Complex measure of empty set.
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Reading the wikipedia pagina, a complex measure $mu$ is a function $mathcal{F} to mathbb{C}$ that is $sigma$-additive. I.e., if $(E_n)_n$ is a set of disjoint sets in the $sigma$-algebra $mathcal{F}$, we have
$$mu(bigcup_{n=1}^infty E_n) = sum_{n=1}^infty mu(E_n)$$
Unlike positive measures, we don't require that $mu(emptyset) = 0$. Why don't we require this last property?
measure-theory definition
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
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add a comment |
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Reading the wikipedia pagina, a complex measure $mu$ is a function $mathcal{F} to mathbb{C}$ that is $sigma$-additive. I.e., if $(E_n)_n$ is a set of disjoint sets in the $sigma$-algebra $mathcal{F}$, we have
$$mu(bigcup_{n=1}^infty E_n) = sum_{n=1}^infty mu(E_n)$$
Unlike positive measures, we don't require that $mu(emptyset) = 0$. Why don't we require this last property?
measure-theory definition
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
$endgroup$
add a comment |
$begingroup$
Reading the wikipedia pagina, a complex measure $mu$ is a function $mathcal{F} to mathbb{C}$ that is $sigma$-additive. I.e., if $(E_n)_n$ is a set of disjoint sets in the $sigma$-algebra $mathcal{F}$, we have
$$mu(bigcup_{n=1}^infty E_n) = sum_{n=1}^infty mu(E_n)$$
Unlike positive measures, we don't require that $mu(emptyset) = 0$. Why don't we require this last property?
measure-theory definition
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
$endgroup$
Reading the wikipedia pagina, a complex measure $mu$ is a function $mathcal{F} to mathbb{C}$ that is $sigma$-additive. I.e., if $(E_n)_n$ is a set of disjoint sets in the $sigma$-algebra $mathcal{F}$, we have
$$mu(bigcup_{n=1}^infty E_n) = sum_{n=1}^infty mu(E_n)$$
Unlike positive measures, we don't require that $mu(emptyset) = 0$. Why don't we require this last property?
measure-theory definition
measure-theory definition
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
asked Dec 1 '18 at 10:14
Math_QEDMath_QED
7,49531450
7,49531450
add a comment |
add a comment |
2 Answers
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oldest
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From $sigma$-additivity, we have $mu(Acup B)=mu(A)+mu(B)$ whenever $Acap B=emptyset$. Take $A=B=emptyset$.
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
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add a comment |
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Note that a complex measure is a bounded function, but a positive measure need not be bounded. So whenever $sigma$ is a positive measure $sigma(phi cup phi)=sigma(phi)+sigma(phi)$ doesn't always imply $sigma(phi)=0$, due to $infty+infty=infty$.
But for a complex measure $mu$ we have $mu(phi)in Bbb C$,hence $mu(phi cup phi)=mu(phi)+mu(phi)implies mu(phi)=0$
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
From $sigma$-additivity, we have $mu(Acup B)=mu(A)+mu(B)$ whenever $Acap B=emptyset$. Take $A=B=emptyset$.
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
$endgroup$
add a comment |
$begingroup$
From $sigma$-additivity, we have $mu(Acup B)=mu(A)+mu(B)$ whenever $Acap B=emptyset$. Take $A=B=emptyset$.
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
$endgroup$
add a comment |
$begingroup$
From $sigma$-additivity, we have $mu(Acup B)=mu(A)+mu(B)$ whenever $Acap B=emptyset$. Take $A=B=emptyset$.
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
$endgroup$
From $sigma$-additivity, we have $mu(Acup B)=mu(A)+mu(B)$ whenever $Acap B=emptyset$. Take $A=B=emptyset$.
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
answered Dec 1 '18 at 10:20
rldiasrldias
3,0101522
3,0101522
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$begingroup$
Note that a complex measure is a bounded function, but a positive measure need not be bounded. So whenever $sigma$ is a positive measure $sigma(phi cup phi)=sigma(phi)+sigma(phi)$ doesn't always imply $sigma(phi)=0$, due to $infty+infty=infty$.
But for a complex measure $mu$ we have $mu(phi)in Bbb C$,hence $mu(phi cup phi)=mu(phi)+mu(phi)implies mu(phi)=0$
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Note that a complex measure is a bounded function, but a positive measure need not be bounded. So whenever $sigma$ is a positive measure $sigma(phi cup phi)=sigma(phi)+sigma(phi)$ doesn't always imply $sigma(phi)=0$, due to $infty+infty=infty$.
But for a complex measure $mu$ we have $mu(phi)in Bbb C$,hence $mu(phi cup phi)=mu(phi)+mu(phi)implies mu(phi)=0$
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Note that a complex measure is a bounded function, but a positive measure need not be bounded. So whenever $sigma$ is a positive measure $sigma(phi cup phi)=sigma(phi)+sigma(phi)$ doesn't always imply $sigma(phi)=0$, due to $infty+infty=infty$.
But for a complex measure $mu$ we have $mu(phi)in Bbb C$,hence $mu(phi cup phi)=mu(phi)+mu(phi)implies mu(phi)=0$
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
$endgroup$
Note that a complex measure is a bounded function, but a positive measure need not be bounded. So whenever $sigma$ is a positive measure $sigma(phi cup phi)=sigma(phi)+sigma(phi)$ doesn't always imply $sigma(phi)=0$, due to $infty+infty=infty$.
But for a complex measure $mu$ we have $mu(phi)in Bbb C$,hence $mu(phi cup phi)=mu(phi)+mu(phi)implies mu(phi)=0$
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
answered Dec 1 '18 at 11:59
UserSUserS
1,5391112
1,5391112
add a comment |
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