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

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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 49 mins 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
            30 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
            30 mins ago











          Your Answer





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          2 Answers
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          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
            30 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, $f_n(x) to 0$ 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
            30 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, $f_n(x) to 0$ 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 13 mins ago

























          answered 44 mins ago









          rolandcyprolandcyp

          1,856315




          1,856315











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
















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















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




          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          30 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
            30 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
            30 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 43 mins ago









          Alex OrtizAlex Ortiz

          11.2k21441




          11.2k21441











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
















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















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




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

















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