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

Won the lottery - how do I keep the money?

Prepend last line of stdin to entire stdin

Why doesn't UK go for the same deal Japan has with EU to resolve Brexit?

What connection does MS Office have to Netscape Navigator?

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

Running a General Election and the European Elections together

How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?

Why is information "lost" when it got into a black hole?

INSERT to a table from a database to other (same SQL Server) using Dynamic SQL

How to avoid supervisors with prejudiced views?

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

Is it ever safe to open a suspicious HTML file (e.g. email attachment)?

"misplaced omit" error when >centering columns

Is it professional to write unrelated content in an almost-empty email?

How do I align (1) and (2)?

Should I tutor a student who I know has cheated on their homework?

A Man With a Stainless Steel Endoskeleton (like The Terminator) Fighting Cloaked Aliens Only He Can See

0 rank tensor vs 1D vector

RigExpert AA-35 - Interpreting The Information

Dominated convergence theorem - what sequence?

I want to delete every two lines after 3rd lines in file contain very large number of lines :

What did we know about the Kessel run before the prequels?

How to count occurrences of text in a file?

Why, when going from special to general relativity, do we just replace partial derivatives with covariant derivatives?



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.
              1 hour ago










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











            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









            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.
                      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 $lim_ntoinftyf_n = f$ 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.
                      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 $lim_ntoinftyf_n = f$ 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.
                      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. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ 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.
                      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. $|f(k,varepsilon)| leq g(k), forall k inmathbbR$ 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.
                    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.
                    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

















                    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

                    Can not update quote_id field of “quote_item” table magento 2Magento 2.1 - We can't remove the item. (Shopping Cart doesnt allow us to remove items before becomes empty)Add value for custom quote item attribute using REST apiREST API endpoint v1/carts/cartId/items always returns error messageCorrect way to save entries to databaseHow to remove all associated quote objects of a customer completelyMagento 2 - Save value from custom input field to quote_itemGet quote_item data using quote id and product id filter in Magento 2How to set additional data to quote_item table from controller in Magento 2?What is the purpose of additional_data column in quote_item table in magento2Set Custom Price to Quote item magento2 from controller

                    How to solve knockout JS error in Magento 2 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?(Magento2) knockout.js:3012 Uncaught ReferenceError: Unable to process bindingUnable to process binding Knockout.js magento 2Cannot read property `scopeLabel` of undefined on Product Detail PageCan't get Customer Data on frontend in Magento 2Magento2 Order Summary - unable to process bindingKO templates are not loading in Magento 2.1 applicationgetting knockout js error magento 2Product grid not load -— Unable to process binding Knockout.js magento 2Product form not loaded in magento2Uncaught ReferenceError: Unable to process binding “if: function()return (isShowLegend()) ” magento 2

                    Nissan Patrol Зміст Перше покоління — 4W60 (1951-1960) | Друге покоління — 60 series (1960-1980) | Третє покоління (1980–2002) | Четверте покоління — Y60 (1987–1998) | П'яте покоління — Y61 (1997–2013) | Шосте покоління — Y62 (2010- ) | Посилання | Зноски | Навігаційне менюОфіційний український сайтТест-драйв Nissan Patrol 2010 7-го поколінняNissan PatrolКак мы тестировали Nissan Patrol 2016рвиправивши або дописавши її