Why is it ok to swap modulus and sum in this case?
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I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.
https://www.youtube.com/watch?v=E8FwgGn1e_o
At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"
That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then
$sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.
real-analysis measure-theory
asked Dec 6 '18 at 22:14
DamoDamo
491210
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add a comment |
$begingroup$
I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.
https://www.youtube.com/watch?v=E8FwgGn1e_o
At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"
That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then
$sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.
real-analysis measure-theory
asked Dec 6 '18 at 22:14
DamoDamo
491210
$endgroup$
add a comment |
$begingroup$
I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.
https://www.youtube.com/watch?v=E8FwgGn1e_o
At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"
That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then
$sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.
real-analysis measure-theory
asked Dec 6 '18 at 22:14
DamoDamo
491210
$endgroup$
I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.
https://www.youtube.com/watch?v=E8FwgGn1e_o
At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"
That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then
$sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.
real-analysis measure-theory
real-analysis measure-theory
asked Dec 6 '18 at 22:14
DamoDamo
491210
asked Dec 6 '18 at 22:14
DamoDamo
491210
asked Dec 6 '18 at 22:14
DamoDamo
491210
asked Dec 6 '18 at 22:14
DamoDamo
491210
asked Dec 6 '18 at 22:14
DamoDamo
491210
491210
add a comment |
add a comment |
1 Answer
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The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:
Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
$$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
Then $|g|leq1$, and
$$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
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1
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I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
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– Damo
Dec 7 '18 at 1:18
1
$begingroup$
Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
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– Damo
Dec 7 '18 at 1:24
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You're completely correct, thanks for pointing that out.
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– Aweygan
Dec 7 '18 at 1:42
add a comment |
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1 Answer
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$begingroup$
The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:
Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
$$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
Then $|g|leq1$, and
$$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
$endgroup$
1
$begingroup$
I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
$endgroup$
– Damo
Dec 7 '18 at 1:18
1
$begingroup$
Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
$endgroup$
– Damo
Dec 7 '18 at 1:24
$begingroup$
You're completely correct, thanks for pointing that out.
$endgroup$
– Aweygan
Dec 7 '18 at 1:42
add a comment |
$begingroup$
The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:
Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
$$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
Then $|g|leq1$, and
$$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
$endgroup$
1
$begingroup$
I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
$endgroup$
– Damo
Dec 7 '18 at 1:18
1
$begingroup$
Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
$endgroup$
– Damo
Dec 7 '18 at 1:24
$begingroup$
You're completely correct, thanks for pointing that out.
$endgroup$
– Aweygan
Dec 7 '18 at 1:42
add a comment |
$begingroup$
The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:
Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
$$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
Then $|g|leq1$, and
$$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
$endgroup$
The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:
Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
$$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
Then $|g|leq1$, and
$$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
edited Dec 7 '18 at 1:41
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
answered Dec 6 '18 at 22:31
AweyganAweygan
14.5k21442
14.5k21442
1
$begingroup$
I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
$endgroup$
– Damo
Dec 7 '18 at 1:18
1
$begingroup$
Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
$endgroup$
– Damo
Dec 7 '18 at 1:24
$begingroup$
You're completely correct, thanks for pointing that out.
$endgroup$
– Aweygan
Dec 7 '18 at 1:42
add a comment |
1
$begingroup$
I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
$endgroup$
– Damo
Dec 7 '18 at 1:18
1
$begingroup$
Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
$endgroup$
– Damo
Dec 7 '18 at 1:24
$begingroup$
You're completely correct, thanks for pointing that out.
$endgroup$
– Aweygan
Dec 7 '18 at 1:42
1
1
$begingroup$
I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
$endgroup$
– Damo
Dec 7 '18 at 1:18
$begingroup$
I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
$endgroup$
– Damo
Dec 7 '18 at 1:18
1
1
$begingroup$
Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
$endgroup$
– Damo
Dec 7 '18 at 1:24
$begingroup$
Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
$endgroup$
– Damo
Dec 7 '18 at 1:24
$begingroup$
You're completely correct, thanks for pointing that out.
$endgroup$
– Aweygan
Dec 7 '18 at 1:42
$begingroup$
You're completely correct, thanks for pointing that out.
$endgroup$
– Aweygan
Dec 7 '18 at 1:42
add a comment |
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