrected, didn't answer my question there, then provided an answer with zero acknowledgement. Not really a cool thing to do.
$endgroup$
– zhw.
Jan 13 at 20:04
$begingroup$
@zhw. I'm so sorry about that if it makes you feel bad. At first I did leave a comment appreciating your pointing out my mistake, but after a while I deleted it only because I thought you probably already read it. Again, I apologize for such a misbehavior.
$endgroup$
– Song
Jan 13 at 20:18
$begingroup$
OK, thanks for your comment here. In the future, at times, you might want to consider not deleting comments of yours even if they're mistaken. Often the resulting discussion can help readers.
$endgroup$
– zhw.
Jan 13 at 20:28
$begingroup$
@zhw Oh, I think it's a really good point. I'll keep it in my mind. Thank you!
$endgroup$
– Song
Jan 13 at 20:31
add a comment |
$begingroup$
(Thanks to @zhw pointing out my mistake) If we allow $Esubsetmathbb{R}^2$ to be a non-measurable set, then obviously it cannot be a countable union of measurable sets of a finite measure. We can construct a non-measurable subset of $[0,1]^2$ in a similar way constructing a Vitali set. This can be an example of a non-$sigma$-finite set that you are looking for. However, we cannot find any examples among measurable sets. The reason is as follows. Note that every measurable set $E$ can be expressed as
$$
E=bigcup_{n=1}^infty E_n
$$ where $E_n =Ecap [-n,n]^2$ and that each $E_n$ has a finite measure $m(E_n)le m([-n,n]^2)= 4n^2$. This shows that there are no measurable, non-$sigma$-finite subset of $mathbb{R}^2$.
answered Jan 12 at 20:16
SongSong
18.6k21651
$endgroup$
(Thanks to @zhw pointing out my mistake) If we allow $Esubsetmathbb{R}^2$ to be a non-measurable set, then obviously it cannot be a countable union of measurable sets of a finite measure. We can construct a non-measurable subset of $[0,1]^2$ in a similar way constructing a Vitali set. This can be an example of a non-$sigma$-finite set that you are looking for. However, we cannot find any examples among measurable sets. The reason is as follows. Note that every measurable set $E$ can be expressed as
$$
E=bigcup_{n=1}^infty E_n
$$ where $E_n =Ecap [-n,n]^2$ and that each $E_n$ has a finite measure $m(E_n)le m([-n,n]^2)= 4n^2$. This shows that there are no measurable, non-$sigma$-finite subset of $mathbb{R}^2$.
answered Jan 12 at 20:16
SongSong
18.6k21651
edited Jan 13 at 20:54
answered Jan 12 at 20:16
SongSong
18.6k21651
answered Jan 12 at 20:16
SongSong
18.6k21651
answered Jan 12 at 20:16
SongSong
18.6k21651
18.6k21651
$begingroup$
You deleted your comments that I corrected, didn't answer my question there, then provided an answer with zero acknowledgement. Not really a cool thing to do.
$endgroup$
– zhw.
Jan 13 at 20:04
$begingroup$
@zhw. I'm so sorry about that if it makes you feel bad. At first I did leave a comment appreciating your pointing out my mistake, but after a while I deleted it only because I thought you probably already read it. Again, I apologize for such a misbehavior.
$endgroup$
– Song
Jan 13 at 20:18
$begingroup$
OK, thanks for your comment here. In the future, at times, you might want to consider not deleting comments of yours even if they're mistaken. Often the resulting discussion can help readers.
$endgroup$
– zhw.
Jan 13 at 20:28
$begingroup$
@zhw Oh, I think it's a really good point. I'll keep it in my mind. Thank you!
$endgroup$
– Song
Jan 13 at 20:31
add a comment |
$begingroup$
You deleted your comments that I corrected, didn't answer my question there, then provided an answer with zero acknowledgement. Not really a cool thing to do.
$endgroup$
– zhw.
Jan 13 at 20:04
$begingroup$
@zhw. I'm so sorry about that if it makes you feel bad. At first I did leave a comment appreciating your pointing out my mistake, but after a while I deleted it only because I thought you probably already read it. Again, I apologize for such a misbehavior.
$endgroup$
– Song
Jan 13 at 20:18
$begingroup$
OK, thanks for your comment here. In the future, at times, you might want to consider not deleting comments of yours even if they're mistaken. Often the resulting discussion can help readers.
$endgroup$
– zhw.
Jan 13 at 20:28
$begingroup$
@zhw Oh, I think it's a really good point. I'll keep it in my mind. Thank you!
$endgroup$
– Song
Jan 13 at 20:31
$begingroup$
You deleted your comments that I corrected, didn't answer my question there, then provided an answer with zero acknowledgement. Not really a cool thing to do.
$endgroup$
– zhw.
Jan 13 at 20:04
$begingroup$
You deleted your comments that I corrected, didn't answer my question there, then provided an answer with zero acknowledgement. Not really a cool thing to do.
$endgroup$
– zhw.
Jan 13 at 20:04
$begingroup$
@zhw. I'm so sorry about that if it makes you feel bad. At first I did leave a comment appreciating your pointing out my mistake, but after a while I deleted it only because I thought you probably already read it. Again, I apologize for such a misbehavior.
$endgroup$
– Song
Jan 13 at 20:18
$begingroup$
@zhw. I'm so sorry about that if it makes you feel bad. At first I did leave a comment appreciating your pointing out my mistake, but after a while I deleted it only because I thought you probably already read it. Again, I apologize for such a misbehavior.
$endgroup$
– Song
Jan 13 at 20:18
$begingroup$
OK, thanks for your comment here. In the future, at times, you might want to consider not deleting comments of yours even if they're mistaken. Often the resulting discussion can help readers.
$endgroup$
– zhw.
Jan 13 at 20:28
$begingroup$
OK, thanks for your comment here. In the future, at times, you might want to consider not deleting comments of yours even if they're mistaken. Often the resulting discussion can help readers.
$endgroup$
– zhw.
Jan 13 at 20:28
$begingroup$
@zhw Oh, I think it's a really good point. I'll keep it in my mind. Thank you!
$endgroup$
– Song
Jan 13 at 20:31
$begingroup$
@zhw Oh, I think it's a really good point. I'll keep it in my mind. Thank you!
$endgroup$
– Song
Jan 13 at 20:31
add a comment |
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$begingroup$
@Song If the subset is Lebesgue measurable, then it is $sigma$-finite.
$endgroup$
– zhw.
Jan 12 at 19:32
$begingroup$
@Song are you confusing $sigma$-compact with $sigma$ -finite?
$endgroup$
– zhw.
Jan 12 at 19:47
$begingroup$
@zhw For some reason, I've read $sigma$-finite as $sigma$-compact. Thanks for pointing out my error!
$endgroup$
– Song
Jan 13 at 20:17