Counterexample for the monotone convergence theoremAbout assumptions in the monotone convergence theoremCounterexample for downward monotone convergence theorem on measurable setAn example related to the Monotone Convergence TheoremWhat are the hypotheses in Levi's monotone convergence theorem?Monotone convergence theorem assuming convergence in measureMonotone Convergence theorem for monotone decreasing sequencesThe meaning of “Need not hold” in Monotone convergence theorem for decreasing sequencesWhy does the Monotone Convergence Theorem require increasing sequence.Counterexample for Monotone convergence theoremAn counterexample for the monotone convergence theorem and dominated convergence theorem
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Counterexample for the monotone convergence theorem
About assumptions in the monotone convergence theoremCounterexample for downward monotone convergence theorem on measurable setAn example related to the Monotone Convergence TheoremWhat are the hypotheses in Levi's monotone convergence theorem?Monotone convergence theorem assuming convergence in measureMonotone Convergence theorem for monotone decreasing sequencesThe meaning of “Need not hold” in Monotone convergence theorem for decreasing sequencesWhy does the Monotone Convergence Theorem require increasing sequence.Counterexample for Monotone convergence theoremAn counterexample for the monotone convergence theorem and dominated convergence theorem
$begingroup$
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 9 hours ago
YuiTo Cheng
2,0532637
asked 10 hours ago
Marine GalantinMarine Galantin
913319
$endgroup$
add a comment |
$begingroup$
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 9 hours ago
YuiTo Cheng
2,0532637
asked 10 hours ago
Marine GalantinMarine Galantin
913319
$endgroup$
$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
10 hours ago
add a comment |
$begingroup$
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 9 hours ago
YuiTo Cheng
2,0532637
asked 10 hours ago
Marine GalantinMarine Galantin
913319
$endgroup$
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 9 hours ago
YuiTo Cheng
2,0532637
asked 10 hours ago
Marine GalantinMarine Galantin
913319
edited 9 hours ago
YuiTo Cheng
2,0532637
asked 10 hours ago
Marine GalantinMarine Galantin
913319
edited 9 hours ago
YuiTo Cheng
2,0532637
edited 9 hours ago
YuiTo Cheng
2,0532637
edited 9 hours ago
YuiTo Cheng
2,0532637
2,0532637
asked 10 hours ago
Marine GalantinMarine Galantin
913319
asked 10 hours ago
Marine GalantinMarine Galantin
913319
asked 10 hours ago
Marine GalantinMarine Galantin
913319
913319
$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
10 hours ago
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
and that the function is strictly positive, I ve understood that without this condition it does not work
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
10 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take $f_n(x)=frac1nboldsymbol 1_[0,n]$. You have that $$lim_nto infty f_n(x)=0,$$ but $$lim_nto infty int_mathbb R f_n=1.$$
answered 10 hours ago
PierrePierre
5610
$endgroup$
$begingroup$
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
10 hours ago
1
$begingroup$
The problem it's $xmapsto fracsin(x)x$ 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 $f(x)=fracsin(x)x$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
10 hours ago
$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
10 hours ago
add a comment |
Your Answer
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1 Answer
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active
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votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Take $f_n(x)=frac1nboldsymbol 1_[0,n]$. You have that $$lim_nto infty f_n(x)=0,$$ but $$lim_nto infty int_mathbb R f_n=1.$$
answered 10 hours ago
PierrePierre
5610
$endgroup$
$begingroup$
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
10 hours ago
1
$begingroup$
The problem it's $xmapsto fracsin(x)x$ 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 $f(x)=fracsin(x)x$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
10 hours ago
$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
10 hours ago
add a comment |
$begingroup$
Take $f_n(x)=frac1nboldsymbol 1_[0,n]$. You have that $$lim_nto infty f_n(x)=0,$$ but $$lim_nto infty int_mathbb R f_n=1.$$
answered 10 hours ago
PierrePierre
5610
$endgroup$
$begingroup$
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
10 hours ago
1
$begingroup$
The problem it's $xmapsto fracsin(x)x$ 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 $f(x)=fracsin(x)x$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
10 hours ago
$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
10 hours ago
add a comment |
$begingroup$
Take $f_n(x)=frac1nboldsymbol 1_[0,n]$. You have that $$lim_nto infty f_n(x)=0,$$ but $$lim_nto infty int_mathbb R f_n=1.$$
answered 10 hours ago
PierrePierre
5610
$endgroup$
Take $f_n(x)=frac1nboldsymbol 1_[0,n]$. You have that $$lim_nto infty f_n(x)=0,$$ but $$lim_nto infty int_mathbb R f_n=1.$$
answered 10 hours ago
PierrePierre
5610
edited 10 hours ago
answered 10 hours ago
PierrePierre
5610
answered 10 hours ago
PierrePierre
5610
answered 10 hours ago
PierrePierre
5610
5610
$begingroup$
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
10 hours ago
1
$begingroup$
The problem it's $xmapsto fracsin(x)x$ 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 $f(x)=fracsin(x)x$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
10 hours ago
$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
10 hours ago
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
10 hours ago
1
$begingroup$
The problem it's $xmapsto fracsin(x)x$ 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 $f(x)=fracsin(x)x$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
10 hours ago
$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
1
1
$begingroup$
The problem it's $xmapsto fracsin(x)x$ 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 $f(x)=fracsin(x)x$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
10 hours ago
$begingroup$
The problem it's $xmapsto fracsin(x)x$ 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 $f(x)=fracsin(x)x$ in $L^1$. @MarineGalantin
$endgroup$
– Pierre
10 hours ago
$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
10 hours ago
$begingroup$
Thank you for taking the time to explain to a neophyte as I am.
$endgroup$
– Marine Galantin
10 hours ago
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
10 hours ago
$begingroup$
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
10 hours ago
$begingroup$
Well here we are talking about measure and integration. This theorem comes with Fatou's Lemma and Dominated convergence theorem.
$endgroup$
– Marine Galantin
10 hours ago