Convergence in $L^1$ implies Convergence in measure using Chebyshev's Inequality












2












$begingroup$


Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25


















2












$begingroup$


Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25
















2












2








2





$begingroup$


Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$




Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure




real-analysis measure-theory lebesgue-integral lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 0:21









InfinityInfinity

384113




384113












  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25




















  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25


















$begingroup$
Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
$endgroup$
– Mark
Jan 5 at 0:25






$begingroup$
Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
$endgroup$
– Mark
Jan 5 at 0:25












1 Answer
1






active

oldest

votes


















0












$begingroup$

There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



$$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
for positive $g$.




Alternatively, notice that



$$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



and apply Case $1$.






share|cite|improve this answer









$endgroup$














    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%2f3062277%2fconvergence-in-l1-implies-convergence-in-measure-using-chebyshevs-inequality%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



    $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
    for positive $g$.




    Alternatively, notice that



    $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



    and apply Case $1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



      $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
      for positive $g$.




      Alternatively, notice that



      $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



      and apply Case $1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



        $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
        for positive $g$.




        Alternatively, notice that



        $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



        and apply Case $1$.






        share|cite|improve this answer









        $endgroup$



        There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



        $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
        for positive $g$.




        Alternatively, notice that



        $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



        and apply Case $1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 0:27









        T. BongersT. Bongers

        23.5k54762




        23.5k54762






























            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%2f3062277%2fconvergence-in-l1-implies-convergence-in-measure-using-chebyshevs-inequality%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

            How do I know what Microsoft account the skydrive app is syncing to?

            Image
            up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
            Read more

            When does type information flow backwards in C++?

            Image
            up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
            Read more

            Grease: Live!

            Image
            A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
            Read more