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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 6 hours ago









      Ivan V.Ivan V.

      931216




      931216






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
          $$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$




          (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
          $$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$




          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 $pmb{a_nto 0^+}$ 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
            3 hours ago










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










          • $begingroup$
            Alright, thank you, much appreciated!
            $endgroup$
            – Ivan V.
            1 hour ago



















          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 $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$



          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









          $endgroup$














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            2 Answers
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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
            $$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$




            (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
            $$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$




            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 $pmb{a_nto 0^+}$ 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
              3 hours ago










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










            • $begingroup$
              Alright, thank you, much appreciated!
              $endgroup$
              – Ivan V.
              1 hour ago
















            3












            $begingroup$

            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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
            $$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$




            (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
            $$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$




            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 $pmb{a_nto 0^+}$ 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
              3 hours ago










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










            • $begingroup$
              Alright, thank you, much appreciated!
              $endgroup$
              – Ivan V.
              1 hour ago














            3












            3








            3





            $begingroup$

            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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
            $$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$




            (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
            $$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$




            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 $pmb{a_nto 0^+}$ 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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
            $$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$




            (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
            $$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$




            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 $pmb{a_nto 0^+}$ 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 5 hours ago

























            answered 5 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
              3 hours ago










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










            • $begingroup$
              Alright, thank you, much appreciated!
              $endgroup$
              – Ivan V.
              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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
              3 hours ago










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










            • $begingroup$
              Alright, thank you, much appreciated!
              $endgroup$
              – Ivan V.
              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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
            3 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
            3 hours ago












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




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












            $begingroup$
            Alright, thank you, much appreciated!
            $endgroup$
            – Ivan V.
            1 hour ago




            $begingroup$
            Alright, thank you, much appreciated!
            $endgroup$
            – Ivan V.
            1 hour ago











            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 $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$



            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 $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$



              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 $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$



                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 $$lim_{ntoinfty}int_0^infty e^{-y/varepsilon_n},dy=0$$



                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



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                answered 5 hours ago









                Saucy O'PathSaucy O'Path

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