Counterexample for the monotone convergence theorem












5












$begingroup$


Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20
















5












$begingroup$


Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20














5












5








5





$begingroup$


Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$




Do you have a counterexample for the monotone convergence theorem when you omit the hypothesis that the sequence is increasing?
I was thinking about the example where the sequence $f_n$ would approach $f$ as $frac {sin(x)} x$ do towards $0$. It appears that the integrals are equal, isn't it?



https://en.wikipedia.org/wiki/Monotone_convergence_theorem






real-analysis integration measure-theory lebesgue-integral lebesgue-measure




real-analysis integration measure-theory lebesgue-integral lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 14:13









YuiTo Cheng

2,1362837




2,1362837










asked Mar 16 at 12:55









Marine GalantinMarine Galantin

940419




940419












  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20


















  • $begingroup$
    and that the function is strictly positive, I ve understood that without this condition it does not work
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:01










  • $begingroup$
    I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
    $endgroup$
    – Henry
    Mar 16 at 13:18










  • $begingroup$
    Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:20
















$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
Mar 16 at 13:01




$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
Mar 16 at 13:01












$begingroup$
I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
$endgroup$
– Henry
Mar 16 at 13:18




$begingroup$
I suspect there may be more than one monotone convergence theorem (I use it for saying that an increasing bounded sequence has a limit which is its supremum)
$endgroup$
– Henry
Mar 16 at 13:18












$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
Mar 16 at 13:20




$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
Mar 16 at 13:20










1 Answer
1






active

oldest

votes


















4












$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18
















4












$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18














4












4








4





$begingroup$

Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$






share|cite|improve this answer











$endgroup$



Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 16 at 13:05

























answered Mar 16 at 13:00









PierrePierre

19511




19511












  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18


















  • $begingroup$
    there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:05






  • 1




    $begingroup$
    The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
    $endgroup$
    – Pierre
    Mar 16 at 13:09












  • $begingroup$
    Thank you for taking the time to explain to a neophyte as I am.
    $endgroup$
    – Marine Galantin
    Mar 16 at 13:18
















$begingroup$
there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
$endgroup$
– Marine Galantin
Mar 16 at 13:05




$begingroup$
there isn't ? can you detail a little bit please ? this function isn't integrable with lebesgue measure ? (I haven't yet integrate / compute anything with this new integral for me. I'm for now only doing the theory)
$endgroup$
– Marine Galantin
Mar 16 at 13:05




1




1




$begingroup$
The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
Mar 16 at 13:09






$begingroup$
The problem it's $xmapsto frac{sin(x)}{x}$ is not Lebesgue integral on $mathbb R$ (even if it's Riemann integrable). Now, if a sequence $(f_n)$ is Cauchy sequence in $L^1$, then it's limit is in $L^1$. Therefore, there are no sequence $(f_n)$ that converges to $f(x)=frac{sin(x)}{x}$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
Mar 16 at 13:09














$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
Mar 16 at 13:18




$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
Mar 16 at 13:18


















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