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

A small doubt about the dominated convergence theorem

Easy to read palindrome checker

Necessary condition on homology group for a set to be contractible

Bartok - Syncopation (1): Meaning of notes in between Grand Staff

How to delete every two lines after 3rd lines in a file contains very large number of lines?

Why isn't the Mueller report being released completely and unredacted?

Is there a difference between "Fahrstuhl" and "Aufzug"

Recycling old answers

Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?

Flying from Cape Town to England and return to another province

Example of a Mathematician/Physicist whose Other Publications during their PhD eclipsed their PhD Thesis

Where do students learn to solve polynomial equations these days?

Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?

Would a completely good Muggle be able to use a wand?

Why didn't Khan get resurrected in the Genesis Explosion?

Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?

Proper way to express "He disappeared them"

Why do airplanes bank sharply to the right after air-to-air refueling?

If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?

Rotate a column

Can Plant Growth be repeatedly cast on the same area to exponentially increase the yield of harvests there (more than twice)?

Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?

Poetry, calligrams and TikZ/PStricks challenge

Legal workarounds for testamentary trust perceived as unfair



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










3












$begingroup$


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









$endgroup$
















    3












    $begingroup$


    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









    $endgroup$














      3












      3








      3





      $begingroup$


      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 9 hours ago









      Ivan V.Ivan V.

      981216




      981216




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          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.
            6 hours ago










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










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


















          3












          $begingroup$

          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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168778%2fdominated-convergence-theorem-what-sequence%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            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.
              6 hours ago










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










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















            4












            $begingroup$

            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.
              6 hours ago










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










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













            4












            4








            4





            $begingroup$

            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 8 hours ago

























            answered 8 hours ago









            Alex OrtizAlex Ortiz

            11.3k21441




            11.3k21441











            • $begingroup$
              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.
              6 hours ago










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










            • $begingroup$
              Alright, thank you, much appreciated!
              $endgroup$
              – Ivan V.
              4 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.
              6 hours ago










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










            • $begingroup$
              Alright, thank you, much appreciated!
              $endgroup$
              – Ivan V.
              4 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.
            6 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.
            6 hours ago












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




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












            $begingroup$
            Alright, thank you, much appreciated!
            $endgroup$
            – Ivan V.
            4 hours ago




            $begingroup$
            Alright, thank you, much appreciated!
            $endgroup$
            – Ivan V.
            4 hours ago











            3












            $begingroup$

            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









            $endgroup$

















              3












              $begingroup$

              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









              $endgroup$















                3












                3








                3





                $begingroup$

                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 9 hours ago









                Saucy O'PathSaucy O'Path

                6,2241627




                6,2241627



























                    draft saved

                    draft discarded
















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168778%2fdominated-convergence-theorem-what-sequence%23new-answer', 'question_page');

                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    How does Billy Russo acquire his 'Jigsaw' mask? Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara Favourite questions and answers from the 1st quarter of 2019Why does Bane wear the mask?Why does Kylo Ren wear a mask?Why did Captain America remove his mask while fighting Batroc the Leaper?How did the OA acquire her wisdom?Is Billy Breckenridge gay?How does Adrian Toomes hide his earnings from the IRS?What is the state of affairs on Nootka Sound by the end of season 1?How did Tia Dalma acquire Captain Barbossa's body?How is one “Deemed Worthy”, to acquire the Greatsword “Dawn”?How did Karen acquire the handgun?

                    Личност Атрибути на личността | Литература и източници | НавигацияРаждането на личносттаредактиратередактирате

                    A sequel to Domino's tragic life Why Christmas is for Friends Cold comfort at Charles' padSad farewell for Lady JanePS Most watched News videos