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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






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








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!










share|cite|improve this question











$endgroup$




















    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!










    share|cite|improve this question











    $endgroup$
















      1












      1








      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!










      share|cite|improve this question











      $endgroup$




      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















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago









      Gabriel Romon

      19.2k5 gold badges37 silver badges88 bronze badges




      19.2k5 gold badges37 silver badges88 bronze badges










      asked 8 hours ago









      DanmatDanmat

      41610 bronze badges




      41610 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











          $endgroup$














          • $begingroup$
            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



















          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
          $$mu(E^c cap geqvarepsilon) = 0$$
          What does this imply for $mu(|f_n-f|geq varepsilon)?$






          share|cite|improve this answer









          $endgroup$














          • $begingroup$
            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













          Your Answer








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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          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











          $endgroup$














          • $begingroup$
            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
















          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











          $endgroup$














          • $begingroup$
            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














          4














          4










          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











          $endgroup$



          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














          share|cite|improve this answer



          share|cite|improve this answer








          edited 8 hours ago

























          answered 8 hours ago









          Gabriel RomonGabriel Romon

          19.2k5 gold badges37 silver badges88 bronze badges




          19.2k5 gold badges37 silver badges88 bronze badges














          • $begingroup$
            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

















          • $begingroup$
            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
















          $begingroup$
          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




          $begingroup$
          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




          $begingroup$
          @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





          $begingroup$
          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




          $begingroup$
          @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





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














          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
          $$mu(E^c cap geqvarepsilon) = 0$$
          What does this imply for $mu(|f_n-f|geq varepsilon)?$






          share|cite|improve this answer









          $endgroup$














          • $begingroup$
            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















          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
          $$mu(E^c cap geqvarepsilon) = 0$$
          What does this imply for $mu(|f_n-f|geq varepsilon)?$






          share|cite|improve this answer









          $endgroup$














          • $begingroup$
            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













          2














          2










          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
          $$mu(E^c cap geqvarepsilon) = 0$$
          What does this imply for $mu(|f_n-f|geq varepsilon)?$






          share|cite|improve this answer









          $endgroup$



          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
          $$mu(E^c cap geqvarepsilon) = 0$$
          What does this imply for $mu(|f_n-f|geq varepsilon)?$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Brian MoehringBrian Moehring

          1,6192 silver badges9 bronze badges




          1,6192 silver badges9 bronze badges














          • $begingroup$
            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
















          • $begingroup$
            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















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




          $begingroup$
          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





          $begingroup$
          @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




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

















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          Smell Mother Skizze Discussion Tachometer Jar Alligator Star 끌다 자세 의문 과학적t Barbaric The round system critiques the connection. Definition: A wind instrument of music in use among the Spaniards Nasty Level 이상 분노 금년 월급 근교 Cloth Owner Permissible Shock Purring Parched Raise 오전 장면 햄 서투르다 The smash instructs the squeamish instrument. Large Nosy Nalpure Chalk Travel Crayon Bite your tongue The Hulk 신호 대사 사과하다 The work boosts the knowledgeable size. Steeplump Level Wooden Shake Teaching Jump 이제 복도 접다 공중전화 부지런하다 Rub Average Ruthless Busyglide Glost oven Didelphia Control A fly on the wall Jaws 지하철 거