Finite measure space suspected property
up vote
0
down vote
favorite
Prove or disprove
In a measure space,
if $mu(X)<infty $ and $lim_{nrightarrowinfty}mu (A_n)=0$ then $mu (bigcap_n bigcup_{kgeq n} A_k)=0$
I have some examples supporting this but could not prove or give any counter examples.
I already know about the theorem that $sum_n mu (A_n)<infty$ implies $mu (bigcap_n bigcup_{kgeq n} A_k)=0$, but it did not help me.
measure-theory
asked Nov 17 at 12:35
Utku Okur
8614
add a comment |
up vote
0
down vote
favorite
Prove or disprove
In a measure space,
if $mu(X)<infty $ and $lim_{nrightarrowinfty}mu (A_n)=0$ then $mu (bigcap_n bigcup_{kgeq n} A_k)=0$
I have some examples supporting this but could not prove or give any counter examples.
I already know about the theorem that $sum_n mu (A_n)<infty$ implies $mu (bigcap_n bigcup_{kgeq n} A_k)=0$, but it did not help me.
measure-theory
asked Nov 17 at 12:35
Utku Okur
8614
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove or disprove
In a measure space,
if $mu(X)<infty $ and $lim_{nrightarrowinfty}mu (A_n)=0$ then $mu (bigcap_n bigcup_{kgeq n} A_k)=0$
I have some examples supporting this but could not prove or give any counter examples.
I already know about the theorem that $sum_n mu (A_n)<infty$ implies $mu (bigcap_n bigcup_{kgeq n} A_k)=0$, but it did not help me.
measure-theory
asked Nov 17 at 12:35
Utku Okur
8614
Prove or disprove
In a measure space,
if $mu(X)<infty $ and $lim_{nrightarrowinfty}mu (A_n)=0$ then $mu (bigcap_n bigcup_{kgeq n} A_k)=0$
I have some examples supporting this but could not prove or give any counter examples.
I already know about the theorem that $sum_n mu (A_n)<infty$ implies $mu (bigcap_n bigcup_{kgeq n} A_k)=0$, but it did not help me.
measure-theory
measure-theory
asked Nov 17 at 12:35
Utku Okur
8614
asked Nov 17 at 12:35
Utku Okur
8614
asked Nov 17 at 12:35
Utku Okur
8614
asked Nov 17 at 12:35
Utku Okur
8614
asked Nov 17 at 12:35
Utku Okur
8614
8614
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $X=[0,1]$ be equipped with Lebesgue measure.
For $A_1,A_2,cdots$ take sequence:$[0,1],[0,frac12],[frac12,0],[0,frac13],[frac13,frac23],[frac23,1],[0,frac14],dots$
Then $lambda(A_n)to0$ but $bigcap_{n=1}^{infty}bigcup_{k=n}^{infty}A_n=[0,1]$ (i.e. for every $xin[0,1]$ we have $xin A_n$ infinitely often) so has measure $1$.
answered Nov 17 at 12:44
drhab
94.9k543125
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $X=[0,1]$ be equipped with Lebesgue measure.
For $A_1,A_2,cdots$ take sequence:$[0,1],[0,frac12],[frac12,0],[0,frac13],[frac13,frac23],[frac23,1],[0,frac14],dots$
Then $lambda(A_n)to0$ but $bigcap_{n=1}^{infty}bigcup_{k=n}^{infty}A_n=[0,1]$ (i.e. for every $xin[0,1]$ we have $xin A_n$ infinitely often) so has measure $1$.
answered Nov 17 at 12:44
drhab
94.9k543125
add a comment |
up vote
2
down vote
accepted
Let $X=[0,1]$ be equipped with Lebesgue measure.
For $A_1,A_2,cdots$ take sequence:$[0,1],[0,frac12],[frac12,0],[0,frac13],[frac13,frac23],[frac23,1],[0,frac14],dots$
Then $lambda(A_n)to0$ but $bigcap_{n=1}^{infty}bigcup_{k=n}^{infty}A_n=[0,1]$ (i.e. for every $xin[0,1]$ we have $xin A_n$ infinitely often) so has measure $1$.
answered Nov 17 at 12:44
drhab
94.9k543125
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $X=[0,1]$ be equipped with Lebesgue measure.
For $A_1,A_2,cdots$ take sequence:$[0,1],[0,frac12],[frac12,0],[0,frac13],[frac13,frac23],[frac23,1],[0,frac14],dots$
Then $lambda(A_n)to0$ but $bigcap_{n=1}^{infty}bigcup_{k=n}^{infty}A_n=[0,1]$ (i.e. for every $xin[0,1]$ we have $xin A_n$ infinitely often) so has measure $1$.
answered Nov 17 at 12:44
drhab
94.9k543125
Let $X=[0,1]$ be equipped with Lebesgue measure.
For $A_1,A_2,cdots$ take sequence:$[0,1],[0,frac12],[frac12,0],[0,frac13],[frac13,frac23],[frac23,1],[0,frac14],dots$
Then $lambda(A_n)to0$ but $bigcap_{n=1}^{infty}bigcup_{k=n}^{infty}A_n=[0,1]$ (i.e. for every $xin[0,1]$ we have $xin A_n$ infinitely often) so has measure $1$.
answered Nov 17 at 12:44
drhab
94.9k543125
answered Nov 17 at 12:44
drhab
94.9k543125
answered Nov 17 at 12:44
drhab
94.9k543125
answered Nov 17 at 12:44
drhab
94.9k543125
94.9k543125
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002318%2ffinite-measure-space-suspected-property%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
