Uniform convergence doesn't preserve improper integrals











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Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence







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  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15















up vote
0
down vote

favorite












Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence







share|cite|improve this question






















  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence







share|cite|improve this question













Given $f_{n} : [0, infty) mapsto mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$



We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$lim_{ntoinfty} int_{a}^{b} f_{n} dx = int_{a}^{b} f(x) dx $$



However when the integral is improper
$$lim_{ntoinfty} int_{a}^{infty} f_{n} dx = lim_{ntoinfty} lim_{btoinfty} int_{a}^{b} f_{n} neq int_{a}^{infty} f(x) dx $$ even though uniform convergence preserves limits?



I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.






real-analysis improper-integrals uniform-convergence




real-analysis improper-integrals uniform-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 17:57









9Sp

223




223












  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15


















  • The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
    – Federico
    Nov 26 at 18:06










  • Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
    – Federico
    Nov 26 at 18:07










  • You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
    – Mark Viola
    Nov 26 at 18:08












  • If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
    – Federico
    Nov 26 at 18:08










  • I haven't studied measure yet, but will look into it! thanks.
    – 9Sp
    Nov 26 at 18:15
















The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
– Federico
Nov 26 at 18:06




The problem is that $[0,infty)$ has infinite measure, so uniform convergence (which is convergence in $L^infty$) does not imply convergence in $L^1$. You may want to look into Lebesgue Theorem
– Federico
Nov 26 at 18:06












Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
– Federico
Nov 26 at 18:07




Note that the same phenomenon happens for series $sum_{k=0}^infty$. Again, the problem is that the underlying measure is not finite
– Federico
Nov 26 at 18:07












You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
– Mark Viola
Nov 26 at 18:08






You're interchanging two limits and interchanging a limit with an integral. If the integral converges uniformly for $nin mathbb{N}$, then you can interchange the limits and proceed
– Mark Viola
Nov 26 at 18:08














If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
– Federico
Nov 26 at 18:08




If it where, then $|int_X f_n,dmu-int_X f,dmu|leq|f_n-f|_infty mu(X)$ and uniform (L^infty) convergence would imply $L^1$ convergence
– Federico
Nov 26 at 18:08












I haven't studied measure yet, but will look into it! thanks.
– 9Sp
Nov 26 at 18:15




I haven't studied measure yet, but will look into it! thanks.
– 9Sp
Nov 26 at 18:15










1 Answer
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If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer





















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20











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If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer





















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20















up vote
0
down vote



accepted










If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer





















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20













up vote
0
down vote



accepted







up vote
0
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accepted






If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.






share|cite|improve this answer












If $(f_n)_{ninmathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $mathbb R$ wich converges uniformly to $fcolon[0,a]longrightarrowmathbb R$, then, for each $varepsilon>0$, you have, if $n$ is large enough,$$(forall xin[0,a]):bigllvert f(x)-f_n(x)bigrrvert<varepsilon,$$and therefore,$$leftlvertint_0^af(x),mathrm dx-int_0^af_n(x),mathrm dxrightrvert<varepsilon a.tag1$$So, if $n$ is large enought, the numbers $int_0^af(x),mathrm dx$ and $int_0^af_n(x),mathrm dx$ are close to each other. But if are integrating over, say $[0,infty)$, then you don't have $(1)$ and so the integrals $int_0^infty f_n(x),mathrm dx$ give you no information about $int_0^infty f(x),mathrm dx$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 18:15









José Carlos Santos

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  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20


















  • If the improper integral converges uniformly, then the interchange is permissible.
    – Mark Viola
    Nov 26 at 18:20
















If the improper integral converges uniformly, then the interchange is permissible.
– Mark Viola
Nov 26 at 18:20




If the improper integral converges uniformly, then the interchange is permissible.
– Mark Viola
Nov 26 at 18:20


















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