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Almost uniform convergence implies convergence in measure

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Almost uniform convergence implies convergence in measure

Almost uniform convergence of $f_n(x) = x^n$ on the interval $[0, 1]$?Uniform convergence and pointwise convergence, understandingQuantifiers in the definition of uniform convergenceDominated a.e. convergence implies almost uniform convergenceEquivalent definition of almost uniform convergenceHow to show that the given sequence of functions converges to $f$ almost everywhere, almost uniformly and in measure?Showing that the almost uniform limit of functions with bounded $L_infty$ norms is in $L_infty$Convergence in measure and almost everywhere proof

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1

$begingroup$

Let $(A,mathcalF,mu)$ be finite measure space and $f_n$ a sequence of finite real measurable functions so that $f_nrightarrow f$ a.e. We say $f_nrightarrow f$ almost uniformly if $epsilon>0$, there is $Esubseteq A$ such that $f_n rightarrow f$ uniformly on $E^c$ and $mu(E)<epsilon$.

I want to show that $f_nrightarrow f$ almost uniformly implies convergence in $mu$. For this, suppose not. Then
$$exists eta,epsilon>0:forall Nin mathbbN:exists n>N:mu(mid f_n-fmidgeqepsilon)geq eta, $$
i.e., for infinitely many points $nin mathbbN$. From the definition of almost uniform convergence, $exists E:mu(E)<eta$ and $f_nrightarrow f$ uniformly on $E^c$. Contradiction.

Question

It seems intuitive to me. But how to deduce this contradiction precisely?
I know that if $xin E$, then it must satify the negation of uniform convergence which is
$$exists epsilon>0:forall Nin mathbbN:exists n>N:mid f_n-fmidgeqepsilon.$$
Now, $x$ may not be in $f_n text does not converge in measure to f $ if $mu(mid f_n(x)-f(x)midgeqepsilon)<eta$. So I conclude that $$f_n text does not converge in measure to f subseteq E$$ implying that $eta>mu(E)geq eta$; a contradiction.

My argument seems right but also very inefficient. How could you express this idea as clean as possible?

Thanks!

real-analysis measure-theory proof-writing

share|cite|improve this question

edited 7 hours ago

Gabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

asked 8 hours ago

DanmatDanmat

4161010 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Let $(A,mathcalF,mu)$ be finite measure space and $f_n$ a sequence of finite real measurable functions so that $f_nrightarrow f$ a.e. We say $f_nrightarrow f$ almost uniformly if $epsilon>0$, there is $Esubseteq A$ such that $f_n rightarrow f$ uniformly on $E^c$ and $mu(E)<epsilon$.

I want to show that $f_nrightarrow f$ almost uniformly implies convergence in $mu$. For this, suppose not. Then
$$exists eta,epsilon>0:forall Nin mathbbN:exists n>N:mu(mid f_n-fmidgeqepsilon)geq eta, $$
i.e., for infinitely many points $nin mathbbN$. From the definition of almost uniform convergence, $exists E:mu(E)<eta$ and $f_nrightarrow f$ uniformly on $E^c$. Contradiction.

Question

It seems intuitive to me. But how to deduce this contradiction precisely?
I know that if $xin E$, then it must satify the negation of uniform convergence which is
$$exists epsilon>0:forall Nin mathbbN:exists n>N:mid f_n-fmidgeqepsilon.$$
Now, $x$ may not be in $f_n text does not converge in measure to f $ if $mu(mid f_n(x)-f(x)midgeqepsilon)<eta$. So I conclude that $$f_n text does not converge in measure to f subseteq E$$ implying that $eta>mu(E)geq eta$; a contradiction.

My argument seems right but also very inefficient. How could you express this idea as clean as possible?

Thanks!

real-analysis measure-theory proof-writing

share|cite|improve this question

edited 7 hours ago

Gabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

asked 8 hours ago

DanmatDanmat

4161010 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $(A,mathcalF,mu)$ be finite measure space and $f_n$ a sequence of finite real measurable functions so that $f_nrightarrow f$ a.e. We say $f_nrightarrow f$ almost uniformly if $epsilon>0$, there is $Esubseteq A$ such that $f_n rightarrow f$ uniformly on $E^c$ and $mu(E)<epsilon$.

I want to show that $f_nrightarrow f$ almost uniformly implies convergence in $mu$. For this, suppose not. Then
$$exists eta,epsilon>0:forall Nin mathbbN:exists n>N:mu(mid f_n-fmidgeqepsilon)geq eta, $$
i.e., for infinitely many points $nin mathbbN$. From the definition of almost uniform convergence, $exists E:mu(E)<eta$ and $f_nrightarrow f$ uniformly on $E^c$. Contradiction.

Question

It seems intuitive to me. But how to deduce this contradiction precisely?
I know that if $xin E$, then it must satify the negation of uniform convergence which is
$$exists epsilon>0:forall Nin mathbbN:exists n>N:mid f_n-fmidgeqepsilon.$$
Now, $x$ may not be in $f_n text does not converge in measure to f $ if $mu(mid f_n(x)-f(x)midgeqepsilon)<eta$. So I conclude that $$f_n text does not converge in measure to f subseteq E$$ implying that $eta>mu(E)geq eta$; a contradiction.

My argument seems right but also very inefficient. How could you express this idea as clean as possible?

Thanks!

real-analysis measure-theory proof-writing

share|cite|improve this question

edited 7 hours ago

Gabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

asked 8 hours ago

DanmatDanmat

4161010 bronze badges

$endgroup$

share|cite|improve this question

edited 7 hours ago

Gabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

asked 8 hours ago

DanmatDanmat

4161010 bronze badges

share|cite|improve this question

edited 7 hours ago

Gabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

asked 8 hours ago

DanmatDanmat

4161010 bronze badges

edited 7 hours ago

Gabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

edited 7 hours ago

Gabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

asked 8 hours ago

DanmatDanmat

4161010 bronze badges

asked 8 hours ago

DanmatDanmat

4161010 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

Here's a way without going for contradiction: Let $epsilon >0$ be fixed and consider some $delta >0$.

By almost uniform convergence, there exists some $E$ with $mu(E)leq delta$ and some $N$ such that $ngeq Nimplies forall xin E^c, |f_n(x)-f(x)|< epsilon$.

For $ngeq N$, note the inclusion $(|f_n-f|geq epsilon) subset E$, hence $mu((|f_n-f|geq epsilon))leq mu(E)leq delta$.

Hence $forall delta>0, exists N, ngeq N implies mu((|f_n-f|geq epsilon))leq delta$. Thus $ mu((|f_n-f|geq epsilon)) to 0$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gabriel RomonGabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! The inclusion $mid f_n-fmidgeq epsilonsubset E$ that I don't understand. What is this infinity-norm that you wrote?
  $endgroup$
  – Danmat
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat I removed the infinity norm (which denotes the sup-norm), do you understand now ?
  $endgroup$
  – Gabriel Romon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It has something to do with the fact that uniform convergence implies pointwise convergence? And then the complement of the latter is contained in the complement of the former.
  $endgroup$
  – Danmat
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat I just wrote the definition of uniform convergence on $E^c$. Pointwise convergence is not used in the proof.
  $endgroup$
  – Gabriel Romon
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But, $E$ is constituted of elements in $A$ such that $f_n rightarrow f$ does not converge uniformly, right?
  $endgroup$
  – Danmat
  7 hours ago
  
  

  
  
  
  

 | 
show 4 more comments

2

$begingroup$

This is proved more succinctly by a direct proof.

Given $varepsilon > 0,$ let $E$ be as in the definition of almost-uniform convergence. Then there is some $N$ such that $n geq N$ implies
$$mu(E^c cap ) = 0$$
What does this imply for $mu(|f_n-f|geq varepsilon)?$

share|cite|improve this answer

answered 8 hours ago

Brian MoehringBrian Moehring

1,61922 silver badges99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How to conclude that $mid f_n-fmid subseteq E$?
  $endgroup$
  – Danmat
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat Forget about $E$ for a second. On $E^c,$ the functions converge uniformly. This means that there will be some $N$ such that $n geq N$ implies $|f_n - f| < varepsilon.$ In particular, on $E^c,$ $$n geq N implies geq varepsilon = emptyset.$$ This directly means $$E^c cap geq varepsilon = emptyset.$$ If you want, then you can use the property of sets that $$A^c cap B = emptyset iff B subseteq A$$
  $endgroup$
  – Brian Moehring
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice! Now, it's all clear to me!
  $endgroup$
  – Danmat
  7 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Here's a way without going for contradiction: Let $epsilon >0$ be fixed and consider some $delta >0$.

By almost uniform convergence, there exists some $E$ with $mu(E)leq delta$ and some $N$ such that $ngeq Nimplies forall xin E^c, |f_n(x)-f(x)|< epsilon$.

For $ngeq N$, note the inclusion $(|f_n-f|geq epsilon) subset E$, hence $mu((|f_n-f|geq epsilon))leq mu(E)leq delta$.

Hence $forall delta>0, exists N, ngeq N implies mu((|f_n-f|geq epsilon))leq delta$. Thus $ mu((|f_n-f|geq epsilon)) to 0$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gabriel RomonGabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! The inclusion $mid f_n-fmidgeq epsilonsubset E$ that I don't understand. What is this infinity-norm that you wrote?
  $endgroup$
  – Danmat
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat I removed the infinity norm (which denotes the sup-norm), do you understand now ?
  $endgroup$
  – Gabriel Romon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It has something to do with the fact that uniform convergence implies pointwise convergence? And then the complement of the latter is contained in the complement of the former.
  $endgroup$
  – Danmat
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat I just wrote the definition of uniform convergence on $E^c$. Pointwise convergence is not used in the proof.
  $endgroup$
  – Gabriel Romon
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But, $E$ is constituted of elements in $A$ such that $f_n rightarrow f$ does not converge uniformly, right?
  $endgroup$
  – Danmat
  7 hours ago
  
  

  
  
  
  

 | 
show 4 more comments

4

$begingroup$

Here's a way without going for contradiction: Let $epsilon >0$ be fixed and consider some $delta >0$.

By almost uniform convergence, there exists some $E$ with $mu(E)leq delta$ and some $N$ such that $ngeq Nimplies forall xin E^c, |f_n(x)-f(x)|< epsilon$.

For $ngeq N$, note the inclusion $(|f_n-f|geq epsilon) subset E$, hence $mu((|f_n-f|geq epsilon))leq mu(E)leq delta$.

Hence $forall delta>0, exists N, ngeq N implies mu((|f_n-f|geq epsilon))leq delta$. Thus $ mu((|f_n-f|geq epsilon)) to 0$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gabriel RomonGabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! The inclusion $mid f_n-fmidgeq epsilonsubset E$ that I don't understand. What is this infinity-norm that you wrote?
  $endgroup$
  – Danmat
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat I removed the infinity norm (which denotes the sup-norm), do you understand now ?
  $endgroup$
  – Gabriel Romon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It has something to do with the fact that uniform convergence implies pointwise convergence? And then the complement of the latter is contained in the complement of the former.
  $endgroup$
  – Danmat
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat I just wrote the definition of uniform convergence on $E^c$. Pointwise convergence is not used in the proof.
  $endgroup$
  – Gabriel Romon
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But, $E$ is constituted of elements in $A$ such that $f_n rightarrow f$ does not converge uniformly, right?
  $endgroup$
  – Danmat
  7 hours ago
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Here's a way without going for contradiction: Let $epsilon >0$ be fixed and consider some $delta >0$.

By almost uniform convergence, there exists some $E$ with $mu(E)leq delta$ and some $N$ such that $ngeq Nimplies forall xin E^c, |f_n(x)-f(x)|< epsilon$.

For $ngeq N$, note the inclusion $(|f_n-f|geq epsilon) subset E$, hence $mu((|f_n-f|geq epsilon))leq mu(E)leq delta$.

Hence $forall delta>0, exists N, ngeq N implies mu((|f_n-f|geq epsilon))leq delta$. Thus $ mu((|f_n-f|geq epsilon)) to 0$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gabriel RomonGabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Gabriel RomonGabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

answered 8 hours ago

Gabriel RomonGabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

answered 8 hours ago

Gabriel RomonGabriel Romon

19.2k55 gold badges3737 silver badges8888 bronze badges

2

$begingroup$

This is proved more succinctly by a direct proof.

Given $varepsilon > 0,$ let $E$ be as in the definition of almost-uniform convergence. Then there is some $N$ such that $n geq N$ implies
$$mu(E^c cap ) = 0$$
What does this imply for $mu(|f_n-f|geq varepsilon)?$

share|cite|improve this answer

answered 8 hours ago

Brian MoehringBrian Moehring

1,61922 silver badges99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How to conclude that $mid f_n-fmid subseteq E$?
  $endgroup$
  – Danmat
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat Forget about $E$ for a second. On $E^c,$ the functions converge uniformly. This means that there will be some $N$ such that $n geq N$ implies $|f_n - f| < varepsilon.$ In particular, on $E^c,$ $$n geq N implies geq varepsilon = emptyset.$$ This directly means $$E^c cap geq varepsilon = emptyset.$$ If you want, then you can use the property of sets that $$A^c cap B = emptyset iff B subseteq A$$
  $endgroup$
  – Brian Moehring
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice! Now, it's all clear to me!
  $endgroup$
  – Danmat
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

This is proved more succinctly by a direct proof.

Given $varepsilon > 0,$ let $E$ be as in the definition of almost-uniform convergence. Then there is some $N$ such that $n geq N$ implies
$$mu(E^c cap ) = 0$$
What does this imply for $mu(|f_n-f|geq varepsilon)?$

share|cite|improve this answer

answered 8 hours ago

Brian MoehringBrian Moehring

1,61922 silver badges99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How to conclude that $mid f_n-fmid subseteq E$?
  $endgroup$
  – Danmat
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Danmat Forget about $E$ for a second. On $E^c,$ the functions converge uniformly. This means that there will be some $N$ such that $n geq N$ implies $|f_n - f| < varepsilon.$ In particular, on $E^c,$ $$n geq N implies geq varepsilon = emptyset.$$ This directly means $$E^c cap geq varepsilon = emptyset.$$ If you want, then you can use the property of sets that $$A^c cap B = emptyset iff B subseteq A$$
  $endgroup$
  – Brian Moehring
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice! Now, it's all clear to me!
  $endgroup$
  – Danmat
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is proved more succinctly by a direct proof.

Given $varepsilon > 0,$ let $E$ be as in the definition of almost-uniform convergence. Then there is some $N$ such that $n geq N$ implies
$$mu(E^c cap ) = 0$$
What does this imply for $mu(|f_n-f|geq varepsilon)?$

share|cite|improve this answer

answered 8 hours ago

Brian MoehringBrian Moehring

1,61922 silver badges99 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Brian MoehringBrian Moehring

1,61922 silver badges99 bronze badges

answered 8 hours ago

Brian MoehringBrian Moehring

1,61922 silver badges99 bronze badges

answered 8 hours ago

Brian MoehringBrian Moehring

1,61922 silver badges99 bronze badges