Prove This Function is Finite Almost Everywhere












6












$begingroup$



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











$endgroup$

















    6












    $begingroup$



    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











    $endgroup$















      6












      6








      6


      3



      $begingroup$



      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











      $endgroup$





      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 Jan 2 at 5:39









      Chase Ryan Taylor

      4,38021530




      4,38021530










      asked Jan 2 at 4:35









      BoboBobo

      975




      975






















          1 Answer
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          5












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            $endgroup$
            – Guacho Perez
            Jan 2 at 6:15










          • $begingroup$
            @GuachoPerez Thanks for the correction!
            $endgroup$
            – Song
            Jan 2 at 6:16











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          1 Answer
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          active

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          5












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            $endgroup$
            – Guacho Perez
            Jan 2 at 6:15










          • $begingroup$
            @GuachoPerez Thanks for the correction!
            $endgroup$
            – Song
            Jan 2 at 6:16
















          5












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            $endgroup$
            – Guacho Perez
            Jan 2 at 6:15










          • $begingroup$
            @GuachoPerez Thanks for the correction!
            $endgroup$
            – Song
            Jan 2 at 6:16














          5












          5








          5





          $begingroup$

          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











          $endgroup$



          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 Jan 2 at 6:16

























          answered Jan 2 at 5:01









          SongSong

          7,563423




          7,563423












          • $begingroup$
            I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            $endgroup$
            – Guacho Perez
            Jan 2 at 6:15










          • $begingroup$
            @GuachoPerez Thanks for the correction!
            $endgroup$
            – Song
            Jan 2 at 6:16


















          • $begingroup$
            I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
            $endgroup$
            – Guacho Perez
            Jan 2 at 6:15










          • $begingroup$
            @GuachoPerez Thanks for the correction!
            $endgroup$
            – Song
            Jan 2 at 6:16
















          $begingroup$
          I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
          $endgroup$
          – Guacho Perez
          Jan 2 at 6:15




          $begingroup$
          I think you mean Tonelli as Fubini requires the integrand to be in $L^1$ already.
          $endgroup$
          – Guacho Perez
          Jan 2 at 6:15












          $begingroup$
          @GuachoPerez Thanks for the correction!
          $endgroup$
          – Song
          Jan 2 at 6:16




          $begingroup$
          @GuachoPerez Thanks for the correction!
          $endgroup$
          – Song
          Jan 2 at 6:16


















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