Show $lim_{a to 0} a cdot mu ({ x in mathbb{R} : |f(x)| > a }) = 0$ for $f in L^1 (mathbb{R})$, $a>0$












1












$begingroup$



Let $f in L^1 (mathbb{R})$, and $a > 0$. Show



$$
lim_{a to 0} a cdot mu (E_a) = 0
$$



where $E_a = { x in mathbb{R} : |f(x)| > a }$.





Try



Since



$$
int_mathbb{R} |f| dmu = int_{E_a} |f| dmu + int_{mathbb{R}setminus E_a } |f| dmu < infty
$$



we have $int_{E_a} |f| dmu < infty$, $forall a$.



But I'm stuck at how I can relate this to find $mu(E_a)$.



Any help will be appreciated.






real-analysis measure-theory







share|cite|improve this question









$endgroup$

















    1












    $begingroup$



    Let $f in L^1 (mathbb{R})$, and $a > 0$. Show



    $$
    lim_{a to 0} a cdot mu (E_a) = 0
    $$



    where $E_a = { x in mathbb{R} : |f(x)| > a }$.





    Try



    Since



    $$
    int_mathbb{R} |f| dmu = int_{E_a} |f| dmu + int_{mathbb{R}setminus E_a } |f| dmu < infty
    $$



    we have $int_{E_a} |f| dmu < infty$, $forall a$.



    But I'm stuck at how I can relate this to find $mu(E_a)$.



    Any help will be appreciated.






    real-analysis measure-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      3



      $begingroup$



      Let $f in L^1 (mathbb{R})$, and $a > 0$. Show



      $$
      lim_{a to 0} a cdot mu (E_a) = 0
      $$



      where $E_a = { x in mathbb{R} : |f(x)| > a }$.





      Try



      Since



      $$
      int_mathbb{R} |f| dmu = int_{E_a} |f| dmu + int_{mathbb{R}setminus E_a } |f| dmu < infty
      $$



      we have $int_{E_a} |f| dmu < infty$, $forall a$.



      But I'm stuck at how I can relate this to find $mu(E_a)$.



      Any help will be appreciated.






      real-analysis measure-theory







      share|cite|improve this question









      $endgroup$





      Let $f in L^1 (mathbb{R})$, and $a > 0$. Show



      $$
      lim_{a to 0} a cdot mu (E_a) = 0
      $$



      where $E_a = { x in mathbb{R} : |f(x)| > a }$.





      Try



      Since



      $$
      int_mathbb{R} |f| dmu = int_{E_a} |f| dmu + int_{mathbb{R}setminus E_a } |f| dmu < infty
      $$



      we have $int_{E_a} |f| dmu < infty$, $forall a$.



      But I'm stuck at how I can relate this to find $mu(E_a)$.



      Any help will be appreciated.






      real-analysis measure-theory




      real-analysis measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 20 '18 at 4:35









      MoreblueMoreblue

      8911217




      8911217






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          It is a consequence of Lebesgue's dominated convergence theorem. Note that
          $$
          acdot 1_{{|f(x)|>a}}leq |f(x)|
          $$           for all $a>0$. Thus for any sequence $a_n$ such that $a_n geq a_{n+1} downarrow 0$, we have
          $$
          lim_{ntoinfty}a_n cdotmu(E_{a_n}) =int lim_{ntoinfty}left(a_ncdot 1_{{|f(x)|>a_n}}right)dmu =int 0dmu = 0.
          $$           This establishes $$lim_{ato 0} acdotmu(E_a) =0.$$



          Note: One can see that $acdot mu(E_a)$ is the area of a rectangle below the graph ${(x,|f(x)|) ;|;xin mathbb{R}}$ of $|f|$ which has finite area by the assumption. This observation is the motivation of this approach.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you explain the connection between the note and the solution? (I know what both say, I just don't see the connection)
            $endgroup$
            – mathworker21
            Dec 20 '18 at 4:56












          • $begingroup$
            In fact, it's just heuristics. Consider, for example, $f(x) = frac{1}{1+x^2}$ and a curve $(t,f(t)),tin mathbb{R}$. Then, $acdotmu(|f|>a)= 2acdot sqrt{frac{1-a}{a}}=2sqrt{a(1-a)}$ is the area of the largest rectangle with height $a$ below the curve, and we can see that as $ato 0$, this area converges to $0$. We can generalize this to general $fin L^1$.
            $endgroup$
            – Song
            Dec 20 '18 at 5:02












          • $begingroup$
            I don't see what that has to do with dominated convergence theorem. your comment just gave an explanation of why the result is true, but it did not shed light on how it was the motivation of your approach, i.e. how it motivated you to use DCT
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:22










          • $begingroup$
            Yes, you're right. Intuition itself does not tell we should use DCT. I selected DCT as an appropriate tool to rigorously prove my intuition. But it somehow tells us that we need some kind of integral limit theorem.
            $endgroup$
            – Song
            Dec 20 '18 at 7:26












          • $begingroup$
            lol just because you used it doesn't tell us that we need some form of it. Problem 15 on page 321 of Carothers real analysis book says that if $f$ is non-negative and in $L^1$, then $sum_{k in mathbb{Z}} 2^kmu({f > 2^k}) < infty$. This kind of result (which is very easy to prove) will also give a proof
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:39











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          It is a consequence of Lebesgue's dominated convergence theorem. Note that
          $$
          acdot 1_{{|f(x)|>a}}leq |f(x)|
          $$           for all $a>0$. Thus for any sequence $a_n$ such that $a_n geq a_{n+1} downarrow 0$, we have
          $$
          lim_{ntoinfty}a_n cdotmu(E_{a_n}) =int lim_{ntoinfty}left(a_ncdot 1_{{|f(x)|>a_n}}right)dmu =int 0dmu = 0.
          $$           This establishes $$lim_{ato 0} acdotmu(E_a) =0.$$



          Note: One can see that $acdot mu(E_a)$ is the area of a rectangle below the graph ${(x,|f(x)|) ;|;xin mathbb{R}}$ of $|f|$ which has finite area by the assumption. This observation is the motivation of this approach.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you explain the connection between the note and the solution? (I know what both say, I just don't see the connection)
            $endgroup$
            – mathworker21
            Dec 20 '18 at 4:56












          • $begingroup$
            In fact, it's just heuristics. Consider, for example, $f(x) = frac{1}{1+x^2}$ and a curve $(t,f(t)),tin mathbb{R}$. Then, $acdotmu(|f|>a)= 2acdot sqrt{frac{1-a}{a}}=2sqrt{a(1-a)}$ is the area of the largest rectangle with height $a$ below the curve, and we can see that as $ato 0$, this area converges to $0$. We can generalize this to general $fin L^1$.
            $endgroup$
            – Song
            Dec 20 '18 at 5:02












          • $begingroup$
            I don't see what that has to do with dominated convergence theorem. your comment just gave an explanation of why the result is true, but it did not shed light on how it was the motivation of your approach, i.e. how it motivated you to use DCT
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:22










          • $begingroup$
            Yes, you're right. Intuition itself does not tell we should use DCT. I selected DCT as an appropriate tool to rigorously prove my intuition. But it somehow tells us that we need some kind of integral limit theorem.
            $endgroup$
            – Song
            Dec 20 '18 at 7:26












          • $begingroup$
            lol just because you used it doesn't tell us that we need some form of it. Problem 15 on page 321 of Carothers real analysis book says that if $f$ is non-negative and in $L^1$, then $sum_{k in mathbb{Z}} 2^kmu({f > 2^k}) < infty$. This kind of result (which is very easy to prove) will also give a proof
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:39
















          5












          $begingroup$

          It is a consequence of Lebesgue's dominated convergence theorem. Note that
          $$
          acdot 1_{{|f(x)|>a}}leq |f(x)|
          $$           for all $a>0$. Thus for any sequence $a_n$ such that $a_n geq a_{n+1} downarrow 0$, we have
          $$
          lim_{ntoinfty}a_n cdotmu(E_{a_n}) =int lim_{ntoinfty}left(a_ncdot 1_{{|f(x)|>a_n}}right)dmu =int 0dmu = 0.
          $$           This establishes $$lim_{ato 0} acdotmu(E_a) =0.$$



          Note: One can see that $acdot mu(E_a)$ is the area of a rectangle below the graph ${(x,|f(x)|) ;|;xin mathbb{R}}$ of $|f|$ which has finite area by the assumption. This observation is the motivation of this approach.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you explain the connection between the note and the solution? (I know what both say, I just don't see the connection)
            $endgroup$
            – mathworker21
            Dec 20 '18 at 4:56












          • $begingroup$
            In fact, it's just heuristics. Consider, for example, $f(x) = frac{1}{1+x^2}$ and a curve $(t,f(t)),tin mathbb{R}$. Then, $acdotmu(|f|>a)= 2acdot sqrt{frac{1-a}{a}}=2sqrt{a(1-a)}$ is the area of the largest rectangle with height $a$ below the curve, and we can see that as $ato 0$, this area converges to $0$. We can generalize this to general $fin L^1$.
            $endgroup$
            – Song
            Dec 20 '18 at 5:02












          • $begingroup$
            I don't see what that has to do with dominated convergence theorem. your comment just gave an explanation of why the result is true, but it did not shed light on how it was the motivation of your approach, i.e. how it motivated you to use DCT
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:22










          • $begingroup$
            Yes, you're right. Intuition itself does not tell we should use DCT. I selected DCT as an appropriate tool to rigorously prove my intuition. But it somehow tells us that we need some kind of integral limit theorem.
            $endgroup$
            – Song
            Dec 20 '18 at 7:26












          • $begingroup$
            lol just because you used it doesn't tell us that we need some form of it. Problem 15 on page 321 of Carothers real analysis book says that if $f$ is non-negative and in $L^1$, then $sum_{k in mathbb{Z}} 2^kmu({f > 2^k}) < infty$. This kind of result (which is very easy to prove) will also give a proof
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:39














          5












          5








          5





          $begingroup$

          It is a consequence of Lebesgue's dominated convergence theorem. Note that
          $$
          acdot 1_{{|f(x)|>a}}leq |f(x)|
          $$           for all $a>0$. Thus for any sequence $a_n$ such that $a_n geq a_{n+1} downarrow 0$, we have
          $$
          lim_{ntoinfty}a_n cdotmu(E_{a_n}) =int lim_{ntoinfty}left(a_ncdot 1_{{|f(x)|>a_n}}right)dmu =int 0dmu = 0.
          $$           This establishes $$lim_{ato 0} acdotmu(E_a) =0.$$



          Note: One can see that $acdot mu(E_a)$ is the area of a rectangle below the graph ${(x,|f(x)|) ;|;xin mathbb{R}}$ of $|f|$ which has finite area by the assumption. This observation is the motivation of this approach.






          share|cite|improve this answer











          $endgroup$



          It is a consequence of Lebesgue's dominated convergence theorem. Note that
          $$
          acdot 1_{{|f(x)|>a}}leq |f(x)|
          $$           for all $a>0$. Thus for any sequence $a_n$ such that $a_n geq a_{n+1} downarrow 0$, we have
          $$
          lim_{ntoinfty}a_n cdotmu(E_{a_n}) =int lim_{ntoinfty}left(a_ncdot 1_{{|f(x)|>a_n}}right)dmu =int 0dmu = 0.
          $$           This establishes $$lim_{ato 0} acdotmu(E_a) =0.$$



          Note: One can see that $acdot mu(E_a)$ is the area of a rectangle below the graph ${(x,|f(x)|) ;|;xin mathbb{R}}$ of $|f|$ which has finite area by the assumption. This observation is the motivation of this approach.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 4:54

























          answered Dec 20 '18 at 4:49









          SongSong

          16.4k1741




          16.4k1741












          • $begingroup$
            Can you explain the connection between the note and the solution? (I know what both say, I just don't see the connection)
            $endgroup$
            – mathworker21
            Dec 20 '18 at 4:56












          • $begingroup$
            In fact, it's just heuristics. Consider, for example, $f(x) = frac{1}{1+x^2}$ and a curve $(t,f(t)),tin mathbb{R}$. Then, $acdotmu(|f|>a)= 2acdot sqrt{frac{1-a}{a}}=2sqrt{a(1-a)}$ is the area of the largest rectangle with height $a$ below the curve, and we can see that as $ato 0$, this area converges to $0$. We can generalize this to general $fin L^1$.
            $endgroup$
            – Song
            Dec 20 '18 at 5:02












          • $begingroup$
            I don't see what that has to do with dominated convergence theorem. your comment just gave an explanation of why the result is true, but it did not shed light on how it was the motivation of your approach, i.e. how it motivated you to use DCT
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:22










          • $begingroup$
            Yes, you're right. Intuition itself does not tell we should use DCT. I selected DCT as an appropriate tool to rigorously prove my intuition. But it somehow tells us that we need some kind of integral limit theorem.
            $endgroup$
            – Song
            Dec 20 '18 at 7:26












          • $begingroup$
            lol just because you used it doesn't tell us that we need some form of it. Problem 15 on page 321 of Carothers real analysis book says that if $f$ is non-negative and in $L^1$, then $sum_{k in mathbb{Z}} 2^kmu({f > 2^k}) < infty$. This kind of result (which is very easy to prove) will also give a proof
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:39


















          • $begingroup$
            Can you explain the connection between the note and the solution? (I know what both say, I just don't see the connection)
            $endgroup$
            – mathworker21
            Dec 20 '18 at 4:56












          • $begingroup$
            In fact, it's just heuristics. Consider, for example, $f(x) = frac{1}{1+x^2}$ and a curve $(t,f(t)),tin mathbb{R}$. Then, $acdotmu(|f|>a)= 2acdot sqrt{frac{1-a}{a}}=2sqrt{a(1-a)}$ is the area of the largest rectangle with height $a$ below the curve, and we can see that as $ato 0$, this area converges to $0$. We can generalize this to general $fin L^1$.
            $endgroup$
            – Song
            Dec 20 '18 at 5:02












          • $begingroup$
            I don't see what that has to do with dominated convergence theorem. your comment just gave an explanation of why the result is true, but it did not shed light on how it was the motivation of your approach, i.e. how it motivated you to use DCT
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:22










          • $begingroup$
            Yes, you're right. Intuition itself does not tell we should use DCT. I selected DCT as an appropriate tool to rigorously prove my intuition. But it somehow tells us that we need some kind of integral limit theorem.
            $endgroup$
            – Song
            Dec 20 '18 at 7:26












          • $begingroup$
            lol just because you used it doesn't tell us that we need some form of it. Problem 15 on page 321 of Carothers real analysis book says that if $f$ is non-negative and in $L^1$, then $sum_{k in mathbb{Z}} 2^kmu({f > 2^k}) < infty$. This kind of result (which is very easy to prove) will also give a proof
            $endgroup$
            – mathworker21
            Dec 20 '18 at 7:39
















          $begingroup$
          Can you explain the connection between the note and the solution? (I know what both say, I just don't see the connection)
          $endgroup$
          – mathworker21
          Dec 20 '18 at 4:56






          $begingroup$
          Can you explain the connection between the note and the solution? (I know what both say, I just don't see the connection)
          $endgroup$
          – mathworker21
          Dec 20 '18 at 4:56














          $begingroup$
          In fact, it's just heuristics. Consider, for example, $f(x) = frac{1}{1+x^2}$ and a curve $(t,f(t)),tin mathbb{R}$. Then, $acdotmu(|f|>a)= 2acdot sqrt{frac{1-a}{a}}=2sqrt{a(1-a)}$ is the area of the largest rectangle with height $a$ below the curve, and we can see that as $ato 0$, this area converges to $0$. We can generalize this to general $fin L^1$.
          $endgroup$
          – Song
          Dec 20 '18 at 5:02






          $begingroup$
          In fact, it's just heuristics. Consider, for example, $f(x) = frac{1}{1+x^2}$ and a curve $(t,f(t)),tin mathbb{R}$. Then, $acdotmu(|f|>a)= 2acdot sqrt{frac{1-a}{a}}=2sqrt{a(1-a)}$ is the area of the largest rectangle with height $a$ below the curve, and we can see that as $ato 0$, this area converges to $0$. We can generalize this to general $fin L^1$.
          $endgroup$
          – Song
          Dec 20 '18 at 5:02














          $begingroup$
          I don't see what that has to do with dominated convergence theorem. your comment just gave an explanation of why the result is true, but it did not shed light on how it was the motivation of your approach, i.e. how it motivated you to use DCT
          $endgroup$
          – mathworker21
          Dec 20 '18 at 7:22




          $begingroup$
          I don't see what that has to do with dominated convergence theorem. your comment just gave an explanation of why the result is true, but it did not shed light on how it was the motivation of your approach, i.e. how it motivated you to use DCT
          $endgroup$
          – mathworker21
          Dec 20 '18 at 7:22












          $begingroup$
          Yes, you're right. Intuition itself does not tell we should use DCT. I selected DCT as an appropriate tool to rigorously prove my intuition. But it somehow tells us that we need some kind of integral limit theorem.
          $endgroup$
          – Song
          Dec 20 '18 at 7:26






          $begingroup$
          Yes, you're right. Intuition itself does not tell we should use DCT. I selected DCT as an appropriate tool to rigorously prove my intuition. But it somehow tells us that we need some kind of integral limit theorem.
          $endgroup$
          – Song
          Dec 20 '18 at 7:26














          $begingroup$
          lol just because you used it doesn't tell us that we need some form of it. Problem 15 on page 321 of Carothers real analysis book says that if $f$ is non-negative and in $L^1$, then $sum_{k in mathbb{Z}} 2^kmu({f > 2^k}) < infty$. This kind of result (which is very easy to prove) will also give a proof
          $endgroup$
          – mathworker21
          Dec 20 '18 at 7:39




          $begingroup$
          lol just because you used it doesn't tell us that we need some form of it. Problem 15 on page 321 of Carothers real analysis book says that if $f$ is non-negative and in $L^1$, then $sum_{k in mathbb{Z}} 2^kmu({f > 2^k}) < infty$. This kind of result (which is very easy to prove) will also give a proof
          $endgroup$
          – mathworker21
          Dec 20 '18 at 7:39


















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