Showing a sequence of functions $f_n$ does not converge uniformly to $f$ on an interval.












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Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.



I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.



Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.



Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.



I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?



Any insight would be much appreciated thanks :)






measure-theory lebesgue-measure







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    $begingroup$


    Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.



    I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.



    Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.



    Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.



    I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?



    Any insight would be much appreciated thanks :)






    measure-theory lebesgue-measure







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.



      I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.



      Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.



      Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.



      I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?



      Any insight would be much appreciated thanks :)






      measure-theory lebesgue-measure







      share|cite|improve this question









      $endgroup$




      Suppose for each $n in mathbb{N}$ we have a function $f_n:[0,1] to [0,1]$ by $f_n(x)=nx$ on the interval $x in [0,frac{1}{n}]$ and $1$ if $x in (frac{1}{n},1]$, and define $f=lim_{n to infty} f_n$.



      I want to show that for any Lebesgue measurable $B subseteq [0,1]$ with Lebesgue measure $lambda(B)=0$ that the functions $f_n$ do not converge to $f$ uniformly on $[0,1] backslash B$.



      Now clearly we have the the limit $f$ is equal to $0$ at $x=0$ and is equal to $1$ elsewhere, but I am struggling to figure out how to implement the fact that we remove the zero measure interval $B$.



      Firstly my thought was to use the uniform convergence theorem, which would work with showing that the functions $f_n$ do not converge uniformly to $f$ as each of the functions $f_n$ is continuous we can use the uniform limit theorem to show that $f_n$ does not converge to $f$.



      I am wondering whether there is a simple result that I am missing that ensures this transfers to the same sort of result when we take away a set of measure $0$, for example does continuity of each $f_n$ still hold in this case in which I can still apply the theorem?



      Any insight would be much appreciated thanks :)






      measure-theory lebesgue-measure




      measure-theory lebesgue-measure






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      asked Dec 6 '18 at 0:21









      MattMatt

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          Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,



          $$
          d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
          $$



          Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.



          Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.






          share|cite|improve this answer











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            $begingroup$

            Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,



            $$
            d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
            $$



            Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.



            Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.






            share|cite|improve this answer











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              2












              $begingroup$

              Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,



              $$
              d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
              $$



              Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.



              Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,



                $$
                d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
                $$



                Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.



                Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.






                share|cite|improve this answer











                $endgroup$



                Take $B subset [0,1]$ of measure zero and set $l = inf (0,1]setminus B$. If it were $l > 0$, then $B$ would contain some neighbourhood of $0$ (maybe without $0$ itself) which contradicts that $|B| = 0$. Hence $l = 0$ and thus we have a sequence $x_n to 0$ contained in $(0,1] setminus B$. Without loss of generality we can assume that $0 < x_n leq frac{1}{2n}$. Thus, $f_n(x_n) leq frac{1}{2}$ and in particular,



                $$
                d_{[0,1] setminus B}(f,f_n) geq |f(x_n) - f_n(x_n)| = f(x_n) - f_n(x_n) geq f(x_n) - frac{1}{2} stackrel{(x_n neq 0)}{=} frac{1}{2}.
                $$



                Taking limits, we get that $lim_n d_{[0,1] setminus B}(f,f_n) geq frac{1}{2} > 0$ as desired.



                Edit: note that we only used that $B$ can't contain a neighbourhood of zero, which is weaker than being of measure zero. Thus uniform convergence fails for a (in some sense) larger family of sets.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 6 '18 at 1:03

























                answered Dec 6 '18 at 0:46









                Guido A.Guido A.

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                7,2861730






























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