If $fin L^1(mu)$ then $f$ is finite $mu$-a.e.












2












$begingroup$



Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).




My proof:



Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$



I intend to show $mu(A)=0$



As $fin L^{1}(mu)$



$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$



Applying Chebyshev inequality:



$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $



Questions:



Is my proof right? If not why not?
What are the alternatives?



Thanks in advance!






measure-theory proof-verification proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
    $endgroup$
    – mathworker21
    Jan 11 at 19:45












  • $begingroup$
    Where does this problem come from?
    $endgroup$
    – Wolfy
    Jan 11 at 21:09










  • $begingroup$
    @Wolfy It comes from an exercise set that a Professor handled me.
    $endgroup$
    – Pedro Gomes
    Jan 11 at 22:38










  • $begingroup$
    I see what book are you using?
    $endgroup$
    – Wolfy
    Jan 11 at 22:39
















2












$begingroup$



Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).




My proof:



Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$



I intend to show $mu(A)=0$



As $fin L^{1}(mu)$



$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$



Applying Chebyshev inequality:



$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $



Questions:



Is my proof right? If not why not?
What are the alternatives?



Thanks in advance!






measure-theory proof-verification proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
    $endgroup$
    – mathworker21
    Jan 11 at 19:45












  • $begingroup$
    Where does this problem come from?
    $endgroup$
    – Wolfy
    Jan 11 at 21:09










  • $begingroup$
    @Wolfy It comes from an exercise set that a Professor handled me.
    $endgroup$
    – Pedro Gomes
    Jan 11 at 22:38










  • $begingroup$
    I see what book are you using?
    $endgroup$
    – Wolfy
    Jan 11 at 22:39














2












2








2





$begingroup$



Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).




My proof:



Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$



I intend to show $mu(A)=0$



As $fin L^{1}(mu)$



$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$



Applying Chebyshev inequality:



$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $



Questions:



Is my proof right? If not why not?
What are the alternatives?



Thanks in advance!






measure-theory proof-verification proof-writing







share|cite|improve this question











$endgroup$





Prove that if $f,gin L^1(mu)$ then $f$ is finite $mu$-a.e. (almost everywhere).




My proof:



Let $A={x:f(x)geqslant n,forall ninmathbb{Z}}$



I intend to show $mu(A)=0$



As $fin L^{1}(mu)$



$int |f|dmuleqslant infty$ then $exists C$ such that $int |f|dmu<C$



Applying Chebyshev inequality:



$mu(A)leqslantfrac{1}{n}int |f|dmu<frac{1}{n}Cimplies mu(A)<frac{1}{n}C\impliesmu(A)<frac{1}{n}Cimplieslim_{ntoinfty}mu(A)<lim_{ntoinfty}frac{1}{n}C=0implies mu(A)=0 $



Questions:



Is my proof right? If not why not?
What are the alternatives?



Thanks in advance!






measure-theory proof-verification proof-writing




measure-theory proof-verification proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 19:54









6005

37.1k752127




37.1k752127










asked Jan 11 at 19:43









Pedro GomesPedro Gomes

2,0062821




2,0062821












  • $begingroup$
    You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
    $endgroup$
    – mathworker21
    Jan 11 at 19:45












  • $begingroup$
    Where does this problem come from?
    $endgroup$
    – Wolfy
    Jan 11 at 21:09










  • $begingroup$
    @Wolfy It comes from an exercise set that a Professor handled me.
    $endgroup$
    – Pedro Gomes
    Jan 11 at 22:38










  • $begingroup$
    I see what book are you using?
    $endgroup$
    – Wolfy
    Jan 11 at 22:39


















  • $begingroup$
    You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
    $endgroup$
    – mathworker21
    Jan 11 at 19:45












  • $begingroup$
    Where does this problem come from?
    $endgroup$
    – Wolfy
    Jan 11 at 21:09










  • $begingroup$
    @Wolfy It comes from an exercise set that a Professor handled me.
    $endgroup$
    – Pedro Gomes
    Jan 11 at 22:38










  • $begingroup$
    I see what book are you using?
    $endgroup$
    – Wolfy
    Jan 11 at 22:39
















$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45






$begingroup$
You should be applying Chebyshev to ${x : f(x) ge n}$ (for some fixed $n$), but since ${x : f(x) ge n} subseteq {x : f(x) ge n, forall n in mathbb{Z}}$, you're fine
$endgroup$
– mathworker21
Jan 11 at 19:45














$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09




$begingroup$
Where does this problem come from?
$endgroup$
– Wolfy
Jan 11 at 21:09












$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38




$begingroup$
@Wolfy It comes from an exercise set that a Professor handled me.
$endgroup$
– Pedro Gomes
Jan 11 at 22:38












$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39




$begingroup$
I see what book are you using?
$endgroup$
– Wolfy
Jan 11 at 22:39










1 Answer
1






active

oldest

votes


















1












$begingroup$

Nice proof—it looks right to me.




What are the alternatives?




Personally I would write the proof a bit simpler.
Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
Now note that $A$ is a measurable set, so
$$
int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
$$
Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.






share|cite|improve this answer











$endgroup$














    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%2f3070274%2fif-f-in-l1-mu-then-f-is-finite-mu-a-e%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









    1












    $begingroup$

    Nice proof—it looks right to me.




    What are the alternatives?




    Personally I would write the proof a bit simpler.
    Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
    Now note that $A$ is a measurable set, so
    $$
    int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
    $$
    Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Nice proof—it looks right to me.




      What are the alternatives?




      Personally I would write the proof a bit simpler.
      Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
      Now note that $A$ is a measurable set, so
      $$
      int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
      $$
      Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Nice proof—it looks right to me.




        What are the alternatives?




        Personally I would write the proof a bit simpler.
        Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
        Now note that $A$ is a measurable set, so
        $$
        int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
        $$
        Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.






        share|cite|improve this answer











        $endgroup$



        Nice proof—it looks right to me.




        What are the alternatives?




        Personally I would write the proof a bit simpler.
        Let $A = {x : f(x) = infty}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!)
        Now note that $A$ is a measurable set, so
        $$
        int |f| , dmu ge int_A |f|, dmu = int_A infty , dmu = infty cdot mu(A).
        $$
        Recall that in measure theory we define $0 cdot infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $mu(A) = 0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 11 at 20:31









        Alex Ortiz

        11.4k21442




        11.4k21442










        answered Jan 11 at 19:50









        60056005

        37.1k752127




        37.1k752127






























            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%2f3070274%2fif-f-in-l1-mu-then-f-is-finite-mu-a-e%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