showing $f=g$ a.e. given $f_nto f$ uniformly and $f_nto g$ in $L^p(Bbb R)$












2












$begingroup$


Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.



Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.



I need to prove that $f=g$ a.e.



So it's enough to show that $||f-g||_p =0$.



By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.



so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.



So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?



I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.



Thanks for helping!










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 15:20










  • $begingroup$
    I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
    $endgroup$
    – gangrene
    Dec 2 '18 at 15:21












  • $begingroup$
    @RhysSteele now i know :)
    $endgroup$
    – user123
    Dec 2 '18 at 19:43
















2












$begingroup$


Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.



Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.



I need to prove that $f=g$ a.e.



So it's enough to show that $||f-g||_p =0$.



By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.



so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.



So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?



I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.



Thanks for helping!










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 15:20










  • $begingroup$
    I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
    $endgroup$
    – gangrene
    Dec 2 '18 at 15:21












  • $begingroup$
    @RhysSteele now i know :)
    $endgroup$
    – user123
    Dec 2 '18 at 19:43














2












2








2





$begingroup$


Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.



Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.



I need to prove that $f=g$ a.e.



So it's enough to show that $||f-g||_p =0$.



By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.



so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.



So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?



I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.



Thanks for helping!










share|cite|improve this question











$endgroup$




Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.



Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.



I need to prove that $f=g$ a.e.



So it's enough to show that $||f-g||_p =0$.



By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.



so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.



So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?



I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.



Thanks for helping!







real-analysis functional-analysis measure-theory lp-spaces






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 15:54









qbert

22.1k32561




22.1k32561










asked Dec 2 '18 at 15:05









user123user123

1,362316




1,362316












  • $begingroup$
    Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 15:20










  • $begingroup$
    I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
    $endgroup$
    – gangrene
    Dec 2 '18 at 15:21












  • $begingroup$
    @RhysSteele now i know :)
    $endgroup$
    – user123
    Dec 2 '18 at 19:43


















  • $begingroup$
    Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 15:20










  • $begingroup$
    I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
    $endgroup$
    – gangrene
    Dec 2 '18 at 15:21












  • $begingroup$
    @RhysSteele now i know :)
    $endgroup$
    – user123
    Dec 2 '18 at 19:43
















$begingroup$
Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
$endgroup$
– Rhys Steele
Dec 2 '18 at 15:20




$begingroup$
Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
$endgroup$
– Rhys Steele
Dec 2 '18 at 15:20












$begingroup$
I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
$endgroup$
– gangrene
Dec 2 '18 at 15:21






$begingroup$
I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
$endgroup$
– gangrene
Dec 2 '18 at 15:21














$begingroup$
@RhysSteele now i know :)
$endgroup$
– user123
Dec 2 '18 at 19:43




$begingroup$
@RhysSteele now i know :)
$endgroup$
– user123
Dec 2 '18 at 19:43










3 Answers
3






active

oldest

votes


















1












$begingroup$

Let $Nin mathbb{N}$. Then,
$$
0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
$$

so $f=g$ almost everywhere on $[-N,N]$.



Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
$$
mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
$$

So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
    $endgroup$
    – user123
    Dec 2 '18 at 19:40





















1












$begingroup$

I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set



$$
E_varepsilon = {|f-g| > varepsilon}
$$



is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives



$$
E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
$$



and taking measures,



$$
|E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
$$



Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.



    Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
      $endgroup$
      – user123
      Dec 2 '18 at 19:40













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let $Nin mathbb{N}$. Then,
    $$
    0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
    $$

    so $f=g$ almost everywhere on $[-N,N]$.



    Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
    $$
    mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
    $$

    So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
      $endgroup$
      – user123
      Dec 2 '18 at 19:40


















    1












    $begingroup$

    Let $Nin mathbb{N}$. Then,
    $$
    0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
    $$

    so $f=g$ almost everywhere on $[-N,N]$.



    Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
    $$
    mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
    $$

    So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
      $endgroup$
      – user123
      Dec 2 '18 at 19:40
















    1












    1








    1





    $begingroup$

    Let $Nin mathbb{N}$. Then,
    $$
    0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
    $$

    so $f=g$ almost everywhere on $[-N,N]$.



    Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
    $$
    mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
    $$

    So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.






    share|cite|improve this answer











    $endgroup$



    Let $Nin mathbb{N}$. Then,
    $$
    0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
    $$

    so $f=g$ almost everywhere on $[-N,N]$.



    Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
    $$
    mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
    $$

    So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 2 '18 at 15:41

























    answered Dec 2 '18 at 15:31









    qbertqbert

    22.1k32561




    22.1k32561












    • $begingroup$
      Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
      $endgroup$
      – user123
      Dec 2 '18 at 19:40




















    • $begingroup$
      Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
      $endgroup$
      – user123
      Dec 2 '18 at 19:40


















    $begingroup$
    Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
    $endgroup$
    – user123
    Dec 2 '18 at 19:40






    $begingroup$
    Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
    $endgroup$
    – user123
    Dec 2 '18 at 19:40













    1












    $begingroup$

    I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set



    $$
    E_varepsilon = {|f-g| > varepsilon}
    $$



    is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives



    $$
    E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
    $$



    and taking measures,



    $$
    |E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
    $$



    Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set



      $$
      E_varepsilon = {|f-g| > varepsilon}
      $$



      is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives



      $$
      E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
      $$



      and taking measures,



      $$
      |E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
      $$



      Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set



        $$
        E_varepsilon = {|f-g| > varepsilon}
        $$



        is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives



        $$
        E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
        $$



        and taking measures,



        $$
        |E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
        $$



        Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).






        share|cite|improve this answer









        $endgroup$



        I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set



        $$
        E_varepsilon = {|f-g| > varepsilon}
        $$



        is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives



        $$
        E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
        $$



        and taking measures,



        $$
        |E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
        $$



        Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 15:20









        Guido A.Guido A.

        7,4271730




        7,4271730























            1












            $begingroup$

            Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.



            Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
              $endgroup$
              – user123
              Dec 2 '18 at 19:40


















            1












            $begingroup$

            Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.



            Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
              $endgroup$
              – user123
              Dec 2 '18 at 19:40
















            1












            1








            1





            $begingroup$

            Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.



            Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.






            share|cite|improve this answer









            $endgroup$



            Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.



            Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 2 '18 at 18:32









            zhw.zhw.

            73.3k43175




            73.3k43175












            • $begingroup$
              Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
              $endgroup$
              – user123
              Dec 2 '18 at 19:40




















            • $begingroup$
              Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
              $endgroup$
              – user123
              Dec 2 '18 at 19:40


















            $begingroup$
            Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
            $endgroup$
            – user123
            Dec 2 '18 at 19:40






            $begingroup$
            Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
            $endgroup$
            – user123
            Dec 2 '18 at 19:40




















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