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 1 hour ago









      Chase Ryan Taylor

      4,38021530




      4,38021530










      asked 3 hours ago









      Bobo

      705




      705






















          1 Answer
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          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











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          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

          4,910317




          4,910317












          • 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


















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