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










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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










    share|cite|improve this question
























      up vote
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      up vote
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      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










      share|cite|improve this question













      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






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      asked Nov 17 at 23:36









      Ricardo Freire

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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].






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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].






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              up vote
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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].






              share|cite|improve this answer























                up vote
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                up vote
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                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].






                share|cite|improve this answer












                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].







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                answered Nov 17 at 23:42









                Kavi Rama Murthy

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                43.5k31751






























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