Dominated convergence theorem - what sequence?












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$begingroup$


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






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      asked 4 hours ago









      Ivan V.Ivan V.

      911216




      911216






















          2 Answers
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          $begingroup$

          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












            $begingroup$

            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.
              1 hour ago










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












            Your Answer





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            2 Answers
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            $begingroup$

            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



                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 $lim_{ntoinfty}f_n = f$ 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.
                      1 hour 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 $lim_{ntoinfty}f_n = f$ 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.
                      1 hour 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 $lim_{ntoinfty}f_n = f$ 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 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
                      1 hour 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
                      1 hour 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. $|f(k,varepsilon)| leq g(k), forall k inmathbb{R}$ 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.
                    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.
                    1 hour 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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