Prove This Function is Finite Almost Everywhere












3















Let $F$ be a closed subset of $[0,1]$ of positive Lebesgue measure. Let $delta(x)$ be defined as $delta(x) = operatorname{dist}(x, F)$. Consider
$$M(x) = int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy$$
Prove that for almost every point $x in F$, $M(x) < infty$.




My thoughts so far are the following: We wish to show that $M in L^{1}(F)$, which is more than sufficient to complete the proof. Thus, consider
$$int_{F} int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy ,dx$$
From here, I would like to proceed by using Fubini Theorem to switch the order of integration to
$$int_{0}^{1}int_{F} frac{delta(y)}{|x -y|^2} , dx ,dy$$
From here, I am not really sure what to do.






measure-theory lebesgue-measure







share|cite|improve this question





























    3















    Let $F$ be a closed subset of $[0,1]$ of positive Lebesgue measure. Let $delta(x)$ be defined as $delta(x) = operatorname{dist}(x, F)$. Consider
    $$M(x) = int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy$$
    Prove that for almost every point $x in F$, $M(x) < infty$.




    My thoughts so far are the following: We wish to show that $M in L^{1}(F)$, which is more than sufficient to complete the proof. Thus, consider
    $$int_{F} int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy ,dx$$
    From here, I would like to proceed by using Fubini Theorem to switch the order of integration to
    $$int_{0}^{1}int_{F} frac{delta(y)}{|x -y|^2} , dx ,dy$$
    From here, I am not really sure what to do.






    measure-theory lebesgue-measure







    share|cite|improve this question



























      3












      3








      3


      3






      Let $F$ be a closed subset of $[0,1]$ of positive Lebesgue measure. Let $delta(x)$ be defined as $delta(x) = operatorname{dist}(x, F)$. Consider
      $$M(x) = int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy$$
      Prove that for almost every point $x in F$, $M(x) < infty$.




      My thoughts so far are the following: We wish to show that $M in L^{1}(F)$, which is more than sufficient to complete the proof. Thus, consider
      $$int_{F} int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy ,dx$$
      From here, I would like to proceed by using Fubini Theorem to switch the order of integration to
      $$int_{0}^{1}int_{F} frac{delta(y)}{|x -y|^2} , dx ,dy$$
      From here, I am not really sure what to do.






      measure-theory lebesgue-measure







      share|cite|improve this question
















      Let $F$ be a closed subset of $[0,1]$ of positive Lebesgue measure. Let $delta(x)$ be defined as $delta(x) = operatorname{dist}(x, F)$. Consider
      $$M(x) = int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy$$
      Prove that for almost every point $x in F$, $M(x) < infty$.




      My thoughts so far are the following: We wish to show that $M in L^{1}(F)$, which is more than sufficient to complete the proof. Thus, consider
      $$int_{F} int_{0}^{1}frac{delta(y)}{|x -y|^2} , dy ,dx$$
      From here, I would like to proceed by using Fubini Theorem to switch the order of integration to
      $$int_{0}^{1}int_{F} frac{delta(y)}{|x -y|^2} , dx ,dy$$
      From here, I am not really sure what to do.






      measure-theory lebesgue-measure




      measure-theory lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Chase Ryan Taylor

      4,38021530




      4,38021530










      asked 3 hours ago









      Bobo

      705




      705






















          1 Answer
          1






          active

          oldest

          votes


















          4














          Note that
          $$
          M(x) = int_{F^ccap I}frac{delta(y)}{|x-y|^2}dy
          $$           where $I=[0,1]$. Hence by Tonelli's theorem, we have
          $$begin{eqnarray}
          int_F M(x)dx &=& int_{F^ccap I}delta(y)left(int_Ffrac{1}{|x-y|^2}dxright)dy\
          &le&int_{F^ccap I}delta(y)left(int_{|z|ge delta(y)}frac{1}{|z|^2}dzright)dy\
          &= &int_{F^ccap I}delta(y)frac{2}{delta(y)}dy= 2|F^c cap I| le 2.
          end{eqnarray}$$






          share|cite|improve this answer























          • I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            – Guacho Perez
            1 hour ago










          • @GuachoPerez Thanks for the correction!
            – Song
            1 hour ago











          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%2f3059145%2fprove-this-function-is-finite-almost-everywhere%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









          4














          Note that
          $$
          M(x) = int_{F^ccap I}frac{delta(y)}{|x-y|^2}dy
          $$           where $I=[0,1]$. Hence by Tonelli's theorem, we have
          $$begin{eqnarray}
          int_F M(x)dx &=& int_{F^ccap I}delta(y)left(int_Ffrac{1}{|x-y|^2}dxright)dy\
          &le&int_{F^ccap I}delta(y)left(int_{|z|ge delta(y)}frac{1}{|z|^2}dzright)dy\
          &= &int_{F^ccap I}delta(y)frac{2}{delta(y)}dy= 2|F^c cap I| le 2.
          end{eqnarray}$$






          share|cite|improve this answer























          • I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            – Guacho Perez
            1 hour ago










          • @GuachoPerez Thanks for the correction!
            – Song
            1 hour ago
















          4














          Note that
          $$
          M(x) = int_{F^ccap I}frac{delta(y)}{|x-y|^2}dy
          $$           where $I=[0,1]$. Hence by Tonelli's theorem, we have
          $$begin{eqnarray}
          int_F M(x)dx &=& int_{F^ccap I}delta(y)left(int_Ffrac{1}{|x-y|^2}dxright)dy\
          &le&int_{F^ccap I}delta(y)left(int_{|z|ge delta(y)}frac{1}{|z|^2}dzright)dy\
          &= &int_{F^ccap I}delta(y)frac{2}{delta(y)}dy= 2|F^c cap I| le 2.
          end{eqnarray}$$






          share|cite|improve this answer























          • I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            – Guacho Perez
            1 hour ago










          • @GuachoPerez Thanks for the correction!
            – Song
            1 hour ago














          4












          4








          4






          Note that
          $$
          M(x) = int_{F^ccap I}frac{delta(y)}{|x-y|^2}dy
          $$           where $I=[0,1]$. Hence by Tonelli's theorem, we have
          $$begin{eqnarray}
          int_F M(x)dx &=& int_{F^ccap I}delta(y)left(int_Ffrac{1}{|x-y|^2}dxright)dy\
          &le&int_{F^ccap I}delta(y)left(int_{|z|ge delta(y)}frac{1}{|z|^2}dzright)dy\
          &= &int_{F^ccap I}delta(y)frac{2}{delta(y)}dy= 2|F^c cap I| le 2.
          end{eqnarray}$$






          share|cite|improve this answer














          Note that
          $$
          M(x) = int_{F^ccap I}frac{delta(y)}{|x-y|^2}dy
          $$           where $I=[0,1]$. Hence by Tonelli's theorem, we have
          $$begin{eqnarray}
          int_F M(x)dx &=& int_{F^ccap I}delta(y)left(int_Ffrac{1}{|x-y|^2}dxright)dy\
          &le&int_{F^ccap I}delta(y)left(int_{|z|ge delta(y)}frac{1}{|z|^2}dzright)dy\
          &= &int_{F^ccap I}delta(y)frac{2}{delta(y)}dy= 2|F^c cap I| le 2.
          end{eqnarray}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          Song

          4,920317




          4,920317












          • I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            – Guacho Perez
            1 hour ago










          • @GuachoPerez Thanks for the correction!
            – Song
            1 hour ago


















          • I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            – Guacho Perez
            1 hour ago










          • @GuachoPerez Thanks for the correction!
            – Song
            1 hour ago
















          I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
          – Guacho Perez
          1 hour ago




          I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
          – Guacho Perez
          1 hour ago












          @GuachoPerez Thanks for the correction!
          – Song
          1 hour ago




          @GuachoPerez Thanks for the correction!
          – Song
          1 hour ago


















          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%2f3059145%2fprove-this-function-is-finite-almost-everywhere%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

          404 Error Contact Form 7 ajax form submitting

          Image
          0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43
          Read more

          How to know if a Active Directory user can login interactively

          Image
          1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor
          Read more

          TypeError: fit_transform() missing 1 required positional argument: 'X'

          Image
          0 I am trying to do Feature Scaling in a dataset, but I get an error and have no idea how to proceed: > Traceback (most recent call last): > > File "<ipython-input-10-71bea414b4d0>", line 22, in <module> > x_train = sc_X.fit_transform(x_train) > > TypeError: fit_transform() missing 1 required positional argument: 'X' and here is my code: import pandas as pd # Importing the dataset dataset = pd.read_csv('Data.csv') X = dataset.iloc[:, :-1].values y = dataset.iloc[:, 3].values # Taking care of missing data from sklearn.preprocessing import Imputer imputer = Imputer(missing_values="NaN", strategy="mean", axis=0) imputer = Imputer.fit(imputer,X[:,1:3]) X[:, 1:3] = Imputer.transform(imputer,X[
          Read more