A basic exercise about Lebesgue's Dominated Convergence Theorem












2












$begingroup$


Consider the sequence $g_n(x)=(1-frac{x}{n})^n$ for $xin[-1,1]$. Calculate the value $$lim_{ntoinfty}int_{-1}^1g_n(x)dx$$
by showing the existence of the limit with the aid of the Lebesgue's dominated convergence theorem (LDCT).



Attempt. For $ninBbb N,xin[-1,1]$, note that $|g_n(x)|=left(1-frac{x}{n}right)^nleq 2$ and the function $h(x)=2$ is integrable on $[-1,1]$. So, we may apply (LDCT) as follows:
$$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 1dx=2.$$
I would be glad if someone could check my attempt. Thanks!






real-analysis measure-theory







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$endgroup$












  • $begingroup$
    No, $|g_n|$ is not bounded by $2$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 16:33
















2












$begingroup$


Consider the sequence $g_n(x)=(1-frac{x}{n})^n$ for $xin[-1,1]$. Calculate the value $$lim_{ntoinfty}int_{-1}^1g_n(x)dx$$
by showing the existence of the limit with the aid of the Lebesgue's dominated convergence theorem (LDCT).



Attempt. For $ninBbb N,xin[-1,1]$, note that $|g_n(x)|=left(1-frac{x}{n}right)^nleq 2$ and the function $h(x)=2$ is integrable on $[-1,1]$. So, we may apply (LDCT) as follows:
$$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 1dx=2.$$
I would be glad if someone could check my attempt. Thanks!






real-analysis measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    No, $|g_n|$ is not bounded by $2$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 16:33














2












2








2





$begingroup$


Consider the sequence $g_n(x)=(1-frac{x}{n})^n$ for $xin[-1,1]$. Calculate the value $$lim_{ntoinfty}int_{-1}^1g_n(x)dx$$
by showing the existence of the limit with the aid of the Lebesgue's dominated convergence theorem (LDCT).



Attempt. For $ninBbb N,xin[-1,1]$, note that $|g_n(x)|=left(1-frac{x}{n}right)^nleq 2$ and the function $h(x)=2$ is integrable on $[-1,1]$. So, we may apply (LDCT) as follows:
$$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 1dx=2.$$
I would be glad if someone could check my attempt. Thanks!






real-analysis measure-theory







share|cite|improve this question









$endgroup$




Consider the sequence $g_n(x)=(1-frac{x}{n})^n$ for $xin[-1,1]$. Calculate the value $$lim_{ntoinfty}int_{-1}^1g_n(x)dx$$
by showing the existence of the limit with the aid of the Lebesgue's dominated convergence theorem (LDCT).



Attempt. For $ninBbb N,xin[-1,1]$, note that $|g_n(x)|=left(1-frac{x}{n}right)^nleq 2$ and the function $h(x)=2$ is integrable on $[-1,1]$. So, we may apply (LDCT) as follows:
$$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 1dx=2.$$
I would be glad if someone could check my attempt. Thanks!






real-analysis measure-theory




real-analysis measure-theory






share|cite|improve this question













share|cite|improve this question











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asked Jan 12 at 16:20









Ergin SuerErgin Suer

1,4631921




1,4631921












  • $begingroup$
    No, $|g_n|$ is not bounded by $2$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 16:33


















  • $begingroup$
    No, $|g_n|$ is not bounded by $2$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 16:33
















$begingroup$
No, $|g_n|$ is not bounded by $2$.
$endgroup$
– David C. Ullrich
Jan 12 at 16:33




$begingroup$
No, $|g_n|$ is not bounded by $2$.
$endgroup$
– David C. Ullrich
Jan 12 at 16:33










3 Answers
3






active

oldest

votes


















3












$begingroup$

Hint. Note that the point-wise limit of sequence of functions $(g_n(x))_n$ is $g(x)=e^{-x}$ (it is not $1$). Hence, by following your approach, we may apply the Lebesgue's dominated convergence theorem (note that $|g_n(x)|$ is bounded by $e$) and
$$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1 e^{-x}dx.$$
What is the final result?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I tried to write a complete solution by using your answer. I would be glad if you could check it. Thanks!
    $endgroup$
    – Ergin Suer
    Jan 12 at 18:23



















0












$begingroup$

It's already been pointed out that the limit of $g_n$ is not what you said. Your inequality $|g_n|le 2$ is also wrong.



To get a correct bound suitable for applying DCT, "recall" that $$log(1+t)le tquad(t>-1).$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I would like to write the complete solution to here by using the answer of Robert Z. We know that the point-wise limit of the sequence $(g_n)$ is $g(x)=e^{-x}$ and we show that $|g_n(x)|<e$ for all $ninBbb N, xin[-1,1]$ to apply LDCT. First, note that
    $$binom{n}{k}frac{1}{n^k}=frac{n!}{(n-k)!k!}frac{1}{n^k}=frac{n(n-1)dots(n-k+1)}{k!}frac{1}{n^k}leq frac{1}{k!}.$$
    So, for any $ninBbb N, xin[-1,1]$, we have
    $$|g_n(x)|=left|left(1-frac{x}{n}right)^nright|=left(1-frac{x}{n}right)^nleq left(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}1^{n-k}left(frac{1}{n}right)^kleq sum_{k=0}^nfrac{1}{k!}<sum_{k=0}^inftyfrac{1}{k!}=e.$$
    By LDCT, we get
    $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 e^{-x}dx=e-frac{1}{e}.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yes, it is correct now.
      $endgroup$
      – Robert Z
      Jan 12 at 18:36











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Hint. Note that the point-wise limit of sequence of functions $(g_n(x))_n$ is $g(x)=e^{-x}$ (it is not $1$). Hence, by following your approach, we may apply the Lebesgue's dominated convergence theorem (note that $|g_n(x)|$ is bounded by $e$) and
    $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1 e^{-x}dx.$$
    What is the final result?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I tried to write a complete solution by using your answer. I would be glad if you could check it. Thanks!
      $endgroup$
      – Ergin Suer
      Jan 12 at 18:23
















    3












    $begingroup$

    Hint. Note that the point-wise limit of sequence of functions $(g_n(x))_n$ is $g(x)=e^{-x}$ (it is not $1$). Hence, by following your approach, we may apply the Lebesgue's dominated convergence theorem (note that $|g_n(x)|$ is bounded by $e$) and
    $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1 e^{-x}dx.$$
    What is the final result?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I tried to write a complete solution by using your answer. I would be glad if you could check it. Thanks!
      $endgroup$
      – Ergin Suer
      Jan 12 at 18:23














    3












    3








    3





    $begingroup$

    Hint. Note that the point-wise limit of sequence of functions $(g_n(x))_n$ is $g(x)=e^{-x}$ (it is not $1$). Hence, by following your approach, we may apply the Lebesgue's dominated convergence theorem (note that $|g_n(x)|$ is bounded by $e$) and
    $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1 e^{-x}dx.$$
    What is the final result?






    share|cite|improve this answer











    $endgroup$



    Hint. Note that the point-wise limit of sequence of functions $(g_n(x))_n$ is $g(x)=e^{-x}$ (it is not $1$). Hence, by following your approach, we may apply the Lebesgue's dominated convergence theorem (note that $|g_n(x)|$ is bounded by $e$) and
    $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1 e^{-x}dx.$$
    What is the final result?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 12 at 16:32

























    answered Jan 12 at 16:25









    Robert ZRobert Z

    101k1069142




    101k1069142












    • $begingroup$
      I tried to write a complete solution by using your answer. I would be glad if you could check it. Thanks!
      $endgroup$
      – Ergin Suer
      Jan 12 at 18:23


















    • $begingroup$
      I tried to write a complete solution by using your answer. I would be glad if you could check it. Thanks!
      $endgroup$
      – Ergin Suer
      Jan 12 at 18:23
















    $begingroup$
    I tried to write a complete solution by using your answer. I would be glad if you could check it. Thanks!
    $endgroup$
    – Ergin Suer
    Jan 12 at 18:23




    $begingroup$
    I tried to write a complete solution by using your answer. I would be glad if you could check it. Thanks!
    $endgroup$
    – Ergin Suer
    Jan 12 at 18:23











    0












    $begingroup$

    It's already been pointed out that the limit of $g_n$ is not what you said. Your inequality $|g_n|le 2$ is also wrong.



    To get a correct bound suitable for applying DCT, "recall" that $$log(1+t)le tquad(t>-1).$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It's already been pointed out that the limit of $g_n$ is not what you said. Your inequality $|g_n|le 2$ is also wrong.



      To get a correct bound suitable for applying DCT, "recall" that $$log(1+t)le tquad(t>-1).$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It's already been pointed out that the limit of $g_n$ is not what you said. Your inequality $|g_n|le 2$ is also wrong.



        To get a correct bound suitable for applying DCT, "recall" that $$log(1+t)le tquad(t>-1).$$






        share|cite|improve this answer









        $endgroup$



        It's already been pointed out that the limit of $g_n$ is not what you said. Your inequality $|g_n|le 2$ is also wrong.



        To get a correct bound suitable for applying DCT, "recall" that $$log(1+t)le tquad(t>-1).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 16:38









        David C. UllrichDavid C. Ullrich

        61.3k43994




        61.3k43994























            0












            $begingroup$

            I would like to write the complete solution to here by using the answer of Robert Z. We know that the point-wise limit of the sequence $(g_n)$ is $g(x)=e^{-x}$ and we show that $|g_n(x)|<e$ for all $ninBbb N, xin[-1,1]$ to apply LDCT. First, note that
            $$binom{n}{k}frac{1}{n^k}=frac{n!}{(n-k)!k!}frac{1}{n^k}=frac{n(n-1)dots(n-k+1)}{k!}frac{1}{n^k}leq frac{1}{k!}.$$
            So, for any $ninBbb N, xin[-1,1]$, we have
            $$|g_n(x)|=left|left(1-frac{x}{n}right)^nright|=left(1-frac{x}{n}right)^nleq left(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}1^{n-k}left(frac{1}{n}right)^kleq sum_{k=0}^nfrac{1}{k!}<sum_{k=0}^inftyfrac{1}{k!}=e.$$
            By LDCT, we get
            $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 e^{-x}dx=e-frac{1}{e}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes, it is correct now.
              $endgroup$
              – Robert Z
              Jan 12 at 18:36
















            0












            $begingroup$

            I would like to write the complete solution to here by using the answer of Robert Z. We know that the point-wise limit of the sequence $(g_n)$ is $g(x)=e^{-x}$ and we show that $|g_n(x)|<e$ for all $ninBbb N, xin[-1,1]$ to apply LDCT. First, note that
            $$binom{n}{k}frac{1}{n^k}=frac{n!}{(n-k)!k!}frac{1}{n^k}=frac{n(n-1)dots(n-k+1)}{k!}frac{1}{n^k}leq frac{1}{k!}.$$
            So, for any $ninBbb N, xin[-1,1]$, we have
            $$|g_n(x)|=left|left(1-frac{x}{n}right)^nright|=left(1-frac{x}{n}right)^nleq left(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}1^{n-k}left(frac{1}{n}right)^kleq sum_{k=0}^nfrac{1}{k!}<sum_{k=0}^inftyfrac{1}{k!}=e.$$
            By LDCT, we get
            $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 e^{-x}dx=e-frac{1}{e}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes, it is correct now.
              $endgroup$
              – Robert Z
              Jan 12 at 18:36














            0












            0








            0





            $begingroup$

            I would like to write the complete solution to here by using the answer of Robert Z. We know that the point-wise limit of the sequence $(g_n)$ is $g(x)=e^{-x}$ and we show that $|g_n(x)|<e$ for all $ninBbb N, xin[-1,1]$ to apply LDCT. First, note that
            $$binom{n}{k}frac{1}{n^k}=frac{n!}{(n-k)!k!}frac{1}{n^k}=frac{n(n-1)dots(n-k+1)}{k!}frac{1}{n^k}leq frac{1}{k!}.$$
            So, for any $ninBbb N, xin[-1,1]$, we have
            $$|g_n(x)|=left|left(1-frac{x}{n}right)^nright|=left(1-frac{x}{n}right)^nleq left(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}1^{n-k}left(frac{1}{n}right)^kleq sum_{k=0}^nfrac{1}{k!}<sum_{k=0}^inftyfrac{1}{k!}=e.$$
            By LDCT, we get
            $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 e^{-x}dx=e-frac{1}{e}.$$






            share|cite|improve this answer











            $endgroup$



            I would like to write the complete solution to here by using the answer of Robert Z. We know that the point-wise limit of the sequence $(g_n)$ is $g(x)=e^{-x}$ and we show that $|g_n(x)|<e$ for all $ninBbb N, xin[-1,1]$ to apply LDCT. First, note that
            $$binom{n}{k}frac{1}{n^k}=frac{n!}{(n-k)!k!}frac{1}{n^k}=frac{n(n-1)dots(n-k+1)}{k!}frac{1}{n^k}leq frac{1}{k!}.$$
            So, for any $ninBbb N, xin[-1,1]$, we have
            $$|g_n(x)|=left|left(1-frac{x}{n}right)^nright|=left(1-frac{x}{n}right)^nleq left(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}1^{n-k}left(frac{1}{n}right)^kleq sum_{k=0}^nfrac{1}{k!}<sum_{k=0}^inftyfrac{1}{k!}=e.$$
            By LDCT, we get
            $$lim_{nto infty}int_{-1}^1g_n(x)dx=int_{-1}^1lim_{nto infty}g_n(x)dx=int_{-1}^1 e^{-x}dx=e-frac{1}{e}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 12 at 21:50

























            answered Jan 12 at 18:21









            Ergin SuerErgin Suer

            1,4631921




            1,4631921












            • $begingroup$
              Yes, it is correct now.
              $endgroup$
              – Robert Z
              Jan 12 at 18:36


















            • $begingroup$
              Yes, it is correct now.
              $endgroup$
              – Robert Z
              Jan 12 at 18:36
















            $begingroup$
            Yes, it is correct now.
            $endgroup$
            – Robert Z
            Jan 12 at 18:36




            $begingroup$
            Yes, it is correct now.
            $endgroup$
            – Robert Z
            Jan 12 at 18:36


















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