Proving that every Cauchy sequence in measure converges in measure
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Let $(X,mathcal{A},mu)$ be a measure space and $(f_n)$ a sequence of real-valued functions on $X$ which is Cauchy in measure; that is, for any $epsilon>0$ there exists $Ninmathbb{N}$ such that for all $m,ngeq N$ we have $mu({xin X | |f_n(x)-f_m(x)|geq epsilon})<epsilon$. Prove that there is a function $f$ on $X$ to which $(f_n)$ converges to $f$ in measure.
(Convergence in measure: For any $epsilon>0,displaystylelim_{n to infty} mu({xin X | |f_n(x)-f(x)|geq epsilon})=0.$)
(This is not a homework problem.) I'm finding this very difficult, probably because the definitions are so complicated. I'm not sure how to approach it. Usually when one tries to prove 'Cauchy implies convergent'-type statements it's just a routine application of the completeness of $mathbb{R}$. One approach may be to establish to existence of a Cauchy subsequence for each $x$ but that's more of a guess rather than an idea inspired by understanding. It's just difficult to get a strong enough intuitive understanding for what's going on to solve it. Fixing an $x$ and constructing $f(x)$ individually is not easy as it seems to depend on all the other $f(x)$ as well (measure being a more global property). Any small hints would be appreciated.
Edit: Not sure if this is common thing on StackExchange but I wanted to make a copy of this question myself in order to get a hint for the problem instead of seeing the answer (and spoiling the problem).
real-analysis measure-theory cauchy-sequences
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
$endgroup$
|
show 2 more comments
$begingroup$
Let $(X,mathcal{A},mu)$ be a measure space and $(f_n)$ a sequence of real-valued functions on $X$ which is Cauchy in measure; that is, for any $epsilon>0$ there exists $Ninmathbb{N}$ such that for all $m,ngeq N$ we have $mu({xin X | |f_n(x)-f_m(x)|geq epsilon})<epsilon$. Prove that there is a function $f$ on $X$ to which $(f_n)$ converges to $f$ in measure.
(Convergence in measure: For any $epsilon>0,displaystylelim_{n to infty} mu({xin X | |f_n(x)-f(x)|geq epsilon})=0.$)
(This is not a homework problem.) I'm finding this very difficult, probably because the definitions are so complicated. I'm not sure how to approach it. Usually when one tries to prove 'Cauchy implies convergent'-type statements it's just a routine application of the completeness of $mathbb{R}$. One approach may be to establish to existence of a Cauchy subsequence for each $x$ but that's more of a guess rather than an idea inspired by understanding. It's just difficult to get a strong enough intuitive understanding for what's going on to solve it. Fixing an $x$ and constructing $f(x)$ individually is not easy as it seems to depend on all the other $f(x)$ as well (measure being a more global property). Any small hints would be appreciated.
Edit: Not sure if this is common thing on StackExchange but I wanted to make a copy of this question myself in order to get a hint for the problem instead of seeing the answer (and spoiling the problem).
real-analysis measure-theory cauchy-sequences
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
$endgroup$
$begingroup$
How do you feel about $infty$?
$endgroup$
– R. Burton
Dec 17 '18 at 14:27
1
$begingroup$
Could you please be more specific?
$endgroup$
– AlephNull
Dec 17 '18 at 14:35
$begingroup$
My first thought is to consider that the function $f_n(x)$ can be rewritten as $g(n,x)$, where $g:mathbb{N}timesmathbb{R}tomathbb{R}$, then consider the value of $g(infty, x)$ - but that requires $g(infty,x)$ to be defined.
$endgroup$
– R. Burton
Dec 17 '18 at 14:44
$begingroup$
Unfortunately, someone deleted their answer giving a nice sequence of hints for this problem and I am not sure why.
$endgroup$
– AlephNull
Dec 17 '18 at 19:18
1
$begingroup$
Convergence almost everywhere implies convergence in measure for finite measures. Your measure is not given to be a finite measure. That may be the reason the answer was deleted.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 23:55
|
show 2 more comments
$begingroup$
Let $(X,mathcal{A},mu)$ be a measure space and $(f_n)$ a sequence of real-valued functions on $X$ which is Cauchy in measure; that is, for any $epsilon>0$ there exists $Ninmathbb{N}$ such that for all $m,ngeq N$ we have $mu({xin X | |f_n(x)-f_m(x)|geq epsilon})<epsilon$. Prove that there is a function $f$ on $X$ to which $(f_n)$ converges to $f$ in measure.
(Convergence in measure: For any $epsilon>0,displaystylelim_{n to infty} mu({xin X | |f_n(x)-f(x)|geq epsilon})=0.$)
(This is not a homework problem.) I'm finding this very difficult, probably because the definitions are so complicated. I'm not sure how to approach it. Usually when one tries to prove 'Cauchy implies convergent'-type statements it's just a routine application of the completeness of $mathbb{R}$. One approach may be to establish to existence of a Cauchy subsequence for each $x$ but that's more of a guess rather than an idea inspired by understanding. It's just difficult to get a strong enough intuitive understanding for what's going on to solve it. Fixing an $x$ and constructing $f(x)$ individually is not easy as it seems to depend on all the other $f(x)$ as well (measure being a more global property). Any small hints would be appreciated.
Edit: Not sure if this is common thing on StackExchange but I wanted to make a copy of this question myself in order to get a hint for the problem instead of seeing the answer (and spoiling the problem).
real-analysis measure-theory cauchy-sequences
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
$endgroup$
Let $(X,mathcal{A},mu)$ be a measure space and $(f_n)$ a sequence of real-valued functions on $X$ which is Cauchy in measure; that is, for any $epsilon>0$ there exists $Ninmathbb{N}$ such that for all $m,ngeq N$ we have $mu({xin X | |f_n(x)-f_m(x)|geq epsilon})<epsilon$. Prove that there is a function $f$ on $X$ to which $(f_n)$ converges to $f$ in measure.
(Convergence in measure: For any $epsilon>0,displaystylelim_{n to infty} mu({xin X | |f_n(x)-f(x)|geq epsilon})=0.$)
(This is not a homework problem.) I'm finding this very difficult, probably because the definitions are so complicated. I'm not sure how to approach it. Usually when one tries to prove 'Cauchy implies convergent'-type statements it's just a routine application of the completeness of $mathbb{R}$. One approach may be to establish to existence of a Cauchy subsequence for each $x$ but that's more of a guess rather than an idea inspired by understanding. It's just difficult to get a strong enough intuitive understanding for what's going on to solve it. Fixing an $x$ and constructing $f(x)$ individually is not easy as it seems to depend on all the other $f(x)$ as well (measure being a more global property). Any small hints would be appreciated.
Edit: Not sure if this is common thing on StackExchange but I wanted to make a copy of this question myself in order to get a hint for the problem instead of seeing the answer (and spoiling the problem).
real-analysis measure-theory cauchy-sequences
real-analysis measure-theory cauchy-sequences
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
edited Dec 17 '18 at 14:15
AlephNull
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
asked Dec 17 '18 at 13:50
AlephNullAlephNull
503110
503110
$begingroup$
How do you feel about $infty$?
$endgroup$
– R. Burton
Dec 17 '18 at 14:27
1
$begingroup$
Could you please be more specific?
$endgroup$
– AlephNull
Dec 17 '18 at 14:35
$begingroup$
My first thought is to consider that the function $f_n(x)$ can be rewritten as $g(n,x)$, where $g:mathbb{N}timesmathbb{R}tomathbb{R}$, then consider the value of $g(infty, x)$ - but that requires $g(infty,x)$ to be defined.
$endgroup$
– R. Burton
Dec 17 '18 at 14:44
$begingroup$
Unfortunately, someone deleted their answer giving a nice sequence of hints for this problem and I am not sure why.
$endgroup$
– AlephNull
Dec 17 '18 at 19:18
1
$begingroup$
Convergence almost everywhere implies convergence in measure for finite measures. Your measure is not given to be a finite measure. That may be the reason the answer was deleted.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 23:55
|
show 2 more comments
$begingroup$
How do you feel about $infty$?
$endgroup$
– R. Burton
Dec 17 '18 at 14:27
1
$begingroup$
Could you please be more specific?
$endgroup$
– AlephNull
Dec 17 '18 at 14:35
$begingroup$
My first thought is to consider that the function $f_n(x)$ can be rewritten as $g(n,x)$, where $g:mathbb{N}timesmathbb{R}tomathbb{R}$, then consider the value of $g(infty, x)$ - but that requires $g(infty,x)$ to be defined.
$endgroup$
– R. Burton
Dec 17 '18 at 14:44
$begingroup$
Unfortunately, someone deleted their answer giving a nice sequence of hints for this problem and I am not sure why.
$endgroup$
– AlephNull
Dec 17 '18 at 19:18
1
$begingroup$
Convergence almost everywhere implies convergence in measure for finite measures. Your measure is not given to be a finite measure. That may be the reason the answer was deleted.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 23:55
$begingroup$
How do you feel about $infty$?
$endgroup$
– R. Burton
Dec 17 '18 at 14:27
$begingroup$
How do you feel about $infty$?
$endgroup$
– R. Burton
Dec 17 '18 at 14:27
1
1
$begingroup$
Could you please be more specific?
$endgroup$
– AlephNull
Dec 17 '18 at 14:35
$begingroup$
Could you please be more specific?
$endgroup$
– AlephNull
Dec 17 '18 at 14:35
$begingroup$
My first thought is to consider that the function $f_n(x)$ can be rewritten as $g(n,x)$, where $g:mathbb{N}timesmathbb{R}tomathbb{R}$, then consider the value of $g(infty, x)$ - but that requires $g(infty,x)$ to be defined.
$endgroup$
– R. Burton
Dec 17 '18 at 14:44
$begingroup$
My first thought is to consider that the function $f_n(x)$ can be rewritten as $g(n,x)$, where $g:mathbb{N}timesmathbb{R}tomathbb{R}$, then consider the value of $g(infty, x)$ - but that requires $g(infty,x)$ to be defined.
$endgroup$
– R. Burton
Dec 17 '18 at 14:44
$begingroup$
Unfortunately, someone deleted their answer giving a nice sequence of hints for this problem and I am not sure why.
$endgroup$
– AlephNull
Dec 17 '18 at 19:18
$begingroup$
Unfortunately, someone deleted their answer giving a nice sequence of hints for this problem and I am not sure why.
$endgroup$
– AlephNull
Dec 17 '18 at 19:18
1
1
$begingroup$
Convergence almost everywhere implies convergence in measure for finite measures. Your measure is not given to be a finite measure. That may be the reason the answer was deleted.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 23:55
$begingroup$
Convergence almost everywhere implies convergence in measure for finite measures. Your measure is not given to be a finite measure. That may be the reason the answer was deleted.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 23:55
|
show 2 more comments
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$begingroup$
How do you feel about $infty$?
$endgroup$
– R. Burton
Dec 17 '18 at 14:27
1
$begingroup$
Could you please be more specific?
$endgroup$
– AlephNull
Dec 17 '18 at 14:35
$begingroup$
My first thought is to consider that the function $f_n(x)$ can be rewritten as $g(n,x)$, where $g:mathbb{N}timesmathbb{R}tomathbb{R}$, then consider the value of $g(infty, x)$ - but that requires $g(infty,x)$ to be defined.
$endgroup$
– R. Burton
Dec 17 '18 at 14:44
$begingroup$
Unfortunately, someone deleted their answer giving a nice sequence of hints for this problem and I am not sure why.
$endgroup$
– AlephNull
Dec 17 '18 at 19:18
1
$begingroup$
Convergence almost everywhere implies convergence in measure for finite measures. Your measure is not given to be a finite measure. That may be the reason the answer was deleted.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 23:55