A small doubt about the dominated convergence theorem The Next CEO of Stack OverflowIs Lebesgue's Dominated Convergence Theorem a logical equivalence?Lebesgue's Dominated Convergence Theorem questionsExample about Dominated Convergence TheoremDominated Convergence TheoremNecessity of generalization of Dominated Convergence theoremSeeking counterexample for Dominated Convergence theoremHypothesis of dominated convergence theoremDominated convergence theorem vs continuityBartle's proof of Lebesgue Dominated Convergence TheoremTheorem similar to dominated convergence theorem

A small doubt about the dominated convergence theorem

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

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

Which one is the true statement?

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

Find non-case sensitive string in a mixed list of elements?

How to edit “Name” property in GCI output?

RigExpert AA-35 - Interpreting The Information

The exact meaning of 'Mom made me a sandwich'

Flying from Cape Town to England and return to another province

Can a Bladesinger Wizard use Bladesong with a Hand Crossbow?

Where does this common spurious transmission come from? Is there a quality difference?

Does Germany produce more waste than the US?

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

Running a General Election and the European Elections together

The past simple of "gaslight" – "gaslighted" or "gaslit"?

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

0 rank tensor vs 1D vector

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

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

Some questions about different axiomatic systems for neighbourhoods

What flight has the highest ratio of time difference to flight time?

What was the first Unix version to run on a microcomputer?

Easy to read palindrome checker



A small doubt about the dominated convergence theorem



The Next CEO of Stack OverflowIs Lebesgue's Dominated Convergence Theorem a logical equivalence?Lebesgue's Dominated Convergence Theorem questionsExample about Dominated Convergence TheoremDominated Convergence TheoremNecessity of generalization of Dominated Convergence theoremSeeking counterexample for Dominated Convergence theoremHypothesis of dominated convergence theoremDominated convergence theorem vs continuityBartle's proof of Lebesgue Dominated Convergence TheoremTheorem similar to dominated convergence theorem










3












$begingroup$



Theorem $mathbfA.2.11$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$




I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?










share|cite|improve this question











$endgroup$
















    3












    $begingroup$



    Theorem $mathbfA.2.11$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$




    I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?










    share|cite|improve this question











    $endgroup$














      3












      3








      3


      1



      $begingroup$



      Theorem $mathbfA.2.11$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$




      I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?










      share|cite|improve this question











      $endgroup$





      Theorem $mathbfA.2.11$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$




      I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?







      measure-theory convergence lebesgue-integral






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Rócherz

      3,0013821




      3,0013821










      asked 1 hour ago









      Ricardo FreireRicardo Freire

      579211




      579211




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
          $$
          f_n(x) := frac1n mathbf1_[0,n](x).
          $$

          Clearly, $f_n in L^1(mathbbR)$ for each $n in mathbbN$. Moreover, $f_n(x) to 0$ as $n to infty$ for each $x in mathbbR$. However,
          beginalign*
          lim_n to infty int_mathbbR f_n,mathrmdm = lim_n to infty int_0^n frac1n,mathrmdx = 1 neq 0.
          endalign*



          Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.




          Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrakM,mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
          $$
          lim_n to infty int_E f_n,mathrmdmu = int_E f,mathrmdmu.
          $$

          In fact, one has $f_n to f$ strongly in $L^1(E)$.




          In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago


















          2












          $begingroup$

          In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_[n,n+1]$ on $mathbf R_ge 0$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_ge 0$, but they are not dominated by an integrable function $g$, and indeed we do not have
          $$
          lim_ntoinfty int f_n = int lim_ntoinftyf_n
          $$

          since in this case, the left-hand side is $1$, but the right-hand side is $0$.




          To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_ge 0$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            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%2f3168945%2fa-small-doubt-about-the-dominated-convergence-theorem%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









          3












          $begingroup$

          This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
          $$
          f_n(x) := frac1n mathbf1_[0,n](x).
          $$

          Clearly, $f_n in L^1(mathbbR)$ for each $n in mathbbN$. Moreover, $f_n(x) to 0$ as $n to infty$ for each $x in mathbbR$. However,
          beginalign*
          lim_n to infty int_mathbbR f_n,mathrmdm = lim_n to infty int_0^n frac1n,mathrmdx = 1 neq 0.
          endalign*



          Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.




          Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrakM,mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
          $$
          lim_n to infty int_E f_n,mathrmdmu = int_E f,mathrmdmu.
          $$

          In fact, one has $f_n to f$ strongly in $L^1(E)$.




          In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago















          3












          $begingroup$

          This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
          $$
          f_n(x) := frac1n mathbf1_[0,n](x).
          $$

          Clearly, $f_n in L^1(mathbbR)$ for each $n in mathbbN$. Moreover, $f_n(x) to 0$ as $n to infty$ for each $x in mathbbR$. However,
          beginalign*
          lim_n to infty int_mathbbR f_n,mathrmdm = lim_n to infty int_0^n frac1n,mathrmdx = 1 neq 0.
          endalign*



          Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.




          Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrakM,mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
          $$
          lim_n to infty int_E f_n,mathrmdmu = int_E f,mathrmdmu.
          $$

          In fact, one has $f_n to f$ strongly in $L^1(E)$.




          In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago













          3












          3








          3





          $begingroup$

          This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
          $$
          f_n(x) := frac1n mathbf1_[0,n](x).
          $$

          Clearly, $f_n in L^1(mathbbR)$ for each $n in mathbbN$. Moreover, $f_n(x) to 0$ as $n to infty$ for each $x in mathbbR$. However,
          beginalign*
          lim_n to infty int_mathbbR f_n,mathrmdm = lim_n to infty int_0^n frac1n,mathrmdx = 1 neq 0.
          endalign*



          Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.




          Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrakM,mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
          $$
          lim_n to infty int_E f_n,mathrmdmu = int_E f,mathrmdmu.
          $$

          In fact, one has $f_n to f$ strongly in $L^1(E)$.




          In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.






          share|cite|improve this answer











          $endgroup$



          This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
          $$
          f_n(x) := frac1n mathbf1_[0,n](x).
          $$

          Clearly, $f_n in L^1(mathbbR)$ for each $n in mathbbN$. Moreover, $f_n(x) to 0$ as $n to infty$ for each $x in mathbbR$. However,
          beginalign*
          lim_n to infty int_mathbbR f_n,mathrmdm = lim_n to infty int_0^n frac1n,mathrmdx = 1 neq 0.
          endalign*



          Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.




          Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrakM,mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
          $$
          lim_n to infty int_E f_n,mathrmdmu = int_E f,mathrmdmu.
          $$

          In fact, one has $f_n to f$ strongly in $L^1(E)$.




          In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          rolandcyprolandcyp

          1,856315




          1,856315











          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago
















          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago















          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          1 hour ago




          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          1 hour ago











          2












          $begingroup$

          In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_[n,n+1]$ on $mathbf R_ge 0$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_ge 0$, but they are not dominated by an integrable function $g$, and indeed we do not have
          $$
          lim_ntoinfty int f_n = int lim_ntoinftyf_n
          $$

          since in this case, the left-hand side is $1$, but the right-hand side is $0$.




          To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_ge 0$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago















          2












          $begingroup$

          In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_[n,n+1]$ on $mathbf R_ge 0$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_ge 0$, but they are not dominated by an integrable function $g$, and indeed we do not have
          $$
          lim_ntoinfty int f_n = int lim_ntoinftyf_n
          $$

          since in this case, the left-hand side is $1$, but the right-hand side is $0$.




          To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_ge 0$.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago













          2












          2








          2





          $begingroup$

          In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_[n,n+1]$ on $mathbf R_ge 0$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_ge 0$, but they are not dominated by an integrable function $g$, and indeed we do not have
          $$
          lim_ntoinfty int f_n = int lim_ntoinftyf_n
          $$

          since in this case, the left-hand side is $1$, but the right-hand side is $0$.




          To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_ge 0$.






          share|cite|improve this answer









          $endgroup$



          In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_[n,n+1]$ on $mathbf R_ge 0$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_ge 0$, but they are not dominated by an integrable function $g$, and indeed we do not have
          $$
          lim_ntoinfty int f_n = int lim_ntoinftyf_n
          $$

          since in this case, the left-hand side is $1$, but the right-hand side is $0$.




          To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_ge 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Alex OrtizAlex Ortiz

          11.2k21441




          11.2k21441











          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago
















          • $begingroup$
            I understood. Thanks a lot for the help
            $endgroup$
            – Ricardo Freire
            1 hour ago















          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          1 hour ago




          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          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%2f3168945%2fa-small-doubt-about-the-dominated-convergence-theorem%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

          Oświęcim Innehåll Historia | Källor | Externa länkar | Navigeringsmeny50°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.2213950°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.221393089658Nordisk familjebok, AuschwitzInsidan tro och existensJewish Community i OświęcimAuschwitz Jewish Center: MuseumAuschwitz Jewish Center

          Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

          Typsetting diagram chases (with TikZ?) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to define the default vertical distance between nodes?Draw edge on arcNumerical conditional within tikz keys?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to place nodes in an absolute coordinate system in tikzCommutative diagram with curve connecting between nodesTikz with standalone: pinning tikz coordinates to page cmDrawing a Decision Diagram with Tikz and layout manager