If $(X, mathbb{X}, mu)$ space of finite measure. $f_nto f$ in measure implies $f_nto f$ almost uniformly.
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If $(X, mathbb{X}, mu)$ space of finite measure. Is true or false the following: $f_nto f$ in measure implies $f_nto f$ almost uniformly.
I know that convergence in measure implies that some subsequence of $ f_n $ converges almost evenly, but I am not able to prove this result, nor do I find a counterexample. Can someone give a tip
measure-theory
asked Nov 17 at 23:36
Ricardo Freire
383110
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If $(X, mathbb{X}, mu)$ space of finite measure. Is true or false the following: $f_nto f$ in measure implies $f_nto f$ almost uniformly.
I know that convergence in measure implies that some subsequence of $ f_n $ converges almost evenly, but I am not able to prove this result, nor do I find a counterexample. Can someone give a tip
measure-theory
asked Nov 17 at 23:36
Ricardo Freire
383110
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $(X, mathbb{X}, mu)$ space of finite measure. Is true or false the following: $f_nto f$ in measure implies $f_nto f$ almost uniformly.
I know that convergence in measure implies that some subsequence of $ f_n $ converges almost evenly, but I am not able to prove this result, nor do I find a counterexample. Can someone give a tip
measure-theory
asked Nov 17 at 23:36
Ricardo Freire
383110
If $(X, mathbb{X}, mu)$ space of finite measure. Is true or false the following: $f_nto f$ in measure implies $f_nto f$ almost uniformly.
I know that convergence in measure implies that some subsequence of $ f_n $ converges almost evenly, but I am not able to prove this result, nor do I find a counterexample. Can someone give a tip
measure-theory
measure-theory
asked Nov 17 at 23:36
Ricardo Freire
383110
asked Nov 17 at 23:36
Ricardo Freire
383110
asked Nov 17 at 23:36
Ricardo Freire
383110
asked Nov 17 at 23:36
Ricardo Freire
383110
asked Nov 17 at 23:36
Ricardo Freire
383110
383110
add a comment |
add a comment |
1 Answer
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False. There is a well known example of sequence that converges in measure but does not converge at any point. [Indicator functions of dyadic intervasl $(frac {i-1} {2^{n}},frac i {2^{n}})$ arranged in sequence with increasing order of $n$ is such a sequence].
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
False. There is a well known example of sequence that converges in measure but does not converge at any point. [Indicator functions of dyadic intervasl $(frac {i-1} {2^{n}},frac i {2^{n}})$ arranged in sequence with increasing order of $n$ is such a sequence].
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
add a comment |
up vote
1
down vote
False. There is a well known example of sequence that converges in measure but does not converge at any point. [Indicator functions of dyadic intervasl $(frac {i-1} {2^{n}},frac i {2^{n}})$ arranged in sequence with increasing order of $n$ is such a sequence].
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
add a comment |
up vote
1
down vote
up vote
1
down vote
False. There is a well known example of sequence that converges in measure but does not converge at any point. [Indicator functions of dyadic intervasl $(frac {i-1} {2^{n}},frac i {2^{n}})$ arranged in sequence with increasing order of $n$ is such a sequence].
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
False. There is a well known example of sequence that converges in measure but does not converge at any point. [Indicator functions of dyadic intervasl $(frac {i-1} {2^{n}},frac i {2^{n}})$ arranged in sequence with increasing order of $n$ is such a sequence].
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
answered Nov 17 at 23:42
Kavi Rama Murthy
41.4k31751
41.4k31751
add a comment |
add a comment |
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