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

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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?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










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 45 mins ago









      Rócherz

      3,0013821




      3,0013821










      asked 58 mins 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
            26 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
            26 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
            26 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
            26 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 8 mins ago

























          answered 40 mins ago









          rolandcyprolandcyp

          1,856315




          1,856315











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
















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















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




          $begingroup$
          I understood. Thanks a lot for the help
          $endgroup$
          – Ricardo Freire
          26 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
            26 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
            26 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 39 mins ago









          Alex OrtizAlex Ortiz

          11.2k21441




          11.2k21441











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
















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















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




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

















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          ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result