Finding a Hahn Decomposition involving a Dirac Measure
$begingroup$
Let $(X,F,mu)$ be a finite measure space, i.e. $mu(X)<infty$. And let $xin X$, and let $delta_x$ be the Dirac measure with respect to $x$, i.e. $delta_x(E)=1$ if $xin E$ and $delta_x(E)=0$ if $xnotin E$. Finally let $nu$ be the signed measure $mu-adelta_x$, where $a=mu(X)$. Find the Hahn decomposition of $nu$.
I'm not really sure how to construct Hahn decompositions. I tried rewriting the proof of the Hahn-Jordan decomposition theorem for the case of this particular signed measure, but it didn't give me a concrete pair of sets.
measure-theory lebesgue-measure dirac-delta signed-measures
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
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add a comment |
$begingroup$
Let $(X,F,mu)$ be a finite measure space, i.e. $mu(X)<infty$. And let $xin X$, and let $delta_x$ be the Dirac measure with respect to $x$, i.e. $delta_x(E)=1$ if $xin E$ and $delta_x(E)=0$ if $xnotin E$. Finally let $nu$ be the signed measure $mu-adelta_x$, where $a=mu(X)$. Find the Hahn decomposition of $nu$.
I'm not really sure how to construct Hahn decompositions. I tried rewriting the proof of the Hahn-Jordan decomposition theorem for the case of this particular signed measure, but it didn't give me a concrete pair of sets.
measure-theory lebesgue-measure dirac-delta signed-measures
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
$endgroup$
add a comment |
$begingroup$
Let $(X,F,mu)$ be a finite measure space, i.e. $mu(X)<infty$. And let $xin X$, and let $delta_x$ be the Dirac measure with respect to $x$, i.e. $delta_x(E)=1$ if $xin E$ and $delta_x(E)=0$ if $xnotin E$. Finally let $nu$ be the signed measure $mu-adelta_x$, where $a=mu(X)$. Find the Hahn decomposition of $nu$.
I'm not really sure how to construct Hahn decompositions. I tried rewriting the proof of the Hahn-Jordan decomposition theorem for the case of this particular signed measure, but it didn't give me a concrete pair of sets.
measure-theory lebesgue-measure dirac-delta signed-measures
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
$endgroup$
Let $(X,F,mu)$ be a finite measure space, i.e. $mu(X)<infty$. And let $xin X$, and let $delta_x$ be the Dirac measure with respect to $x$, i.e. $delta_x(E)=1$ if $xin E$ and $delta_x(E)=0$ if $xnotin E$. Finally let $nu$ be the signed measure $mu-adelta_x$, where $a=mu(X)$. Find the Hahn decomposition of $nu$.
I'm not really sure how to construct Hahn decompositions. I tried rewriting the proof of the Hahn-Jordan decomposition theorem for the case of this particular signed measure, but it didn't give me a concrete pair of sets.
measure-theory lebesgue-measure dirac-delta signed-measures
measure-theory lebesgue-measure dirac-delta signed-measures
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
asked Nov 28 '18 at 6:24
Keshav SrinivasanKeshav Srinivasan
2,06111443
2,06111443
add a comment |
add a comment |
1 Answer
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Since the $sigma$-algebra isn't specified, you cannot give an explicit choice for the Hahn-decomposition. (For example $F= {X, emptyset}$ gives only a trivial decomposition. One other example, is $F={X, A,A^c,emptyset})$ with $X= [0,1]$, $A= [0,1/2]$ and $mu = delta_0$ and $x=1$. Then $X= A cup A^c$ is the Hahn-decomposition.)
Assume that ${x} in F$, then the decomposition can be determined. If $mu({x}) -a ge 0$, then $nu := mu -a delta_x$ is a non-negative measure and $nu^{-} =0$ is trivial. (Thus the Hahn-decomposition is $X=P cup N$ with $N = emptyset$.) On the other hand, if $mu({x}) -a <0$. Then the Hahn-decompisition is $X =P cup N$ with $N= {x}$ and $P = X setminus {x}$.
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since the $sigma$-algebra isn't specified, you cannot give an explicit choice for the Hahn-decomposition. (For example $F= {X, emptyset}$ gives only a trivial decomposition. One other example, is $F={X, A,A^c,emptyset})$ with $X= [0,1]$, $A= [0,1/2]$ and $mu = delta_0$ and $x=1$. Then $X= A cup A^c$ is the Hahn-decomposition.)
Assume that ${x} in F$, then the decomposition can be determined. If $mu({x}) -a ge 0$, then $nu := mu -a delta_x$ is a non-negative measure and $nu^{-} =0$ is trivial. (Thus the Hahn-decomposition is $X=P cup N$ with $N = emptyset$.) On the other hand, if $mu({x}) -a <0$. Then the Hahn-decompisition is $X =P cup N$ with $N= {x}$ and $P = X setminus {x}$.
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
$endgroup$
add a comment |
$begingroup$
Since the $sigma$-algebra isn't specified, you cannot give an explicit choice for the Hahn-decomposition. (For example $F= {X, emptyset}$ gives only a trivial decomposition. One other example, is $F={X, A,A^c,emptyset})$ with $X= [0,1]$, $A= [0,1/2]$ and $mu = delta_0$ and $x=1$. Then $X= A cup A^c$ is the Hahn-decomposition.)
Assume that ${x} in F$, then the decomposition can be determined. If $mu({x}) -a ge 0$, then $nu := mu -a delta_x$ is a non-negative measure and $nu^{-} =0$ is trivial. (Thus the Hahn-decomposition is $X=P cup N$ with $N = emptyset$.) On the other hand, if $mu({x}) -a <0$. Then the Hahn-decompisition is $X =P cup N$ with $N= {x}$ and $P = X setminus {x}$.
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
$endgroup$
add a comment |
$begingroup$
Since the $sigma$-algebra isn't specified, you cannot give an explicit choice for the Hahn-decomposition. (For example $F= {X, emptyset}$ gives only a trivial decomposition. One other example, is $F={X, A,A^c,emptyset})$ with $X= [0,1]$, $A= [0,1/2]$ and $mu = delta_0$ and $x=1$. Then $X= A cup A^c$ is the Hahn-decomposition.)
Assume that ${x} in F$, then the decomposition can be determined. If $mu({x}) -a ge 0$, then $nu := mu -a delta_x$ is a non-negative measure and $nu^{-} =0$ is trivial. (Thus the Hahn-decomposition is $X=P cup N$ with $N = emptyset$.) On the other hand, if $mu({x}) -a <0$. Then the Hahn-decompisition is $X =P cup N$ with $N= {x}$ and $P = X setminus {x}$.
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
$endgroup$
Since the $sigma$-algebra isn't specified, you cannot give an explicit choice for the Hahn-decomposition. (For example $F= {X, emptyset}$ gives only a trivial decomposition. One other example, is $F={X, A,A^c,emptyset})$ with $X= [0,1]$, $A= [0,1/2]$ and $mu = delta_0$ and $x=1$. Then $X= A cup A^c$ is the Hahn-decomposition.)
Assume that ${x} in F$, then the decomposition can be determined. If $mu({x}) -a ge 0$, then $nu := mu -a delta_x$ is a non-negative measure and $nu^{-} =0$ is trivial. (Thus the Hahn-decomposition is $X=P cup N$ with $N = emptyset$.) On the other hand, if $mu({x}) -a <0$. Then the Hahn-decompisition is $X =P cup N$ with $N= {x}$ and $P = X setminus {x}$.
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
answered Nov 28 '18 at 9:45
p4schp4sch
5,190217
5,190217
add a comment |
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