Dominated convergence theorem - what sequence? The Next CEO of Stack OverflowWhat are some good integration problems where you can use some of the function convergence theorem of Lesbegue integrals?Find Limit Using Lebesgue Dominated ConvergenceSolving these types of integrals, using Monotone convergence theorem and Dominated convergence theorem.Applications of Dominated/Monotone convergence theoremLebesgue Dominated Convergence Theorem exampleDominated convergence theorem for log-integrable rational functionsuniform or dominated convergence of sequence of functions which are boundedBartle's proof of Lebesgue Dominated Convergence TheoremCalculate the limit using dominated or monotone convergence theoremUsing dominated convergence theorem to move limit inside the integral

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Dominated convergence theorem - what sequence?



The Next CEO of Stack OverflowWhat are some good integration problems where you can use some of the function convergence theorem of Lesbegue integrals?Find Limit Using Lebesgue Dominated ConvergenceSolving these types of integrals, using Monotone convergence theorem and Dominated convergence theorem.Applications of Dominated/Monotone convergence theoremLebesgue Dominated Convergence Theorem exampleDominated convergence theorem for log-integrable rational functionsuniform or dominated convergence of sequence of functions which are boundedBartle's proof of Lebesgue Dominated Convergence TheoremCalculate the limit using dominated or monotone convergence theoremUsing dominated convergence theorem to move limit inside the integral










2












$begingroup$


Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)



P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
    $$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
    Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)



    P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!










    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
      $$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
      Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)



      P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!










      share|cite|improve this question









      $endgroup$




      Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
      $$lim_varepsilonto0^+ int_-infty^infty dk f(k,varepsilon) overset?= int_-infty^infty dklim_varepsilonto0^+ f(k,varepsilon).$$
      Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)



      P.S. Before someone marks it as a duplicate, please take a minute to understand (not saying that you won't) what it is that I'm asking here. Thank you!







      integration limits






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      Ivan V.Ivan V.

      911216




      911216




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Let's look at it in a silly case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$



          This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$



          And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            The statement of the dominated convergence theorem (DCT) is as follows:




            "Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
            $$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$




            (The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)



            As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:




            Proposition. If $f$ is a function, then
            $$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$




            With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):




            "Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
            $$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$




            The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
              $endgroup$
              – Ivan V.
              2 hours ago










            • $begingroup$
              @IvanV.: Yes, that's correct!
              $endgroup$
              – Alex Ortiz
              1 hour ago











            Your Answer





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            2 Answers
            2






            active

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            2 Answers
            2






            active

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            active

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            active

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            2












            $begingroup$

            Let's look at it in a silly case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$



            This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$



            And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              Let's look at it in a silly case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$



              This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$



              And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                Let's look at it in a silly case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$



                This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$



                And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.






                share|cite|improve this answer









                $endgroup$



                Let's look at it in a silly case. We want to prove by DCT that $$lim_varepsilonto0^+ int_0^infty e^-y/varepsilon,dy=0$$



                This is the case if and only if for all sequence $varepsilon_nto 0^+$ it holds $$lim_ntoinftyint_0^infty e^-y/varepsilon_n,dy=0$$



                And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^-x$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Saucy O'PathSaucy O'Path

                6,2141627




                6,2141627





















                    2












                    $begingroup$

                    The statement of the dominated convergence theorem (DCT) is as follows:




                    "Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
                    $$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$




                    (The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)



                    As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:




                    Proposition. If $f$ is a function, then
                    $$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$




                    With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):




                    "Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
                    $$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$




                    The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
                      $endgroup$
                      – Ivan V.
                      2 hours ago










                    • $begingroup$
                      @IvanV.: Yes, that's correct!
                      $endgroup$
                      – Alex Ortiz
                      1 hour ago















                    2












                    $begingroup$

                    The statement of the dominated convergence theorem (DCT) is as follows:




                    "Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
                    $$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$




                    (The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)



                    As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:




                    Proposition. If $f$ is a function, then
                    $$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$




                    With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):




                    "Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
                    $$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$




                    The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
                      $endgroup$
                      – Ivan V.
                      2 hours ago










                    • $begingroup$
                      @IvanV.: Yes, that's correct!
                      $endgroup$
                      – Alex Ortiz
                      1 hour ago













                    2












                    2








                    2





                    $begingroup$

                    The statement of the dominated convergence theorem (DCT) is as follows:




                    "Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
                    $$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$




                    (The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)



                    As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:




                    Proposition. If $f$ is a function, then
                    $$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$




                    With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):




                    "Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
                    $$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$




                    The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.






                    share|cite|improve this answer











                    $endgroup$



                    The statement of the dominated convergence theorem (DCT) is as follows:




                    "Discrete" DCT. Suppose $f_n_n=1^infty$ is a sequence of (measurable) functions such that $|f_n| le g$ for some integrable function $g$ and all $n$, and $lim_ntoinftyf_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_ntoinftyint f_n = int f$ (by the triangle inequality). This can be written as
                    $$ lim_ntoinftyint f_n = int lim_ntoinfty f_n.$$




                    (The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)



                    As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $f_n_n=1^infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $f_epsilon_0<epsilon<epsilon_0$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:




                    Proposition. If $f$ is a function, then
                    $$lim_epsilonto0^+f(epsilon) = L iff lim_ntoinftyf(a_n) = Lquad textfor $mathbfall$ sequences $a_nto 0^+$.$$




                    With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):




                    "Continuous" DCT. Suppose $f_epsilon_0<epsilon<epsilon_0$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_epsilon| le g$ for some integrable function $g$ and all $0<epsilon<epsilon_0$, and $lim_epsilonto0^+f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_epsilonto 0^+int f_epsilon = int f$. This can be written as
                    $$ lim_epsilonto0^+int f_epsilon = int lim_epsilonto0^+ f_epsilon.$$




                    The way we use the continuous DCT in practice is by picking an arbitrary sequence $pmba_nto 0^+$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_nto 0^+$ and properties of the family $f_epsilon$ that are known to us.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 4 hours ago

























                    answered 4 hours ago









                    Alex OrtizAlex Ortiz

                    11.2k21441




                    11.2k21441











                    • $begingroup$
                      Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
                      $endgroup$
                      – Ivan V.
                      2 hours ago










                    • $begingroup$
                      @IvanV.: Yes, that's correct!
                      $endgroup$
                      – Alex Ortiz
                      1 hour ago
















                    • $begingroup$
                      Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
                      $endgroup$
                      – Ivan V.
                      2 hours ago










                    • $begingroup$
                      @IvanV.: Yes, that's correct!
                      $endgroup$
                      – Alex Ortiz
                      1 hour ago















                    $begingroup$
                    Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
                    $endgroup$
                    – Ivan V.
                    2 hours ago




                    $begingroup$
                    Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ and all $varepsilon$ between $0$ and some positive $varepsilon_0$. Then I check if $f(k,varepsilon) to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right?
                    $endgroup$
                    – Ivan V.
                    2 hours ago












                    $begingroup$
                    @IvanV.: Yes, that's correct!
                    $endgroup$
                    – Alex Ortiz
                    1 hour ago




                    $begingroup$
                    @IvanV.: Yes, that's correct!
                    $endgroup$
                    – Alex Ortiz
                    1 hour ago

















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