infinite sum as integral for the counting measure
$begingroup$
Let $u: mathbb{N} to mathbb{R}^+$ be a positive sequence. Then it is true that
$$
int u dmu = sum_{n=1}^infty u(n).
$$
For the measure space $(mathbb{N}, mathcal{P}(mathbb{N}),mu)$, where $mu$ is the counting measure. The proof uses monotone convergence theorem, and goes as follows:
Take the measure space $(mathbb{N},mathcal{P}(mathbb{N}), underbrace{sum_{j=1}^infty delta_j}_{mu})$.
Let $u: mathbb{N} to mathbb{R}^+$ be a positive function ($u geq 0$).
We define
$$
u_n(k) =
begin{cases}
u(k) & k leq n, \
0 & k > n.
end{cases}
$$
For all $n in mathbb{N}$, $u_n$ is a normal function since we have $u_n = sum_{k=1}^n u(k)1_{{k}}$. Thus, the approximation cuts of $u$ after $n$. Moreover, we have
$$
int u_n dmu = I_mu(f) = sum_{k=1}^n u(k) mu({k}) = sum_{k=1}^n u(k).
$$
Since we have $0 leq u_n uparrow u$, then it follows by MCT that
$$
int_{mathbb{N}} u dmu stackrel{MTC}{=} lim_{n to infty} int u_n dmu = sum_{k=1}^infty u(k).
$$
Now can someone tell me under what condition this is true if $u: mathbb{N} to mathbb{R}$ is not necessarily positive?
integration sequences-and-series measure-theory
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
$endgroup$
|
show 11 more comments
$begingroup$
Let $u: mathbb{N} to mathbb{R}^+$ be a positive sequence. Then it is true that
$$
int u dmu = sum_{n=1}^infty u(n).
$$
For the measure space $(mathbb{N}, mathcal{P}(mathbb{N}),mu)$, where $mu$ is the counting measure. The proof uses monotone convergence theorem, and goes as follows:
Take the measure space $(mathbb{N},mathcal{P}(mathbb{N}), underbrace{sum_{j=1}^infty delta_j}_{mu})$.
Let $u: mathbb{N} to mathbb{R}^+$ be a positive function ($u geq 0$).
We define
$$
u_n(k) =
begin{cases}
u(k) & k leq n, \
0 & k > n.
end{cases}
$$
For all $n in mathbb{N}$, $u_n$ is a normal function since we have $u_n = sum_{k=1}^n u(k)1_{{k}}$. Thus, the approximation cuts of $u$ after $n$. Moreover, we have
$$
int u_n dmu = I_mu(f) = sum_{k=1}^n u(k) mu({k}) = sum_{k=1}^n u(k).
$$
Since we have $0 leq u_n uparrow u$, then it follows by MCT that
$$
int_{mathbb{N}} u dmu stackrel{MTC}{=} lim_{n to infty} int u_n dmu = sum_{k=1}^infty u(k).
$$
Now can someone tell me under what condition this is true if $u: mathbb{N} to mathbb{R}$ is not necessarily positive?
integration sequences-and-series measure-theory
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
$endgroup$
$begingroup$
I think you confuses the definitions. The equality holds by the definition of the integral and has nothing to do with the monotone convergence theorem. Also if $u:mathbb{N}rightarrowmathbb{R}^{+}$ then it is necessarily positive (again by definition).
$endgroup$
– Yanko
Dec 3 '18 at 18:37
$begingroup$
Why it holds by definition?
$endgroup$
– Sigurd
Dec 3 '18 at 18:38
$begingroup$
Which definition of integration do you use?
$endgroup$
– Yanko
Dec 3 '18 at 18:40
$begingroup$
For step functions / simple functions $f = sum_{i=1}^N alpha_i 1_{A_i}$: $I_mu(f) := sum_{i=1}^N alpha_i mu({A_i})$. For positive functions: $int u dmu := sup{I_mu(g): g leq u, g $ simple $} in [0,infty]$. For general functions: $int u dmu = int u^+dmu - int u^- dmu$.
$endgroup$
– Sigurd
Dec 3 '18 at 18:43
$begingroup$
Alright so a simple function in this case is a function that is supported only on finitely many points. It is only left to understand why $sup {I_u(g):gleq u, g text{ simple} } = lim_{nrightarrowinfty} I_u(u_n)$ where $u_n$ is $u$ restricted to the first $n$ points (i.e. $u_n(k)=u(k)$ if $kleq n$ and $0$ otherwise). Does it explain?
$endgroup$
– Yanko
Dec 3 '18 at 18:46
|
show 11 more comments
$begingroup$
Let $u: mathbb{N} to mathbb{R}^+$ be a positive sequence. Then it is true that
$$
int u dmu = sum_{n=1}^infty u(n).
$$
For the measure space $(mathbb{N}, mathcal{P}(mathbb{N}),mu)$, where $mu$ is the counting measure. The proof uses monotone convergence theorem, and goes as follows:
Take the measure space $(mathbb{N},mathcal{P}(mathbb{N}), underbrace{sum_{j=1}^infty delta_j}_{mu})$.
Let $u: mathbb{N} to mathbb{R}^+$ be a positive function ($u geq 0$).
We define
$$
u_n(k) =
begin{cases}
u(k) & k leq n, \
0 & k > n.
end{cases}
$$
For all $n in mathbb{N}$, $u_n$ is a normal function since we have $u_n = sum_{k=1}^n u(k)1_{{k}}$. Thus, the approximation cuts of $u$ after $n$. Moreover, we have
$$
int u_n dmu = I_mu(f) = sum_{k=1}^n u(k) mu({k}) = sum_{k=1}^n u(k).
$$
Since we have $0 leq u_n uparrow u$, then it follows by MCT that
$$
int_{mathbb{N}} u dmu stackrel{MTC}{=} lim_{n to infty} int u_n dmu = sum_{k=1}^infty u(k).
$$
Now can someone tell me under what condition this is true if $u: mathbb{N} to mathbb{R}$ is not necessarily positive?
integration sequences-and-series measure-theory
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
$endgroup$
Let $u: mathbb{N} to mathbb{R}^+$ be a positive sequence. Then it is true that
$$
int u dmu = sum_{n=1}^infty u(n).
$$
For the measure space $(mathbb{N}, mathcal{P}(mathbb{N}),mu)$, where $mu$ is the counting measure. The proof uses monotone convergence theorem, and goes as follows:
Take the measure space $(mathbb{N},mathcal{P}(mathbb{N}), underbrace{sum_{j=1}^infty delta_j}_{mu})$.
Let $u: mathbb{N} to mathbb{R}^+$ be a positive function ($u geq 0$).
We define
$$
u_n(k) =
begin{cases}
u(k) & k leq n, \
0 & k > n.
end{cases}
$$
For all $n in mathbb{N}$, $u_n$ is a normal function since we have $u_n = sum_{k=1}^n u(k)1_{{k}}$. Thus, the approximation cuts of $u$ after $n$. Moreover, we have
$$
int u_n dmu = I_mu(f) = sum_{k=1}^n u(k) mu({k}) = sum_{k=1}^n u(k).
$$
Since we have $0 leq u_n uparrow u$, then it follows by MCT that
$$
int_{mathbb{N}} u dmu stackrel{MTC}{=} lim_{n to infty} int u_n dmu = sum_{k=1}^infty u(k).
$$
Now can someone tell me under what condition this is true if $u: mathbb{N} to mathbb{R}$ is not necessarily positive?
integration sequences-and-series measure-theory
integration sequences-and-series measure-theory
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
edited Dec 3 '18 at 18:54
Sigurd
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
asked Dec 3 '18 at 18:35
SigurdSigurd
480211
480211
$begingroup$
I think you confuses the definitions. The equality holds by the definition of the integral and has nothing to do with the monotone convergence theorem. Also if $u:mathbb{N}rightarrowmathbb{R}^{+}$ then it is necessarily positive (again by definition).
$endgroup$
– Yanko
Dec 3 '18 at 18:37
$begingroup$
Why it holds by definition?
$endgroup$
– Sigurd
Dec 3 '18 at 18:38
$begingroup$
Which definition of integration do you use?
$endgroup$
– Yanko
Dec 3 '18 at 18:40
$begingroup$
For step functions / simple functions $f = sum_{i=1}^N alpha_i 1_{A_i}$: $I_mu(f) := sum_{i=1}^N alpha_i mu({A_i})$. For positive functions: $int u dmu := sup{I_mu(g): g leq u, g $ simple $} in [0,infty]$. For general functions: $int u dmu = int u^+dmu - int u^- dmu$.
$endgroup$
– Sigurd
Dec 3 '18 at 18:43
$begingroup$
Alright so a simple function in this case is a function that is supported only on finitely many points. It is only left to understand why $sup {I_u(g):gleq u, g text{ simple} } = lim_{nrightarrowinfty} I_u(u_n)$ where $u_n$ is $u$ restricted to the first $n$ points (i.e. $u_n(k)=u(k)$ if $kleq n$ and $0$ otherwise). Does it explain?
$endgroup$
– Yanko
Dec 3 '18 at 18:46
|
show 11 more comments
$begingroup$
I think you confuses the definitions. The equality holds by the definition of the integral and has nothing to do with the monotone convergence theorem. Also if $u:mathbb{N}rightarrowmathbb{R}^{+}$ then it is necessarily positive (again by definition).
$endgroup$
– Yanko
Dec 3 '18 at 18:37
$begingroup$
Why it holds by definition?
$endgroup$
– Sigurd
Dec 3 '18 at 18:38
$begingroup$
Which definition of integration do you use?
$endgroup$
– Yanko
Dec 3 '18 at 18:40
$begingroup$
For step functions / simple functions $f = sum_{i=1}^N alpha_i 1_{A_i}$: $I_mu(f) := sum_{i=1}^N alpha_i mu({A_i})$. For positive functions: $int u dmu := sup{I_mu(g): g leq u, g $ simple $} in [0,infty]$. For general functions: $int u dmu = int u^+dmu - int u^- dmu$.
$endgroup$
– Sigurd
Dec 3 '18 at 18:43
$begingroup$
Alright so a simple function in this case is a function that is supported only on finitely many points. It is only left to understand why $sup {I_u(g):gleq u, g text{ simple} } = lim_{nrightarrowinfty} I_u(u_n)$ where $u_n$ is $u$ restricted to the first $n$ points (i.e. $u_n(k)=u(k)$ if $kleq n$ and $0$ otherwise). Does it explain?
$endgroup$
– Yanko
Dec 3 '18 at 18:46
$begingroup$
I think you confuses the definitions. The equality holds by the definition of the integral and has nothing to do with the monotone convergence theorem. Also if $u:mathbb{N}rightarrowmathbb{R}^{+}$ then it is necessarily positive (again by definition).
$endgroup$
– Yanko
Dec 3 '18 at 18:37
$begingroup$
I think you confuses the definitions. The equality holds by the definition of the integral and has nothing to do with the monotone convergence theorem. Also if $u:mathbb{N}rightarrowmathbb{R}^{+}$ then it is necessarily positive (again by definition).
$endgroup$
– Yanko
Dec 3 '18 at 18:37
$begingroup$
Why it holds by definition?
$endgroup$
– Sigurd
Dec 3 '18 at 18:38
$begingroup$
Why it holds by definition?
$endgroup$
– Sigurd
Dec 3 '18 at 18:38
$begingroup$
Which definition of integration do you use?
$endgroup$
– Yanko
Dec 3 '18 at 18:40
$begingroup$
Which definition of integration do you use?
$endgroup$
– Yanko
Dec 3 '18 at 18:40
$begingroup$
For step functions / simple functions $f = sum_{i=1}^N alpha_i 1_{A_i}$: $I_mu(f) := sum_{i=1}^N alpha_i mu({A_i})$. For positive functions: $int u dmu := sup{I_mu(g): g leq u, g $ simple $} in [0,infty]$. For general functions: $int u dmu = int u^+dmu - int u^- dmu$.
$endgroup$
– Sigurd
Dec 3 '18 at 18:43
$begingroup$
For step functions / simple functions $f = sum_{i=1}^N alpha_i 1_{A_i}$: $I_mu(f) := sum_{i=1}^N alpha_i mu({A_i})$. For positive functions: $int u dmu := sup{I_mu(g): g leq u, g $ simple $} in [0,infty]$. For general functions: $int u dmu = int u^+dmu - int u^- dmu$.
$endgroup$
– Sigurd
Dec 3 '18 at 18:43
$begingroup$
Alright so a simple function in this case is a function that is supported only on finitely many points. It is only left to understand why $sup {I_u(g):gleq u, g text{ simple} } = lim_{nrightarrowinfty} I_u(u_n)$ where $u_n$ is $u$ restricted to the first $n$ points (i.e. $u_n(k)=u(k)$ if $kleq n$ and $0$ otherwise). Does it explain?
$endgroup$
– Yanko
Dec 3 '18 at 18:46
$begingroup$
Alright so a simple function in this case is a function that is supported only on finitely many points. It is only left to understand why $sup {I_u(g):gleq u, g text{ simple} } = lim_{nrightarrowinfty} I_u(u_n)$ where $u_n$ is $u$ restricted to the first $n$ points (i.e. $u_n(k)=u(k)$ if $kleq n$ and $0$ otherwise). Does it explain?
$endgroup$
– Yanko
Dec 3 '18 at 18:46
|
show 11 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I prove
If $u(n)$ is a sequence such that $sum_n |u(n)| <infty$ then $int u dmu = sum_n u(n)$
By the definition of integral we have $int u dmu = int u^+ dmu - int u^- dmu$
By what you just showed for $u^+,u^-$ we have $u^pm = sum_n u^pm (n)$
Since the sum is uniformly convergence it is the same to sum from $1,2,...$ or to sum first the $n$ for which $u(n)>0$ and only then the $n$ for which $u(n)<0$.
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
$endgroup$
add a comment |
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$begingroup$
I prove
If $u(n)$ is a sequence such that $sum_n |u(n)| <infty$ then $int u dmu = sum_n u(n)$
By the definition of integral we have $int u dmu = int u^+ dmu - int u^- dmu$
By what you just showed for $u^+,u^-$ we have $u^pm = sum_n u^pm (n)$
Since the sum is uniformly convergence it is the same to sum from $1,2,...$ or to sum first the $n$ for which $u(n)>0$ and only then the $n$ for which $u(n)<0$.
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
$endgroup$
add a comment |
$begingroup$
I prove
If $u(n)$ is a sequence such that $sum_n |u(n)| <infty$ then $int u dmu = sum_n u(n)$
By the definition of integral we have $int u dmu = int u^+ dmu - int u^- dmu$
By what you just showed for $u^+,u^-$ we have $u^pm = sum_n u^pm (n)$
Since the sum is uniformly convergence it is the same to sum from $1,2,...$ or to sum first the $n$ for which $u(n)>0$ and only then the $n$ for which $u(n)<0$.
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
$endgroup$
add a comment |
$begingroup$
I prove
If $u(n)$ is a sequence such that $sum_n |u(n)| <infty$ then $int u dmu = sum_n u(n)$
By the definition of integral we have $int u dmu = int u^+ dmu - int u^- dmu$
By what you just showed for $u^+,u^-$ we have $u^pm = sum_n u^pm (n)$
Since the sum is uniformly convergence it is the same to sum from $1,2,...$ or to sum first the $n$ for which $u(n)>0$ and only then the $n$ for which $u(n)<0$.
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
$endgroup$
I prove
If $u(n)$ is a sequence such that $sum_n |u(n)| <infty$ then $int u dmu = sum_n u(n)$
By the definition of integral we have $int u dmu = int u^+ dmu - int u^- dmu$
By what you just showed for $u^+,u^-$ we have $u^pm = sum_n u^pm (n)$
Since the sum is uniformly convergence it is the same to sum from $1,2,...$ or to sum first the $n$ for which $u(n)>0$ and only then the $n$ for which $u(n)<0$.
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
answered Dec 3 '18 at 19:10
YankoYanko
6,4411528
6,4411528
add a comment |
add a comment |
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$begingroup$
I think you confuses the definitions. The equality holds by the definition of the integral and has nothing to do with the monotone convergence theorem. Also if $u:mathbb{N}rightarrowmathbb{R}^{+}$ then it is necessarily positive (again by definition).
$endgroup$
– Yanko
Dec 3 '18 at 18:37
$begingroup$
Why it holds by definition?
$endgroup$
– Sigurd
Dec 3 '18 at 18:38
$begingroup$
Which definition of integration do you use?
$endgroup$
– Yanko
Dec 3 '18 at 18:40
$begingroup$
For step functions / simple functions $f = sum_{i=1}^N alpha_i 1_{A_i}$: $I_mu(f) := sum_{i=1}^N alpha_i mu({A_i})$. For positive functions: $int u dmu := sup{I_mu(g): g leq u, g $ simple $} in [0,infty]$. For general functions: $int u dmu = int u^+dmu - int u^- dmu$.
$endgroup$
– Sigurd
Dec 3 '18 at 18:43
$begingroup$
Alright so a simple function in this case is a function that is supported only on finitely many points. It is only left to understand why $sup {I_u(g):gleq u, g text{ simple} } = lim_{nrightarrowinfty} I_u(u_n)$ where $u_n$ is $u$ restricted to the first $n$ points (i.e. $u_n(k)=u(k)$ if $kleq n$ and $0$ otherwise). Does it explain?
$endgroup$
– Yanko
Dec 3 '18 at 18:46