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If $f_m$ is increasing, show that $lim_mtoinftysum_ninmathbb Nf_m(n)=sum_nin mathbb Nf(n)$
Clash Royale CLAN TAG#URR8PPP
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Let $f_m:mathbb Nto mathbb R^+cup+infty $ an increasing sequence. Set $$f(n):=lim_mto infty f_m(n)$$
(it can possibly be $infty $).
I want to prove that $$lim_mto infty sum_ninmathbb N f_m(n)=sum_ninmathbb Nf(n).$$
First, $$sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n)$$
is clear, and thus $$lim_mto infty sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n).$$
I have problem for the reverse inequality. I tried to use the fact that $f(n)=sup_minmathbb Nf_m(n).$ So if $varepsilon>0$, there is $M_n$ s.t. $$f(n)-varepsilonleq f_m(n)leq f(n),$$
for all $mgeq M_n$. Now since $sum_ninmathbb Nvarepsilon$ doesn't converge, I can't conclude.
Context
On $(mathbb N,mathbb P(mathbb N),mu)$ where $mu(A)=# A$, I want to show that $$int_mathbb Nf(n)dmu(n)=sum_ninmathbb Nf(n).$$
If $f$ is simple it work. Now let $f$ measurable. There is a sequence of simple function s.t. $f_m(n)nearrow f(n)$. Then, using Monotone convergence, $$int_mathbb N f(n)dmu(n)=lim_mto infty int_mathbb Nf_m(n)dmu(n)=lim_mto infty sum_ninmathbb Nf_m(n).$$
So now, I have to prove that $$lim_mto infty sum_ninmathbb Nf_m(n)=sum_ninmathbb Nf(n),$$
to conclude.
real-analysis sequences-and-series
asked Jul 22 at 14:20
user380364
927214
 |Â
show 4 more comments
up vote
0
down vote
favorite
Let $f_m:mathbb Nto mathbb R^+cup+infty $ an increasing sequence. Set $$f(n):=lim_mto infty f_m(n)$$
(it can possibly be $infty $).
I want to prove that $$lim_mto infty sum_ninmathbb N f_m(n)=sum_ninmathbb Nf(n).$$
First, $$sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n)$$
is clear, and thus $$lim_mto infty sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n).$$
I have problem for the reverse inequality. I tried to use the fact that $f(n)=sup_minmathbb Nf_m(n).$ So if $varepsilon>0$, there is $M_n$ s.t. $$f(n)-varepsilonleq f_m(n)leq f(n),$$
for all $mgeq M_n$. Now since $sum_ninmathbb Nvarepsilon$ doesn't converge, I can't conclude.
Context
On $(mathbb N,mathbb P(mathbb N),mu)$ where $mu(A)=# A$, I want to show that $$int_mathbb Nf(n)dmu(n)=sum_ninmathbb Nf(n).$$
If $f$ is simple it work. Now let $f$ measurable. There is a sequence of simple function s.t. $f_m(n)nearrow f(n)$. Then, using Monotone convergence, $$int_mathbb N f(n)dmu(n)=lim_mto infty int_mathbb Nf_m(n)dmu(n)=lim_mto infty sum_ninmathbb Nf_m(n).$$
So now, I have to prove that $$lim_mto infty sum_ninmathbb Nf_m(n)=sum_ninmathbb Nf(n),$$
to conclude.
real-analysis sequences-and-series
asked Jul 22 at 14:20
user380364
927214
2
It's a quick result from Lebesgue's monotone convergence theorem.
– Rumpelstiltskin
Jul 22 at 14:24
@Adam: I add more context. So as you see, I can't use it like this.
– user380364
Jul 22 at 14:31
If you want an inequality from below you can truncate the sum: $sum_n<Mf_m(n)leqsum_ninmathbbNf_m(n)$. Take limit $lim_mtoinfty$ and get $sum_n<Mf(n)leq lim_mtoinftysum_ninmathbbNf_m(n)leqsum_ninmathbbNf(n)$. Finally, take limit $Mtoinfty$.
– user574889
Jul 22 at 14:32
Use the fact that Lebesgue's integral is countably additive, you will have $int_mathbbNf(n)dmu(n)=sum_iinmathbbN int_i f(n)dmu(n) = sum_iinmathbbN f(i) $
– Rumpelstiltskin
Jul 22 at 14:34
@DavidC.Ullrich: Yes, sorry, I wanted to write $mathbb Nto mathbb Rcuppminfty $
– user380364
Jul 22 at 15:43
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f_m:mathbb Nto mathbb R^+cup+infty $ an increasing sequence. Set $$f(n):=lim_mto infty f_m(n)$$
(it can possibly be $infty $).
I want to prove that $$lim_mto infty sum_ninmathbb N f_m(n)=sum_ninmathbb Nf(n).$$
First, $$sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n)$$
is clear, and thus $$lim_mto infty sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n).$$
I have problem for the reverse inequality. I tried to use the fact that $f(n)=sup_minmathbb Nf_m(n).$ So if $varepsilon>0$, there is $M_n$ s.t. $$f(n)-varepsilonleq f_m(n)leq f(n),$$
for all $mgeq M_n$. Now since $sum_ninmathbb Nvarepsilon$ doesn't converge, I can't conclude.
Context
On $(mathbb N,mathbb P(mathbb N),mu)$ where $mu(A)=# A$, I want to show that $$int_mathbb Nf(n)dmu(n)=sum_ninmathbb Nf(n).$$
If $f$ is simple it work. Now let $f$ measurable. There is a sequence of simple function s.t. $f_m(n)nearrow f(n)$. Then, using Monotone convergence, $$int_mathbb N f(n)dmu(n)=lim_mto infty int_mathbb Nf_m(n)dmu(n)=lim_mto infty sum_ninmathbb Nf_m(n).$$
So now, I have to prove that $$lim_mto infty sum_ninmathbb Nf_m(n)=sum_ninmathbb Nf(n),$$
to conclude.
real-analysis sequences-and-series
asked Jul 22 at 14:20
user380364
927214
Let $f_m:mathbb Nto mathbb R^+cup+infty $ an increasing sequence. Set $$f(n):=lim_mto infty f_m(n)$$
(it can possibly be $infty $).
I want to prove that $$lim_mto infty sum_ninmathbb N f_m(n)=sum_ninmathbb Nf(n).$$
First, $$sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n)$$
is clear, and thus $$lim_mto infty sum_ninmathbb Nf_m(n)leq sum_ninmathbb Nf(n).$$
I have problem for the reverse inequality. I tried to use the fact that $f(n)=sup_minmathbb Nf_m(n).$ So if $varepsilon>0$, there is $M_n$ s.t. $$f(n)-varepsilonleq f_m(n)leq f(n),$$
for all $mgeq M_n$. Now since $sum_ninmathbb Nvarepsilon$ doesn't converge, I can't conclude.
Context
On $(mathbb N,mathbb P(mathbb N),mu)$ where $mu(A)=# A$, I want to show that $$int_mathbb Nf(n)dmu(n)=sum_ninmathbb Nf(n).$$
If $f$ is simple it work. Now let $f$ measurable. There is a sequence of simple function s.t. $f_m(n)nearrow f(n)$. Then, using Monotone convergence, $$int_mathbb N f(n)dmu(n)=lim_mto infty int_mathbb Nf_m(n)dmu(n)=lim_mto infty sum_ninmathbb Nf_m(n).$$
So now, I have to prove that $$lim_mto infty sum_ninmathbb Nf_m(n)=sum_ninmathbb Nf(n),$$
to conclude.
real-analysis sequences-and-series
asked Jul 22 at 14:20
user380364
927214
edited Jul 25 at 17:42
asked Jul 22 at 14:20
user380364
927214
asked Jul 22 at 14:20
user380364
927214
asked Jul 22 at 14:20
user380364
927214
927214
2
It's a quick result from Lebesgue's monotone convergence theorem.
– Rumpelstiltskin
Jul 22 at 14:24
@Adam: I add more context. So as you see, I can't use it like this.
– user380364
Jul 22 at 14:31
If you want an inequality from below you can truncate the sum: $sum_n<Mf_m(n)leqsum_ninmathbbNf_m(n)$. Take limit $lim_mtoinfty$ and get $sum_n<Mf(n)leq lim_mtoinftysum_ninmathbbNf_m(n)leqsum_ninmathbbNf(n)$. Finally, take limit $Mtoinfty$.
– user574889
Jul 22 at 14:32
Use the fact that Lebesgue's integral is countably additive, you will have $int_mathbbNf(n)dmu(n)=sum_iinmathbbN int_i f(n)dmu(n) = sum_iinmathbbN f(i) $
– Rumpelstiltskin
Jul 22 at 14:34
@DavidC.Ullrich: Yes, sorry, I wanted to write $mathbb Nto mathbb Rcuppminfty $
– user380364
Jul 22 at 15:43
 |Â
show 4 more comments
2
It's a quick result from Lebesgue's monotone convergence theorem.
– Rumpelstiltskin
Jul 22 at 14:24
@Adam: I add more context. So as you see, I can't use it like this.
– user380364
Jul 22 at 14:31
If you want an inequality from below you can truncate the sum: $sum_n<Mf_m(n)leqsum_ninmathbbNf_m(n)$. Take limit $lim_mtoinfty$ and get $sum_n<Mf(n)leq lim_mtoinftysum_ninmathbbNf_m(n)leqsum_ninmathbbNf(n)$. Finally, take limit $Mtoinfty$.
– user574889
Jul 22 at 14:32
Use the fact that Lebesgue's integral is countably additive, you will have $int_mathbbNf(n)dmu(n)=sum_iinmathbbN int_i f(n)dmu(n) = sum_iinmathbbN f(i) $
– Rumpelstiltskin
Jul 22 at 14:34
@DavidC.Ullrich: Yes, sorry, I wanted to write $mathbb Nto mathbb Rcuppminfty $
– user380364
Jul 22 at 15:43
2
2
It's a quick result from Lebesgue's monotone convergence theorem.
– Rumpelstiltskin
Jul 22 at 14:24
It's a quick result from Lebesgue's monotone convergence theorem.
– Rumpelstiltskin
Jul 22 at 14:24
@Adam: I add more context. So as you see, I can't use it like this.
– user380364
Jul 22 at 14:31
@Adam: I add more context. So as you see, I can't use it like this.
– user380364
Jul 22 at 14:31
If you want an inequality from below you can truncate the sum: $sum_n<Mf_m(n)leqsum_ninmathbbNf_m(n)$. Take limit $lim_mtoinfty$ and get $sum_n<Mf(n)leq lim_mtoinftysum_ninmathbbNf_m(n)leqsum_ninmathbbNf(n)$. Finally, take limit $Mtoinfty$.
– user574889
Jul 22 at 14:32
If you want an inequality from below you can truncate the sum: $sum_n<Mf_m(n)leqsum_ninmathbbNf_m(n)$. Take limit $lim_mtoinfty$ and get $sum_n<Mf(n)leq lim_mtoinftysum_ninmathbbNf_m(n)leqsum_ninmathbbNf(n)$. Finally, take limit $Mtoinfty$.
– user574889
Jul 22 at 14:32
Use the fact that Lebesgue's integral is countably additive, you will have $int_mathbbNf(n)dmu(n)=sum_iinmathbbN int_i f(n)dmu(n) = sum_iinmathbbN f(i) $
– Rumpelstiltskin
Jul 22 at 14:34
Use the fact that Lebesgue's integral is countably additive, you will have $int_mathbbNf(n)dmu(n)=sum_iinmathbbN int_i f(n)dmu(n) = sum_iinmathbbN f(i) $
– Rumpelstiltskin
Jul 22 at 14:34
@DavidC.Ullrich: Yes, sorry, I wanted to write $mathbb Nto mathbb Rcuppminfty $
– user380364
Jul 22 at 15:43
@DavidC.Ullrich: Yes, sorry, I wanted to write $mathbb Nto mathbb Rcuppminfty $
– user380364
Jul 22 at 15:43
 |Â
show 4 more comments
1 Answer
1
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oldest
votes
up vote
1
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First, the last section seems a little silly. If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if $f_m+1ge f_mge0$ then $limint f_m=intlim_m f_m$, qed. (Hint for showing that if $mu$ is counting measure and $fge0$ then $int_Bbb Nf,dmu=sum_nf(n)$: By definition $sum_nf(n)=lim_Ntoinftysum_n=1^Nf(n)$.)
There's a very simple elementary proof. Let $$sum_n f(n)=sin[0,infty].$$
Monotonicity makes it clear that the limit exists and that $$lim_msum_nf_m(n)le s.$$
For the other direction, suppose that $alpha<s$. (Note I'm talking about a general $alpha<s$ instead of talking about $s-epsilon$ in order to include the possibility that $s=infty$.) Choose $N$ so that $$sum_n=1^Nf(n)>alpha.$$
Then it's clear that there exists $M$ so $$sum_n=1^Nf_m(n)>alphaquad(m>M).$$Hence $$lim_msum_nf_m(n)>alpha,$$and since this holds for every $alpha<s$ we must have $$lim_msum_nf_m(n)ge s.$$
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$lim_mto infty int_mathbb Nf_m=int_mathbb Nf.$$ But to prove that $int_mathbb Nf(n)dmu=sum_ninmathbb Nf(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ?
– user380364
Jul 22 at 15:48
Let's get the hypothesis straight first. The post says $f_mto mathbb Nto mathbb Rcuppminfty $ , which is gibberish. If we change that to $f_m:mathbb Nto mathbb Rcuppminfty $ it makes sense, but then none of the sums you mention need exist.
– David C. Ullrich
Jul 22 at 15:59
@user380364 Under the current hypothesis it's impossible to prove that $int f(n),dmu=sum f(n)$, in fact neither the integral nor the sum need even exist.
– David C. Ullrich
Jul 22 at 16:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First, the last section seems a little silly. If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if $f_m+1ge f_mge0$ then $limint f_m=intlim_m f_m$, qed. (Hint for showing that if $mu$ is counting measure and $fge0$ then $int_Bbb Nf,dmu=sum_nf(n)$: By definition $sum_nf(n)=lim_Ntoinftysum_n=1^Nf(n)$.)
There's a very simple elementary proof. Let $$sum_n f(n)=sin[0,infty].$$
Monotonicity makes it clear that the limit exists and that $$lim_msum_nf_m(n)le s.$$
For the other direction, suppose that $alpha<s$. (Note I'm talking about a general $alpha<s$ instead of talking about $s-epsilon$ in order to include the possibility that $s=infty$.) Choose $N$ so that $$sum_n=1^Nf(n)>alpha.$$
Then it's clear that there exists $M$ so $$sum_n=1^Nf_m(n)>alphaquad(m>M).$$Hence $$lim_msum_nf_m(n)>alpha,$$and since this holds for every $alpha<s$ we must have $$lim_msum_nf_m(n)ge s.$$
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$lim_mto infty int_mathbb Nf_m=int_mathbb Nf.$$ But to prove that $int_mathbb Nf(n)dmu=sum_ninmathbb Nf(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ?
– user380364
Jul 22 at 15:48
Let's get the hypothesis straight first. The post says $f_mto mathbb Nto mathbb Rcuppminfty $ , which is gibberish. If we change that to $f_m:mathbb Nto mathbb Rcuppminfty $ it makes sense, but then none of the sums you mention need exist.
– David C. Ullrich
Jul 22 at 15:59
@user380364 Under the current hypothesis it's impossible to prove that $int f(n),dmu=sum f(n)$, in fact neither the integral nor the sum need even exist.
– David C. Ullrich
Jul 22 at 16:36
add a comment |Â
up vote
1
down vote
First, the last section seems a little silly. If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if $f_m+1ge f_mge0$ then $limint f_m=intlim_m f_m$, qed. (Hint for showing that if $mu$ is counting measure and $fge0$ then $int_Bbb Nf,dmu=sum_nf(n)$: By definition $sum_nf(n)=lim_Ntoinftysum_n=1^Nf(n)$.)
There's a very simple elementary proof. Let $$sum_n f(n)=sin[0,infty].$$
Monotonicity makes it clear that the limit exists and that $$lim_msum_nf_m(n)le s.$$
For the other direction, suppose that $alpha<s$. (Note I'm talking about a general $alpha<s$ instead of talking about $s-epsilon$ in order to include the possibility that $s=infty$.) Choose $N$ so that $$sum_n=1^Nf(n)>alpha.$$
Then it's clear that there exists $M$ so $$sum_n=1^Nf_m(n)>alphaquad(m>M).$$Hence $$lim_msum_nf_m(n)>alpha,$$and since this holds for every $alpha<s$ we must have $$lim_msum_nf_m(n)ge s.$$
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$lim_mto infty int_mathbb Nf_m=int_mathbb Nf.$$ But to prove that $int_mathbb Nf(n)dmu=sum_ninmathbb Nf(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ?
– user380364
Jul 22 at 15:48
Let's get the hypothesis straight first. The post says $f_mto mathbb Nto mathbb Rcuppminfty $ , which is gibberish. If we change that to $f_m:mathbb Nto mathbb Rcuppminfty $ it makes sense, but then none of the sums you mention need exist.
– David C. Ullrich
Jul 22 at 15:59
@user380364 Under the current hypothesis it's impossible to prove that $int f(n),dmu=sum f(n)$, in fact neither the integral nor the sum need even exist.
– David C. Ullrich
Jul 22 at 16:36
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First, the last section seems a little silly. If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if $f_m+1ge f_mge0$ then $limint f_m=intlim_m f_m$, qed. (Hint for showing that if $mu$ is counting measure and $fge0$ then $int_Bbb Nf,dmu=sum_nf(n)$: By definition $sum_nf(n)=lim_Ntoinftysum_n=1^Nf(n)$.)
There's a very simple elementary proof. Let $$sum_n f(n)=sin[0,infty].$$
Monotonicity makes it clear that the limit exists and that $$lim_msum_nf_m(n)le s.$$
For the other direction, suppose that $alpha<s$. (Note I'm talking about a general $alpha<s$ instead of talking about $s-epsilon$ in order to include the possibility that $s=infty$.) Choose $N$ so that $$sum_n=1^Nf(n)>alpha.$$
Then it's clear that there exists $M$ so $$sum_n=1^Nf_m(n)>alphaquad(m>M).$$Hence $$lim_msum_nf_m(n)>alpha,$$and since this holds for every $alpha<s$ we must have $$lim_msum_nf_m(n)ge s.$$
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
First, the last section seems a little silly. If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if $f_m+1ge f_mge0$ then $limint f_m=intlim_m f_m$, qed. (Hint for showing that if $mu$ is counting measure and $fge0$ then $int_Bbb Nf,dmu=sum_nf(n)$: By definition $sum_nf(n)=lim_Ntoinftysum_n=1^Nf(n)$.)
There's a very simple elementary proof. Let $$sum_n f(n)=sin[0,infty].$$
Monotonicity makes it clear that the limit exists and that $$lim_msum_nf_m(n)le s.$$
For the other direction, suppose that $alpha<s$. (Note I'm talking about a general $alpha<s$ instead of talking about $s-epsilon$ in order to include the possibility that $s=infty$.) Choose $N$ so that $$sum_n=1^Nf(n)>alpha.$$
Then it's clear that there exists $M$ so $$sum_n=1^Nf_m(n)>alphaquad(m>M).$$Hence $$lim_msum_nf_m(n)>alpha,$$and since this holds for every $alpha<s$ we must have $$lim_msum_nf_m(n)ge s.$$
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
edited Jul 25 at 17:49
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
answered Jul 22 at 15:00
David C. Ullrich
54.1k33481
54.1k33481
Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$lim_mto infty int_mathbb Nf_m=int_mathbb Nf.$$ But to prove that $int_mathbb Nf(n)dmu=sum_ninmathbb Nf(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ?
– user380364
Jul 22 at 15:48
Let's get the hypothesis straight first. The post says $f_mto mathbb Nto mathbb Rcuppminfty $ , which is gibberish. If we change that to $f_m:mathbb Nto mathbb Rcuppminfty $ it makes sense, but then none of the sums you mention need exist.
– David C. Ullrich
Jul 22 at 15:59
@user380364 Under the current hypothesis it's impossible to prove that $int f(n),dmu=sum f(n)$, in fact neither the integral nor the sum need even exist.
– David C. Ullrich
Jul 22 at 16:36
add a comment |Â
Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$lim_mto infty int_mathbb Nf_m=int_mathbb Nf.$$ But to prove that $int_mathbb Nf(n)dmu=sum_ninmathbb Nf(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ?
– user380364
Jul 22 at 15:48
Let's get the hypothesis straight first. The post says $f_mto mathbb Nto mathbb Rcuppminfty $ , which is gibberish. If we change that to $f_m:mathbb Nto mathbb Rcuppminfty $ it makes sense, but then none of the sums you mention need exist.
– David C. Ullrich
Jul 22 at 15:59
@user380364 Under the current hypothesis it's impossible to prove that $int f(n),dmu=sum f(n)$, in fact neither the integral nor the sum need even exist.
– David C. Ullrich
Jul 22 at 16:36
Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$lim_mto infty int_mathbb Nf_m=int_mathbb Nf.$$ But to prove that $int_mathbb Nf(n)dmu=sum_ninmathbb Nf(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ?
– user380364
Jul 22 at 15:48
Thank you. I don't understand your "If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if...". I know that $$lim_mto infty int_mathbb Nf_m=int_mathbb Nf.$$ But to prove that $int_mathbb Nf(n)dmu=sum_ninmathbb Nf(n)$ it's not MCT, is it ? Your proof looks perfect, and you didn't use MCT, did you ?
– user380364
Jul 22 at 15:48
Let's get the hypothesis straight first. The post says $f_mto mathbb Nto mathbb Rcuppminfty $ , which is gibberish. If we change that to $f_m:mathbb Nto mathbb Rcuppminfty $ it makes sense, but then none of the sums you mention need exist.
– David C. Ullrich
Jul 22 at 15:59
Let's get the hypothesis straight first. The post says $f_mto mathbb Nto mathbb Rcuppminfty $ , which is gibberish. If we change that to $f_m:mathbb Nto mathbb Rcuppminfty $ it makes sense, but then none of the sums you mention need exist.
– David C. Ullrich
Jul 22 at 15:59
@user380364 Under the current hypothesis it's impossible to prove that $int f(n),dmu=sum f(n)$, in fact neither the integral nor the sum need even exist.
– David C. Ullrich
Jul 22 at 16:36
@user380364 Under the current hypothesis it's impossible to prove that $int f(n),dmu=sum f(n)$, in fact neither the integral nor the sum need even exist.
– David C. Ullrich
Jul 22 at 16:36
add a comment |Â
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2
It's a quick result from Lebesgue's monotone convergence theorem.
– Rumpelstiltskin
Jul 22 at 14:24
@Adam: I add more context. So as you see, I can't use it like this.
– user380364
Jul 22 at 14:31
If you want an inequality from below you can truncate the sum: $sum_n<Mf_m(n)leqsum_ninmathbbNf_m(n)$. Take limit $lim_mtoinfty$ and get $sum_n<Mf(n)leq lim_mtoinftysum_ninmathbbNf_m(n)leqsum_ninmathbbNf(n)$. Finally, take limit $Mtoinfty$.
– user574889
Jul 22 at 14:32
Use the fact that Lebesgue's integral is countably additive, you will have $int_mathbbNf(n)dmu(n)=sum_iinmathbbN int_i f(n)dmu(n) = sum_iinmathbbN f(i) $
– Rumpelstiltskin
Jul 22 at 14:34
@DavidC.Ullrich: Yes, sorry, I wanted to write $mathbb Nto mathbb Rcuppminfty $
– user380364
Jul 22 at 15:43