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







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












$begingroup$


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











share|cite|improve this question




share|cite|improve this question










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












    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071071%2fa-basic-exercise-about-lebesgues-dominated-convergence-theorem%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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

    102k1072145




    102k1072145












    • $begingroup$
      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

        61.8k44095




        61.8k44095























            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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071071%2fa-basic-exercise-about-lebesgues-dominated-convergence-theorem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more