Lebesgue measure: $limlimits_{n to infty} n cdot m(S_n)=0$ where $S_n={x in E mid |f(x)|geq n} $












2














Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$



We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.






measure-theory lebesgue-integral lebesgue-measure







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  • Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
    – Martin Sleziak
    Nov 27 at 12:27
















2














Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$



We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.






measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question
























  • Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
    – Martin Sleziak
    Nov 27 at 12:27














2












2








2







Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$



We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.






measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question















Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n={x in E mid |f(x)|geq n} $$ Then $$lim_{n to infty} n cdot m(S_n)=0 $$



We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.






measure-theory lebesgue-integral lebesgue-measure




measure-theory lebesgue-integral lebesgue-measure






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edited Nov 27 at 12:25









Martin Sleziak

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asked Nov 27 at 6:04









ICanMakeYouHateME

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154












  • Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
    – Martin Sleziak
    Nov 27 at 12:27


















  • Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
    – Martin Sleziak
    Nov 27 at 12:27
















Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27




Searching in Approach0 gives a few questions which seem similar: Lebesgue integrable function and limit: $lim_{ntoinfty} ncdot m(A_n)=0$ or Evaluate the limit of $lim_{n to infty}n.mu(X_n)$.
– Martin Sleziak
Nov 27 at 12:27










2 Answers
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Observe that
$$begin{aligned}
int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
&geq int_{S_n} n + int_{E setminus S_n} |f| \
&= ncdot m(S_n) + int_{E setminus S_n} |f| \
end{aligned}$$
As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
$$begin{aligned}
ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
&= int_{S_n} |f| \
&= int_E |f| chi_{S_n} \
end{aligned}$$
where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
$$begin{aligned}
lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
&= int_E lim_{n to infty} |f| chi_{S_n} \
&= 0
end{aligned}$$
As the LHS is nonnegative, this gives us the desired result.






share|cite|improve this answer





























    3














    Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.






    share|cite|improve this answer





















    • What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
      – ICanMakeYouHateME
      Nov 27 at 6:18








    • 1




      Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
      – Dante Grevino
      Nov 27 at 6:21











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    3














    Observe that
    $$begin{aligned}
    int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
    &geq int_{S_n} n + int_{E setminus S_n} |f| \
    &= ncdot m(S_n) + int_{E setminus S_n} |f| \
    end{aligned}$$
    As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
    $$begin{aligned}
    ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
    &= int_{S_n} |f| \
    &= int_E |f| chi_{S_n} \
    end{aligned}$$
    where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
    $$begin{aligned}
    lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
    &= int_E lim_{n to infty} |f| chi_{S_n} \
    &= 0
    end{aligned}$$
    As the LHS is nonnegative, this gives us the desired result.






    share|cite|improve this answer


























      3














      Observe that
      $$begin{aligned}
      int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
      &geq int_{S_n} n + int_{E setminus S_n} |f| \
      &= ncdot m(S_n) + int_{E setminus S_n} |f| \
      end{aligned}$$
      As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
      $$begin{aligned}
      ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
      &= int_{S_n} |f| \
      &= int_E |f| chi_{S_n} \
      end{aligned}$$
      where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
      $$begin{aligned}
      lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
      &= int_E lim_{n to infty} |f| chi_{S_n} \
      &= 0
      end{aligned}$$
      As the LHS is nonnegative, this gives us the desired result.






      share|cite|improve this answer
























        3












        3








        3






        Observe that
        $$begin{aligned}
        int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
        &geq int_{S_n} n + int_{E setminus S_n} |f| \
        &= ncdot m(S_n) + int_{E setminus S_n} |f| \
        end{aligned}$$
        As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
        $$begin{aligned}
        ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
        &= int_{S_n} |f| \
        &= int_E |f| chi_{S_n} \
        end{aligned}$$
        where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
        $$begin{aligned}
        lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
        &= int_E lim_{n to infty} |f| chi_{S_n} \
        &= 0
        end{aligned}$$
        As the LHS is nonnegative, this gives us the desired result.






        share|cite|improve this answer












        Observe that
        $$begin{aligned}
        int_E |f| &= int_{S_n} |f| + int_{E setminus S_n} |f| \
        &geq int_{S_n} n + int_{E setminus S_n} |f| \
        &= ncdot m(S_n) + int_{E setminus S_n} |f| \
        end{aligned}$$
        As $int_E |f|$ is finite, so is $int_{E setminus S_n} |f|$, so we can subtract the latter from both sides to obtain
        $$begin{aligned}
        ncdot m(S_n) &leq int_E |f| - int_{E setminus S_n} |f| \
        &= int_{S_n} |f| \
        &= int_E |f| chi_{S_n} \
        end{aligned}$$
        where $chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have
        $$begin{aligned}
        lim_{n to infty}ncdot m(S_n) &leq lim_{n to infty}int_E |f| chi_{S_n} \
        &= int_E lim_{n to infty} |f| chi_{S_n} \
        &= 0
        end{aligned}$$
        As the LHS is nonnegative, this gives us the desired result.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 6:16









        Bungo

        13.6k22148




        13.6k22148























            3














            Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.






            share|cite|improve this answer





















            • What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
              – ICanMakeYouHateME
              Nov 27 at 6:18








            • 1




              Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
              – Dante Grevino
              Nov 27 at 6:21
















            3














            Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.






            share|cite|improve this answer





















            • What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
              – ICanMakeYouHateME
              Nov 27 at 6:18








            • 1




              Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
              – Dante Grevino
              Nov 27 at 6:21














            3












            3








            3






            Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.






            share|cite|improve this answer












            Note that $n.m(S_n)=int_{S_n}nleqint_{S_n}|f|$ for every $n$ and $lim_{nto+infty}int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 at 6:12









            Dante Grevino

            90819




            90819












            • What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
              – ICanMakeYouHateME
              Nov 27 at 6:18








            • 1




              Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
              – Dante Grevino
              Nov 27 at 6:21


















            • What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
              – ICanMakeYouHateME
              Nov 27 at 6:18








            • 1




              Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
              – Dante Grevino
              Nov 27 at 6:21
















            What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
            – ICanMakeYouHateME
            Nov 27 at 6:18






            What theorem do we need to know to show that $lim_{n to infty} int_{S_n} |f|=0$? Is this because the measure of the $S_n$ is going to zero?
            – ICanMakeYouHateME
            Nov 27 at 6:18






            1




            1




            Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
            – Dante Grevino
            Nov 27 at 6:21




            Lebesgue's Dominated Convergence Theorem: Note that $int_{S_n}|f|=int_E|f|chi_{S_n}$, $|f|chi_{S_n}$ converges to $0$ a.e. and it is dominated by $|f|$, which is integrable by hypothesis.
            – Dante Grevino
            Nov 27 at 6:21


















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