A small doubt about the dominated convergence theorem The Next CEO of Stack OverflowIs Lebesgue's Dominated Convergence Theorem a logical equivalence?Generalisation of Dominated Convergence TheoremLebesgue Convergence using The General Lebesgue Dominated Convergence TheoremVariant of dominated convergence theoremExample about Dominated Convergence TheoremDominated Convergence TheoremHypothesis of dominated convergence theoremBartle's proof of Lebesgue Dominated Convergence TheoremAn counterexample for the monotone convergence theorem and dominated convergence theoremTheorem similar to dominated convergence theorem

Is there always a complete, orthogonal set of unitary matrices?

How many extra stops do monopods offer for tele photographs?

Chain wire methods together in Lightning Web Components

Make solar eclipses exceedingly rare, but still have new moons

Would be okay to drive on this tire?

RigExpert AA-35 - Interpreting The Information

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

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

Method for adding error messages to a dictionary given a key

Why is quantifier elimination desirable for a given theory?

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

Why don't programming languages automatically manage the synchronous/asynchronous problem?

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

How to avoid supervisors with prejudiced views?

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

Does increasing your ability score affect your main stat?

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

Domestic-to-international connection at Orlando (MCO)

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

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

Math-accent symbol over parentheses enclosing accented symbol (amsmath)

The exact meaning of 'Mom made me a sandwich'

Is it possible to replace duplicates of a character with one character using tr

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



A small doubt about the dominated convergence theorem



The Next CEO of Stack OverflowIs Lebesgue's Dominated Convergence Theorem a logical equivalence?Generalisation of Dominated Convergence TheoremLebesgue Convergence using The General Lebesgue Dominated Convergence TheoremVariant of dominated convergence theoremExample about Dominated Convergence TheoremDominated Convergence TheoremHypothesis of dominated convergence theoremBartle's proof of Lebesgue Dominated Convergence TheoremAn counterexample for the monotone convergence theorem and dominated convergence theoremTheorem similar to dominated convergence theorem










2












$begingroup$



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











$endgroup$
















    2












    $begingroup$



    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











    $endgroup$














      2












      2








      2





      $begingroup$



      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 34 mins ago









      Rócherz

      3,0013821




      3,0013821










      asked 46 mins ago









      Ricardo FreireRicardo Freire

      574211




      574211




















          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, $f_n to 0$ 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
            15 mins 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
            15 mins 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, $f_n to 0$ 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
            15 mins 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, $f_n to 0$ 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
            15 mins 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, $f_n to 0$ 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, $f_n to 0$ 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 22 mins ago

























          answered 29 mins ago









          rolandcyprolandcyp

          1,856315




          1,856315











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
















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















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




          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          15 mins 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
            15 mins 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
            15 mins 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 28 mins ago









          Alex OrtizAlex Ortiz

          11.2k21441




          11.2k21441











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
















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















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




          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          15 mins 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

          Are there any AGPL-style licences that require source code modifications to be public? 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?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

          2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

          Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e