Convergence in $L^1$ implies Convergence in measure using Chebyshev's Inequality












2












$begingroup$


Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25


















2












$begingroup$


Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25
















2












2








2





$begingroup$


Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$




Let ${f_n}$ be a sequence of Lebesgue integrable functions that converges to $f$ in $L^1$. Then, I have to show that ${f_n}$ converges in measure.



Here is my approach:



So Claim: $mu(x:|f_n(x)-f(x)|geqepsilon) to 0$



for any $epsilon >0$, by Chebyshev's Inequality, we have



Case 1:
Consider $mu(x:|f_n(x)-f(x)|> epsilon)<frac{1}{epsilon}int{|f_n(x)-f(x)|dmu} to0$ (as $f_n to f$ in $L^1)$



Case 2: Consider $mu(x:|f_n(x)-f(x)|=epsilon)=mu(x:-epsilon +f(x)leq f_n(x)leqepsilon+f(x)) to 0(???)$ for any arbitrary $epsilon$.



I am not sure about Case 2.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral lebesgue-measure




real-analysis measure-theory lebesgue-integral lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 0:21









InfinityInfinity

389113




389113












  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25




















  • $begingroup$
    Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
    $endgroup$
    – Mark
    Jan 5 at 0:25


















$begingroup$
Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
$endgroup$
– Mark
Jan 5 at 0:25






$begingroup$
Not sure why do you need case $2$. If you proved that $mu(x: |f_n(x)-f(x)|>epsilon)to 0$ for any $epsilon>0$ then you proved convergence in measure.
$endgroup$
– Mark
Jan 5 at 0:25












1 Answer
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$begingroup$

There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



$$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
for positive $g$.




Alternatively, notice that



$$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



and apply Case $1$.






share|cite|improve this answer









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    $begingroup$

    There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



    $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
    for positive $g$.




    Alternatively, notice that



    $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



    and apply Case $1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



      $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
      for positive $g$.




      Alternatively, notice that



      $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



      and apply Case $1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



        $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
        for positive $g$.




        Alternatively, notice that



        $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



        and apply Case $1$.






        share|cite|improve this answer









        $endgroup$



        There's not really any reason to have case 2, because Chebyshev's inequality also works with $le$ replacing $<$. If you're dead set on using it, then you haven't got the setup quite right (your first equality is wrong). The relevant estimate would be that



        $$mu{x : g(x) = lambda} = frac 1 {lambda} int_{{x : g(x) = lambda}} g, dmu le frac{1}{lambda} int g , dmu$$
        for positive $g$.




        Alternatively, notice that



        $$mu{|f_n - f| ge epsilon} le muleft{|f_n - f| > frac{epsilon}{2}right}$$



        and apply Case $1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 0:27









        T. BongersT. Bongers

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        23.5k54762






























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