Mixing
Show that $lim _{ntoinfty}sum_{i=1}^{N}u_n(A_i)=sum_{i=1}^{N}lim _{ntoinfty}u_n(A_i)$
-2
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I want to show that $lim _{ntoinfty}sum_{i=1}^{N}u_n(A_i)=sum_{i=1}^{N}lim _{ntoinfty}u_n(A_i)$ , where $(u_n)_{n=1}^{infty}$ is a sequence of increasing positive measures on a measurable space. Here increasing means that $u_n(A)leq u_{n+1} (A)$ for any measurable set A. Any help is greatly appreciated since I don't know where to start;(
real-analysis measure-theory
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
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add a comment |
-2
$begingroup$
I want to show that $lim _{ntoinfty}sum_{i=1}^{N}u_n(A_i)=sum_{i=1}^{N}lim _{ntoinfty}u_n(A_i)$ , where $(u_n)_{n=1}^{infty}$ is a sequence of increasing positive measures on a measurable space. Here increasing means that $u_n(A)leq u_{n+1} (A)$ for any measurable set A. Any help is greatly appreciated since I don't know where to start;(
real-analysis measure-theory
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
$endgroup$
1
$begingroup$
If it's a finite sum, then there shouldn't be a problem. That they're increasing guarantees convergence
$endgroup$
– Jakobian
Jan 2 at 12:18
$begingroup$
Thanks. Yes it is finite, but why its no problem? Is there any formal proof for this?
$endgroup$
– Michael Maier
Jan 2 at 12:20
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You are asking an elementary question in Calculus: limit of a (finite) sum is the sum of the limits. What has this to do with measure theory?
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:32
1
$begingroup$
What we're asking of basically is that we have $N$ increasing sequences, $x_{n, 1}, x_{n, 2}, ..., x_{n, N}$, and we want to check if $sum_i lim_n x_{n, i} = lim_n sum_i x_{n, i}$. We know that for every $1leq i leq N$, $lim_n x_{n, i}$ exists, being possible $infty$, because they are monotone. Even if for one of the $i$, $lim_n x_{n, i} = infty$, both sides would be infinite, and the equality holds. If they are all finite, then we use elementary result in calculus, as Kavi Rama Murthy said, limit of a sum of sequences is sum of limits, provided everything makes sense (so they exist)
$endgroup$
– Jakobian
Jan 2 at 14:01
$begingroup$
Got it now, thanks!
$endgroup$
– Michael Maier
Jan 2 at 14:36
add a comment |
-2
-2
-2
$begingroup$
I want to show that $lim _{ntoinfty}sum_{i=1}^{N}u_n(A_i)=sum_{i=1}^{N}lim _{ntoinfty}u_n(A_i)$ , where $(u_n)_{n=1}^{infty}$ is a sequence of increasing positive measures on a measurable space. Here increasing means that $u_n(A)leq u_{n+1} (A)$ for any measurable set A. Any help is greatly appreciated since I don't know where to start;(
real-analysis measure-theory
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
$endgroup$
I want to show that $lim _{ntoinfty}sum_{i=1}^{N}u_n(A_i)=sum_{i=1}^{N}lim _{ntoinfty}u_n(A_i)$ , where $(u_n)_{n=1}^{infty}$ is a sequence of increasing positive measures on a measurable space. Here increasing means that $u_n(A)leq u_{n+1} (A)$ for any measurable set A. Any help is greatly appreciated since I don't know where to start;(
real-analysis measure-theory
real-analysis measure-theory
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
asked Jan 2 at 12:15
Michael MaierMichael Maier
698
698
1
$begingroup$
If it's a finite sum, then there shouldn't be a problem. That they're increasing guarantees convergence
$endgroup$
– Jakobian
Jan 2 at 12:18
$begingroup$
Thanks. Yes it is finite, but why its no problem? Is there any formal proof for this?
$endgroup$
– Michael Maier
Jan 2 at 12:20
$begingroup$
You are asking an elementary question in Calculus: limit of a (finite) sum is the sum of the limits. What has this to do with measure theory?
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:32
1
$begingroup$
What we're asking of basically is that we have $N$ increasing sequences, $x_{n, 1}, x_{n, 2}, ..., x_{n, N}$, and we want to check if $sum_i lim_n x_{n, i} = lim_n sum_i x_{n, i}$. We know that for every $1leq i leq N$, $lim_n x_{n, i}$ exists, being possible $infty$, because they are monotone. Even if for one of the $i$, $lim_n x_{n, i} = infty$, both sides would be infinite, and the equality holds. If they are all finite, then we use elementary result in calculus, as Kavi Rama Murthy said, limit of a sum of sequences is sum of limits, provided everything makes sense (so they exist)
$endgroup$
– Jakobian
Jan 2 at 14:01
$begingroup$
Got it now, thanks!
$endgroup$
– Michael Maier
Jan 2 at 14:36
add a comment |
1
$begingroup$
If it's a finite sum, then there shouldn't be a problem. That they're increasing guarantees convergence
$endgroup$
– Jakobian
Jan 2 at 12:18
$begingroup$
Thanks. Yes it is finite, but why its no problem? Is there any formal proof for this?
$endgroup$
– Michael Maier
Jan 2 at 12:20
$begingroup$
You are asking an elementary question in Calculus: limit of a (finite) sum is the sum of the limits. What has this to do with measure theory?
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:32
1
$begingroup$
What we're asking of basically is that we have $N$ increasing sequences, $x_{n, 1}, x_{n, 2}, ..., x_{n, N}$, and we want to check if $sum_i lim_n x_{n, i} = lim_n sum_i x_{n, i}$. We know that for every $1leq i leq N$, $lim_n x_{n, i}$ exists, being possible $infty$, because they are monotone. Even if for one of the $i$, $lim_n x_{n, i} = infty$, both sides would be infinite, and the equality holds. If they are all finite, then we use elementary result in calculus, as Kavi Rama Murthy said, limit of a sum of sequences is sum of limits, provided everything makes sense (so they exist)
$endgroup$
– Jakobian
Jan 2 at 14:01
$begingroup$
Got it now, thanks!
$endgroup$
– Michael Maier
Jan 2 at 14:36
1
1
$begingroup$
If it's a finite sum, then there shouldn't be a problem. That they're increasing guarantees convergence
$endgroup$
– Jakobian
Jan 2 at 12:18
$begingroup$
If it's a finite sum, then there shouldn't be a problem. That they're increasing guarantees convergence
$endgroup$
– Jakobian
Jan 2 at 12:18
$begingroup$
Thanks. Yes it is finite, but why its no problem? Is there any formal proof for this?
$endgroup$
– Michael Maier
Jan 2 at 12:20
$begingroup$
Thanks. Yes it is finite, but why its no problem? Is there any formal proof for this?
$endgroup$
– Michael Maier
Jan 2 at 12:20
$begingroup$
You are asking an elementary question in Calculus: limit of a (finite) sum is the sum of the limits. What has this to do with measure theory?
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:32
$begingroup$
You are asking an elementary question in Calculus: limit of a (finite) sum is the sum of the limits. What has this to do with measure theory?
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:32
1
1
$begingroup$
What we're asking of basically is that we have $N$ increasing sequences, $x_{n, 1}, x_{n, 2}, ..., x_{n, N}$, and we want to check if $sum_i lim_n x_{n, i} = lim_n sum_i x_{n, i}$. We know that for every $1leq i leq N$, $lim_n x_{n, i}$ exists, being possible $infty$, because they are monotone. Even if for one of the $i$, $lim_n x_{n, i} = infty$, both sides would be infinite, and the equality holds. If they are all finite, then we use elementary result in calculus, as Kavi Rama Murthy said, limit of a sum of sequences is sum of limits, provided everything makes sense (so they exist)
$endgroup$
– Jakobian
Jan 2 at 14:01
$begingroup$
What we're asking of basically is that we have $N$ increasing sequences, $x_{n, 1}, x_{n, 2}, ..., x_{n, N}$, and we want to check if $sum_i lim_n x_{n, i} = lim_n sum_i x_{n, i}$. We know that for every $1leq i leq N$, $lim_n x_{n, i}$ exists, being possible $infty$, because they are monotone. Even if for one of the $i$, $lim_n x_{n, i} = infty$, both sides would be infinite, and the equality holds. If they are all finite, then we use elementary result in calculus, as Kavi Rama Murthy said, limit of a sum of sequences is sum of limits, provided everything makes sense (so they exist)
$endgroup$
– Jakobian
Jan 2 at 14:01
$begingroup$
Got it now, thanks!
$endgroup$
– Michael Maier
Jan 2 at 14:36
$begingroup$
Got it now, thanks!
$endgroup$
– Michael Maier
Jan 2 at 14:36
add a comment |
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1
$begingroup$
If it's a finite sum, then there shouldn't be a problem. That they're increasing guarantees convergence
$endgroup$
– Jakobian
Jan 2 at 12:18
$begingroup$
Thanks. Yes it is finite, but why its no problem? Is there any formal proof for this?
$endgroup$
– Michael Maier
Jan 2 at 12:20
$begingroup$
You are asking an elementary question in Calculus: limit of a (finite) sum is the sum of the limits. What has this to do with measure theory?
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:32
1
$begingroup$
What we're asking of basically is that we have $N$ increasing sequences, $x_{n, 1}, x_{n, 2}, ..., x_{n, N}$, and we want to check if $sum_i lim_n x_{n, i} = lim_n sum_i x_{n, i}$. We know that for every $1leq i leq N$, $lim_n x_{n, i}$ exists, being possible $infty$, because they are monotone. Even if for one of the $i$, $lim_n x_{n, i} = infty$, both sides would be infinite, and the equality holds. If they are all finite, then we use elementary result in calculus, as Kavi Rama Murthy said, limit of a sum of sequences is sum of limits, provided everything makes sense (so they exist)
$endgroup$
– Jakobian
Jan 2 at 14:01
$begingroup$
Got it now, thanks!
$endgroup$
– Michael Maier
Jan 2 at 14:36