“Almost uniformly convergent” Implies “Uniformly convergent almost everywhere” . Is there something...












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I find a tricky proof shows "almost uniformly convergent" implies "uniformly convergent almost everywhere". I know it is wrong, for there is a counterexample. Can anyone help me why this proof is wrong?



I got the inspiration from the proof that shows "almost uniformly convergent" implies "convergent almost everywhere":



It says $forall epsilon, exists B_epsilon, mu(B_epsilon)<epsilon$ outside of which $f_n$ converges uniformly to $f$ and prove $Bequivbigcap_{ninmathbb{N}}B_{frac{1}{n}}$ is zero-measure. I think in this case, outside $B$, $f_n to f$ no only converges pointwise (as the proof says) but also converges uniformly (for it is true outside any $B_frac{1}{n}$). Since $mu(B)=0$, it implies $f_n to f$ converges uniformly almost everywhere.






real-analysis measure-theory almost-everywhere







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    1












    $begingroup$


    I find a tricky proof shows "almost uniformly convergent" implies "uniformly convergent almost everywhere". I know it is wrong, for there is a counterexample. Can anyone help me why this proof is wrong?



    I got the inspiration from the proof that shows "almost uniformly convergent" implies "convergent almost everywhere":



    It says $forall epsilon, exists B_epsilon, mu(B_epsilon)<epsilon$ outside of which $f_n$ converges uniformly to $f$ and prove $Bequivbigcap_{ninmathbb{N}}B_{frac{1}{n}}$ is zero-measure. I think in this case, outside $B$, $f_n to f$ no only converges pointwise (as the proof says) but also converges uniformly (for it is true outside any $B_frac{1}{n}$). Since $mu(B)=0$, it implies $f_n to f$ converges uniformly almost everywhere.






    real-analysis measure-theory almost-everywhere







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I find a tricky proof shows "almost uniformly convergent" implies "uniformly convergent almost everywhere". I know it is wrong, for there is a counterexample. Can anyone help me why this proof is wrong?



      I got the inspiration from the proof that shows "almost uniformly convergent" implies "convergent almost everywhere":



      It says $forall epsilon, exists B_epsilon, mu(B_epsilon)<epsilon$ outside of which $f_n$ converges uniformly to $f$ and prove $Bequivbigcap_{ninmathbb{N}}B_{frac{1}{n}}$ is zero-measure. I think in this case, outside $B$, $f_n to f$ no only converges pointwise (as the proof says) but also converges uniformly (for it is true outside any $B_frac{1}{n}$). Since $mu(B)=0$, it implies $f_n to f$ converges uniformly almost everywhere.






      real-analysis measure-theory almost-everywhere







      share|cite|improve this question









      $endgroup$




      I find a tricky proof shows "almost uniformly convergent" implies "uniformly convergent almost everywhere". I know it is wrong, for there is a counterexample. Can anyone help me why this proof is wrong?



      I got the inspiration from the proof that shows "almost uniformly convergent" implies "convergent almost everywhere":



      It says $forall epsilon, exists B_epsilon, mu(B_epsilon)<epsilon$ outside of which $f_n$ converges uniformly to $f$ and prove $Bequivbigcap_{ninmathbb{N}}B_{frac{1}{n}}$ is zero-measure. I think in this case, outside $B$, $f_n to f$ no only converges pointwise (as the proof says) but also converges uniformly (for it is true outside any $B_frac{1}{n}$). Since $mu(B)=0$, it implies $f_n to f$ converges uniformly almost everywhere.






      real-analysis measure-theory almost-everywhere




      real-analysis measure-theory almost-everywhere






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 28 '18 at 13:59









      XuchuangXuchuang

      82




      82






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Consider a simple example:



          $x^n to 0$ almost uniformly on $[0,1]$. Indeed, for every $varepsilon > 0$ we have $x^n to 0$ uniformly on $B_varepsilon^c = [0,1-varepsilon]$ where $B_varepsilon = [1-varepsilon, 1]$ has measure $varepsilon$.



          However on $$B^c = left(bigcap_{n=1}^infty B_{frac1n}right)^c = left(bigcap_{n=1}^infty left[1-frac1n,1right]right)^c = {1}^c = [0,1rangle$$ the convergence $x^n to 0$ is not uniform. It is only uniform on segments $B_{frac1n}^c=left[0,1-frac1nright]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:41












          • $begingroup$
            @Xuchuang On $B_{frac1n}^c$ we have $f_n to f$ uniformly, and in particular pointwise. Then for every $x in B^c$, we have $x in B_{frac1n}^c$ for some $n in mathbb{N}$ so $f_n(x) to f(x)$. Since $x$ was arbitrary, we conclude $f_n to f$ pointwise on $B^c$. We cannot conclude $f_n to f$ uniformly as the example shows.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:12












          • $begingroup$
            It's sort of like how the function $x mapsto frac1x$ is uniformly continuous on all segments $[a,b] subseteq langle 0, inftyrangle$ but it is not uniformly continuous on entire $langle 0,inftyrangle$.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:14












          • $begingroup$
            It makes sense. $x$ and $epsilon$ are both arbitrary in uniformly convergence. Thank you very much.
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:02



















          0












          $begingroup$

          You have that ${f_n}$ converges uniformly on each set $B_{frac 1n}^c$. There is no reason why it should converge uniformly on $displaystyle bigcup_n B_{frac 1n}^c$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $bigcup_n B_{frac{1}{n}}^c$. Is there a difference between them.
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:16










          • $begingroup$
            That isn't to hard to see. If a point $x$ belongs to $bigcup_n B^c_{frac 1n}$, then there exists (at least one) index $n$ with $x in B_{frac 1n}^c$. This means that $f_n(x) to f(x)$.
            $endgroup$
            – Umberto P.
            Dec 28 '18 at 14:18












          • $begingroup$
            But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:27










          • $begingroup$
            You are right. The wrong place is when unionizing those $B_{frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you!
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:08













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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Consider a simple example:



          $x^n to 0$ almost uniformly on $[0,1]$. Indeed, for every $varepsilon > 0$ we have $x^n to 0$ uniformly on $B_varepsilon^c = [0,1-varepsilon]$ where $B_varepsilon = [1-varepsilon, 1]$ has measure $varepsilon$.



          However on $$B^c = left(bigcap_{n=1}^infty B_{frac1n}right)^c = left(bigcap_{n=1}^infty left[1-frac1n,1right]right)^c = {1}^c = [0,1rangle$$ the convergence $x^n to 0$ is not uniform. It is only uniform on segments $B_{frac1n}^c=left[0,1-frac1nright]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:41












          • $begingroup$
            @Xuchuang On $B_{frac1n}^c$ we have $f_n to f$ uniformly, and in particular pointwise. Then for every $x in B^c$, we have $x in B_{frac1n}^c$ for some $n in mathbb{N}$ so $f_n(x) to f(x)$. Since $x$ was arbitrary, we conclude $f_n to f$ pointwise on $B^c$. We cannot conclude $f_n to f$ uniformly as the example shows.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:12












          • $begingroup$
            It's sort of like how the function $x mapsto frac1x$ is uniformly continuous on all segments $[a,b] subseteq langle 0, inftyrangle$ but it is not uniformly continuous on entire $langle 0,inftyrangle$.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:14












          • $begingroup$
            It makes sense. $x$ and $epsilon$ are both arbitrary in uniformly convergence. Thank you very much.
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:02
















          1












          $begingroup$

          Consider a simple example:



          $x^n to 0$ almost uniformly on $[0,1]$. Indeed, for every $varepsilon > 0$ we have $x^n to 0$ uniformly on $B_varepsilon^c = [0,1-varepsilon]$ where $B_varepsilon = [1-varepsilon, 1]$ has measure $varepsilon$.



          However on $$B^c = left(bigcap_{n=1}^infty B_{frac1n}right)^c = left(bigcap_{n=1}^infty left[1-frac1n,1right]right)^c = {1}^c = [0,1rangle$$ the convergence $x^n to 0$ is not uniform. It is only uniform on segments $B_{frac1n}^c=left[0,1-frac1nright]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:41












          • $begingroup$
            @Xuchuang On $B_{frac1n}^c$ we have $f_n to f$ uniformly, and in particular pointwise. Then for every $x in B^c$, we have $x in B_{frac1n}^c$ for some $n in mathbb{N}$ so $f_n(x) to f(x)$. Since $x$ was arbitrary, we conclude $f_n to f$ pointwise on $B^c$. We cannot conclude $f_n to f$ uniformly as the example shows.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:12












          • $begingroup$
            It's sort of like how the function $x mapsto frac1x$ is uniformly continuous on all segments $[a,b] subseteq langle 0, inftyrangle$ but it is not uniformly continuous on entire $langle 0,inftyrangle$.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:14












          • $begingroup$
            It makes sense. $x$ and $epsilon$ are both arbitrary in uniformly convergence. Thank you very much.
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:02














          1












          1








          1





          $begingroup$

          Consider a simple example:



          $x^n to 0$ almost uniformly on $[0,1]$. Indeed, for every $varepsilon > 0$ we have $x^n to 0$ uniformly on $B_varepsilon^c = [0,1-varepsilon]$ where $B_varepsilon = [1-varepsilon, 1]$ has measure $varepsilon$.



          However on $$B^c = left(bigcap_{n=1}^infty B_{frac1n}right)^c = left(bigcap_{n=1}^infty left[1-frac1n,1right]right)^c = {1}^c = [0,1rangle$$ the convergence $x^n to 0$ is not uniform. It is only uniform on segments $B_{frac1n}^c=left[0,1-frac1nright]$.






          share|cite|improve this answer









          $endgroup$



          Consider a simple example:



          $x^n to 0$ almost uniformly on $[0,1]$. Indeed, for every $varepsilon > 0$ we have $x^n to 0$ uniformly on $B_varepsilon^c = [0,1-varepsilon]$ where $B_varepsilon = [1-varepsilon, 1]$ has measure $varepsilon$.



          However on $$B^c = left(bigcap_{n=1}^infty B_{frac1n}right)^c = left(bigcap_{n=1}^infty left[1-frac1n,1right]right)^c = {1}^c = [0,1rangle$$ the convergence $x^n to 0$ is not uniform. It is only uniform on segments $B_{frac1n}^c=left[0,1-frac1nright]$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 14:13









          mechanodroidmechanodroid

          27.8k62447




          27.8k62447












          • $begingroup$
            Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:41












          • $begingroup$
            @Xuchuang On $B_{frac1n}^c$ we have $f_n to f$ uniformly, and in particular pointwise. Then for every $x in B^c$, we have $x in B_{frac1n}^c$ for some $n in mathbb{N}$ so $f_n(x) to f(x)$. Since $x$ was arbitrary, we conclude $f_n to f$ pointwise on $B^c$. We cannot conclude $f_n to f$ uniformly as the example shows.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:12












          • $begingroup$
            It's sort of like how the function $x mapsto frac1x$ is uniformly continuous on all segments $[a,b] subseteq langle 0, inftyrangle$ but it is not uniformly continuous on entire $langle 0,inftyrangle$.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:14












          • $begingroup$
            It makes sense. $x$ and $epsilon$ are both arbitrary in uniformly convergence. Thank you very much.
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:02


















          • $begingroup$
            Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:41












          • $begingroup$
            @Xuchuang On $B_{frac1n}^c$ we have $f_n to f$ uniformly, and in particular pointwise. Then for every $x in B^c$, we have $x in B_{frac1n}^c$ for some $n in mathbb{N}$ so $f_n(x) to f(x)$. Since $x$ was arbitrary, we conclude $f_n to f$ pointwise on $B^c$. We cannot conclude $f_n to f$ uniformly as the example shows.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:12












          • $begingroup$
            It's sort of like how the function $x mapsto frac1x$ is uniformly continuous on all segments $[a,b] subseteq langle 0, inftyrangle$ but it is not uniformly continuous on entire $langle 0,inftyrangle$.
            $endgroup$
            – mechanodroid
            Dec 28 '18 at 15:14












          • $begingroup$
            It makes sense. $x$ and $epsilon$ are both arbitrary in uniformly convergence. Thank you very much.
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:02
















          $begingroup$
          Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference?
          $endgroup$
          – Xuchuang
          Dec 28 '18 at 14:41






          $begingroup$
          Thanks for the clear example. It helps. As Umberto says in another answer, that is the wrong place. But I am a little confused about why combining all $B_{frac{1}{n}}$ is right for pointwise convergence but not for uniformly convergence. Can you tell me their difference?
          $endgroup$
          – Xuchuang
          Dec 28 '18 at 14:41














          $begingroup$
          @Xuchuang On $B_{frac1n}^c$ we have $f_n to f$ uniformly, and in particular pointwise. Then for every $x in B^c$, we have $x in B_{frac1n}^c$ for some $n in mathbb{N}$ so $f_n(x) to f(x)$. Since $x$ was arbitrary, we conclude $f_n to f$ pointwise on $B^c$. We cannot conclude $f_n to f$ uniformly as the example shows.
          $endgroup$
          – mechanodroid
          Dec 28 '18 at 15:12






          $begingroup$
          @Xuchuang On $B_{frac1n}^c$ we have $f_n to f$ uniformly, and in particular pointwise. Then for every $x in B^c$, we have $x in B_{frac1n}^c$ for some $n in mathbb{N}$ so $f_n(x) to f(x)$. Since $x$ was arbitrary, we conclude $f_n to f$ pointwise on $B^c$. We cannot conclude $f_n to f$ uniformly as the example shows.
          $endgroup$
          – mechanodroid
          Dec 28 '18 at 15:12














          $begingroup$
          It's sort of like how the function $x mapsto frac1x$ is uniformly continuous on all segments $[a,b] subseteq langle 0, inftyrangle$ but it is not uniformly continuous on entire $langle 0,inftyrangle$.
          $endgroup$
          – mechanodroid
          Dec 28 '18 at 15:14






          $begingroup$
          It's sort of like how the function $x mapsto frac1x$ is uniformly continuous on all segments $[a,b] subseteq langle 0, inftyrangle$ but it is not uniformly continuous on entire $langle 0,inftyrangle$.
          $endgroup$
          – mechanodroid
          Dec 28 '18 at 15:14














          $begingroup$
          It makes sense. $x$ and $epsilon$ are both arbitrary in uniformly convergence. Thank you very much.
          $endgroup$
          – Xuchuang
          Dec 29 '18 at 3:02




          $begingroup$
          It makes sense. $x$ and $epsilon$ are both arbitrary in uniformly convergence. Thank you very much.
          $endgroup$
          – Xuchuang
          Dec 29 '18 at 3:02











          0












          $begingroup$

          You have that ${f_n}$ converges uniformly on each set $B_{frac 1n}^c$. There is no reason why it should converge uniformly on $displaystyle bigcup_n B_{frac 1n}^c$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $bigcup_n B_{frac{1}{n}}^c$. Is there a difference between them.
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:16










          • $begingroup$
            That isn't to hard to see. If a point $x$ belongs to $bigcup_n B^c_{frac 1n}$, then there exists (at least one) index $n$ with $x in B_{frac 1n}^c$. This means that $f_n(x) to f(x)$.
            $endgroup$
            – Umberto P.
            Dec 28 '18 at 14:18












          • $begingroup$
            But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:27










          • $begingroup$
            You are right. The wrong place is when unionizing those $B_{frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you!
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:08


















          0












          $begingroup$

          You have that ${f_n}$ converges uniformly on each set $B_{frac 1n}^c$. There is no reason why it should converge uniformly on $displaystyle bigcup_n B_{frac 1n}^c$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $bigcup_n B_{frac{1}{n}}^c$. Is there a difference between them.
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:16










          • $begingroup$
            That isn't to hard to see. If a point $x$ belongs to $bigcup_n B^c_{frac 1n}$, then there exists (at least one) index $n$ with $x in B_{frac 1n}^c$. This means that $f_n(x) to f(x)$.
            $endgroup$
            – Umberto P.
            Dec 28 '18 at 14:18












          • $begingroup$
            But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:27










          • $begingroup$
            You are right. The wrong place is when unionizing those $B_{frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you!
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:08
















          0












          0








          0





          $begingroup$

          You have that ${f_n}$ converges uniformly on each set $B_{frac 1n}^c$. There is no reason why it should converge uniformly on $displaystyle bigcup_n B_{frac 1n}^c$.






          share|cite|improve this answer









          $endgroup$



          You have that ${f_n}$ converges uniformly on each set $B_{frac 1n}^c$. There is no reason why it should converge uniformly on $displaystyle bigcup_n B_{frac 1n}^c$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 14:07









          Umberto P.Umberto P.

          39.4k13166




          39.4k13166












          • $begingroup$
            Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $bigcup_n B_{frac{1}{n}}^c$. Is there a difference between them.
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:16










          • $begingroup$
            That isn't to hard to see. If a point $x$ belongs to $bigcup_n B^c_{frac 1n}$, then there exists (at least one) index $n$ with $x in B_{frac 1n}^c$. This means that $f_n(x) to f(x)$.
            $endgroup$
            – Umberto P.
            Dec 28 '18 at 14:18












          • $begingroup$
            But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:27










          • $begingroup$
            You are right. The wrong place is when unionizing those $B_{frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you!
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:08




















          • $begingroup$
            Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $bigcup_n B_{frac{1}{n}}^c$. Is there a difference between them.
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:16










          • $begingroup$
            That isn't to hard to see. If a point $x$ belongs to $bigcup_n B^c_{frac 1n}$, then there exists (at least one) index $n$ with $x in B_{frac 1n}^c$. This means that $f_n(x) to f(x)$.
            $endgroup$
            – Umberto P.
            Dec 28 '18 at 14:18












          • $begingroup$
            But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right?
            $endgroup$
            – Xuchuang
            Dec 28 '18 at 14:27










          • $begingroup$
            You are right. The wrong place is when unionizing those $B_{frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you!
            $endgroup$
            – Xuchuang
            Dec 29 '18 at 3:08


















          $begingroup$
          Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $bigcup_n B_{frac{1}{n}}^c$. Is there a difference between them.
          $endgroup$
          – Xuchuang
          Dec 28 '18 at 14:16




          $begingroup$
          Thanks, I also think that may be the wrong place. But the original proof also don't show why it should converge pointwise on $bigcup_n B_{frac{1}{n}}^c$. Is there a difference between them.
          $endgroup$
          – Xuchuang
          Dec 28 '18 at 14:16












          $begingroup$
          That isn't to hard to see. If a point $x$ belongs to $bigcup_n B^c_{frac 1n}$, then there exists (at least one) index $n$ with $x in B_{frac 1n}^c$. This means that $f_n(x) to f(x)$.
          $endgroup$
          – Umberto P.
          Dec 28 '18 at 14:18






          $begingroup$
          That isn't to hard to see. If a point $x$ belongs to $bigcup_n B^c_{frac 1n}$, then there exists (at least one) index $n$ with $x in B_{frac 1n}^c$. This means that $f_n(x) to f(x)$.
          $endgroup$
          – Umberto P.
          Dec 28 '18 at 14:18














          $begingroup$
          But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right?
          $endgroup$
          – Xuchuang
          Dec 28 '18 at 14:27




          $begingroup$
          But the converges pointwise is derived from converges uniformly in the proof. If there exists a single index $n$ for pointwise, it must already have an index $n$ for uniformly to support the derivation. Am I right?
          $endgroup$
          – Xuchuang
          Dec 28 '18 at 14:27












          $begingroup$
          You are right. The wrong place is when unionizing those $B_{frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you!
          $endgroup$
          – Xuchuang
          Dec 29 '18 at 3:08






          $begingroup$
          You are right. The wrong place is when unionizing those $B_{frac{1}{n}}^c$, it may change interval's bound (like close to open). Pointwise convergence is OK with that but uniformly convergence is not. Thank you!
          $endgroup$
          – Xuchuang
          Dec 29 '18 at 3:08




















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