About the Lebesgue points of a product of functions












1














We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral







share|cite|improve this question






















  • Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    – p4sch
    Nov 30 '18 at 13:29












  • @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    – cob
    Nov 30 '18 at 14:18










  • @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    – cob
    Nov 30 '18 at 14:21










  • @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    – mathworker21
    Nov 30 '18 at 14:38










  • If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    – p4sch
    Nov 30 '18 at 15:03


















1














We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral







share|cite|improve this question






















  • Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    – p4sch
    Nov 30 '18 at 13:29












  • @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    – cob
    Nov 30 '18 at 14:18










  • @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    – cob
    Nov 30 '18 at 14:21










  • @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    – mathworker21
    Nov 30 '18 at 14:38










  • If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    – p4sch
    Nov 30 '18 at 15:03
















1












1








1


1





We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral







share|cite|improve this question













We say that a measurable function $g:mathbb R to mathbb R$ has a Lebesgue point at $x in mathbb R$ if



$$ frac1{2s}int_{x-s}^{x+s} |g(y)-g(x)| , dy to 0 quad text{as } s to 0.$$



If $g$ is continuous, then every $x in mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by



$$ g(x) = sum_{k=0}^infty (-1)^kchi_{(2^{-k-1},2^{-k}]} $$



is an example of a function that does not have a Lebesgue point at $0$.



Now suppose that $g: mathbb R to mathbb R$ and $h: mathbb R to mathbb R$ have Lebesgue points at every $x in mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?






real-analysis measure-theory lebesgue-integral




real-analysis measure-theory lebesgue-integral






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 13:03









cobcob

1063




1063












  • Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    – p4sch
    Nov 30 '18 at 13:29












  • @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    – cob
    Nov 30 '18 at 14:18










  • @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    – cob
    Nov 30 '18 at 14:21










  • @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    – mathworker21
    Nov 30 '18 at 14:38










  • If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    – p4sch
    Nov 30 '18 at 15:03




















  • Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
    – p4sch
    Nov 30 '18 at 13:29












  • @p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
    – cob
    Nov 30 '18 at 14:18










  • @mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
    – cob
    Nov 30 '18 at 14:21










  • @cob hm I thought you had said "must $gh$ have non-Lebesgue points"
    – mathworker21
    Nov 30 '18 at 14:38










  • If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
    – p4sch
    Nov 30 '18 at 15:03


















Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
– p4sch
Nov 30 '18 at 13:29






Since $gh$ is not-necessary locally-integrable, the question does not make sense. (Take $h=g = sqrt{x}^{-1}$, then $h$ and $g$ are locally-integrable, but not $gh$.)
– p4sch
Nov 30 '18 at 13:29














@p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
– cob
Nov 30 '18 at 14:18




@p4sch Note that $sqrt{x}^{-1}$ does not have a Lebesgue point at $0$.
– cob
Nov 30 '18 at 14:18












@mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
– cob
Nov 30 '18 at 14:21




@mathworker21 If $g=h=0$, then $gh=0$ has a Lebesgue point at every $x in mathbb R$.
– cob
Nov 30 '18 at 14:21












@cob hm I thought you had said "must $gh$ have non-Lebesgue points"
– mathworker21
Nov 30 '18 at 14:38




@cob hm I thought you had said "must $gh$ have non-Lebesgue points"
– mathworker21
Nov 30 '18 at 14:38












If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
– p4sch
Nov 30 '18 at 15:03






If $(q_n)_{n in mathbb{N}}$ is an enumeration of $mathbb{Q}$, we can define $h(x) = g(x) = sum_{n=1}^infty 2^{-n} (max {sqrt{|q_n}|,1})^{-1} sqrt{|x-q_n|}^{-1}$. This function is locally-integrable, but $hcdot g$ is not locally-integrable in any point of $mathbb{R}$.
– p4sch
Nov 30 '18 at 15:03












1 Answer
1






active

oldest

votes


















0














Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






share|cite|improve this answer





















    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%2f3020066%2fabout-the-lebesgue-points-of-a-product-of-functions%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














    Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



    However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






    share|cite|improve this answer


























      0














      Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



      However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






      share|cite|improve this answer
























        0












        0








        0






        Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



        However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.






        share|cite|improve this answer












        Take $$g(x)=sum_{k=0}^infty 2^{k/2} chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$frac1{2^{-n+1}}int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| , dy = 2^{n-1}sum_{k=n}^infty 2^{-3k/2} = 2^{-n/2-1}sum_{k=0}^infty 2^{-3k/2}$$ which goes to $0$ as $n to infty$. This also implies $$frac1{2s}int_{-s}^{s} |g(y)-g(0)| , dy to 0 quad text{as $s to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.



        However, $$g^2(x)=sum_{k=0}^infty 2^k chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 17:47









        cobcob

        1063




        1063






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3020066%2fabout-the-lebesgue-points-of-a-product-of-functions%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more