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

Solving system of ODEs with extra parameter

Where do students learn to solve polynomial equations these days?

How to avoid supervisors with prejudiced views?

Is there a way to save my career from absolute disaster?

Why the difference in type-inference over the as-pattern in two similar function definitions?

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

Is it my responsibility to learn a new technology in my own time my employer wants to implement?

Domestic-to-international connection at Orlando (MCO)

How many extra stops do monopods offer for tele photographs?

Is a distribution that is normal, but highly skewed considered Gaussian?

Why do remote US companies require working in the US?

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

Does increasing your ability score affect your main stat?

WOW air has ceased operation, can I get my tickets refunded?

Does soap repel water?

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

I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin

"misplaced omit" error when >centering columns

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

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

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

Writing differences on a blackboard

How to get from Geneva Airport to Metabief?

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



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

                    How should I use the fbox command correctly to avoid producing a Bad Box message?How to put a long piece of text in a box?How to specify height and width of fboxIs there an arrayrulecolor-like command to change the rule color of fbox?What is the command to highlight bad boxes in pdf?Why does fbox sometimes place the box *over* the graphic image?how to put the text in the boxHow to create command for a box where text inside the box can automatically adjust?how can I make an fbox like command with certain color, shape and width of border?how to use fbox in align modeFbox increase the spacing between the box and it content (inner margin)how to change the box height of an equationWhat is the use of the hbox in a newcommand command?

                    Doxepinum Nexus interni Notae | Tabula navigationis3158DB01142WHOa682390"Structural Analysis of the Histamine H1 Receptor""Transdermal and Topical Drug Administration in the Treatment of Pain""Antidepressants as antipruritic agents: A review"

                    inputenc: Unicode character … not set up for use with LaTeX The Next CEO of Stack OverflowEntering Unicode characters in LaTeXHow to solve the `Package inputenc Error: Unicode char not set up for use with LaTeX` problem?solve “Unicode char is not set up for use with LaTeX” without special handling of every new interesting UTF-8 characterPackage inputenc Error: Unicode character ² (U+B2)(inputenc) not set up for use with LaTeX. acroI2C[I²C]package inputenc error unicode char (u + 190) not set up for use with latexPackage inputenc Error: Unicode char u8:′ not set up for use with LaTeX. 3′inputenc Error: Unicode char u8: not set up for use with LaTeX with G-BriefPackage Inputenc Error: Unicode char u8: not set up for use with LaTeXPackage inputenc Error: Unicode char ́ (U+301)(inputenc) not set up for use with LaTeX. includePackage inputenc Error: Unicode char ̂ (U+302)(inputenc) not set up for use with LaTeX. … $widehatleft (OA,AA' right )$Package inputenc Error: Unicode char â„¡ (U+2121)(inputenc) not set up for use with LaTeX. printbibliography[heading=bibintoc]Package inputenc Error: Unicode char − (U+2212)(inputenc) not set up for use with LaTeXPackage inputenc Error: Unicode character α (U+3B1) not set up for use with LaTeXPackage inputenc Error: Unicode characterError: ! Package inputenc Error: Unicode char ⊘ (U+2298)(inputenc) not set up for use with LaTeX