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
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
asked Mar 16 at 12:55
Marine GalantinMarine Galantin
940419
$endgroup$
add a comment |
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
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
asked Mar 16 at 12:55
Marine GalantinMarine Galantin
940419
$endgroup$
$begingroup$
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
add a comment |
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
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
asked Mar 16 at 12:55
Marine GalantinMarine Galantin
940419
$endgroup$
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
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
asked Mar 16 at 12:55
Marine GalantinMarine Galantin
940419
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
asked Mar 16 at 12:55
Marine GalantinMarine Galantin
940419
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
edited Mar 16 at 14:13
YuiTo Cheng
2,1362837
2,1362837
asked Mar 16 at 12:55
Marine GalantinMarine Galantin
940419
asked Mar 16 at 12:55
Marine GalantinMarine Galantin
940419
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
add a comment |
$begingroup$
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
add a comment |
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 $$lim_{nto infty }int_{mathbb R} f_n=1.$$
answered Mar 16 at 13:00
PierrePierre
19511
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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4
$begingroup$
Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$
answered Mar 16 at 13:00
PierrePierre
19511
$endgroup$
$begingroup$
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
add a comment |
4
$begingroup$
Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$
answered Mar 16 at 13:00
PierrePierre
19511
$endgroup$
$begingroup$
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
add a comment |
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 $$lim_{nto infty }int_{mathbb R} f_n=1.$$
answered Mar 16 at 13:00
PierrePierre
19511
$endgroup$
Take $f_n(x)=frac{1}{n}boldsymbol 1_{[0,n]}$. You have that $$lim_{nto infty }f_n(x)=0,$$ but $$lim_{nto infty }int_{mathbb R} f_n=1.$$
answered Mar 16 at 13:00
PierrePierre
19511
edited Mar 16 at 13:05
answered Mar 16 at 13:00
PierrePierre
19511
answered Mar 16 at 13:00
PierrePierre
19511
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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