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.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    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

    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


















            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

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
            Read more