Sigma Algebra Generated by Subset and Radon-Nikodym derivative
$begingroup$
So this is a question to study for qualifying exams, not a homework question!
Let (X,M,μ) be a measure space with μ(X) < ∞. Suppose A ∈ M and 0 < μ(A) < μ(X).
Let N be the σ-algebra generated by {A}, and denote the the restriction of μ to N by μ|N .
(a) Describe N and N-measurable functions from X to R.
(b) Let f be an M-measurable, μ-integrable function from X to R. Define
ν to be the signed measure on (X,N) given by ν : E →$int$ fdμ for every E ∈ N.
Determine, on (X, N ), the Radon-Nikodym derivative of ν with respect to μ|N .
a) The sigma algebra is N = {A, $A^c$,
$mathbb{R}$, $emptyset$}. If B is a Borel set, and f is a measurable function, then:
$f(A) subseteq B$
$f(A^c) subseteq B^c$
b) I'm really not sure how to approach this question. Also, if my understanding for part a is incorrect then I will be on the wrong track here too. Any suggestions for how to approach this? Also, is my understanding of part A correct?
Thank you for any and all help. Again, studying for quals, not doing homework!
real-analysis measure-theory radon-nikodym
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
$endgroup$
add a comment |
$begingroup$
So this is a question to study for qualifying exams, not a homework question!
Let (X,M,μ) be a measure space with μ(X) < ∞. Suppose A ∈ M and 0 < μ(A) < μ(X).
Let N be the σ-algebra generated by {A}, and denote the the restriction of μ to N by μ|N .
(a) Describe N and N-measurable functions from X to R.
(b) Let f be an M-measurable, μ-integrable function from X to R. Define
ν to be the signed measure on (X,N) given by ν : E →$int$ fdμ for every E ∈ N.
Determine, on (X, N ), the Radon-Nikodym derivative of ν with respect to μ|N .
a) The sigma algebra is N = {A, $A^c$,
$mathbb{R}$, $emptyset$}. If B is a Borel set, and f is a measurable function, then:
$f(A) subseteq B$
$f(A^c) subseteq B^c$
b) I'm really not sure how to approach this question. Also, if my understanding for part a is incorrect then I will be on the wrong track here too. Any suggestions for how to approach this? Also, is my understanding of part A correct?
Thank you for any and all help. Again, studying for quals, not doing homework!
real-analysis measure-theory radon-nikodym
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
$endgroup$
add a comment |
$begingroup$
So this is a question to study for qualifying exams, not a homework question!
Let (X,M,μ) be a measure space with μ(X) < ∞. Suppose A ∈ M and 0 < μ(A) < μ(X).
Let N be the σ-algebra generated by {A}, and denote the the restriction of μ to N by μ|N .
(a) Describe N and N-measurable functions from X to R.
(b) Let f be an M-measurable, μ-integrable function from X to R. Define
ν to be the signed measure on (X,N) given by ν : E →$int$ fdμ for every E ∈ N.
Determine, on (X, N ), the Radon-Nikodym derivative of ν with respect to μ|N .
a) The sigma algebra is N = {A, $A^c$,
$mathbb{R}$, $emptyset$}. If B is a Borel set, and f is a measurable function, then:
$f(A) subseteq B$
$f(A^c) subseteq B^c$
b) I'm really not sure how to approach this question. Also, if my understanding for part a is incorrect then I will be on the wrong track here too. Any suggestions for how to approach this? Also, is my understanding of part A correct?
Thank you for any and all help. Again, studying for quals, not doing homework!
real-analysis measure-theory radon-nikodym
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
$endgroup$
So this is a question to study for qualifying exams, not a homework question!
Let (X,M,μ) be a measure space with μ(X) < ∞. Suppose A ∈ M and 0 < μ(A) < μ(X).
Let N be the σ-algebra generated by {A}, and denote the the restriction of μ to N by μ|N .
(a) Describe N and N-measurable functions from X to R.
(b) Let f be an M-measurable, μ-integrable function from X to R. Define
ν to be the signed measure on (X,N) given by ν : E →$int$ fdμ for every E ∈ N.
Determine, on (X, N ), the Radon-Nikodym derivative of ν with respect to μ|N .
a) The sigma algebra is N = {A, $A^c$,
$mathbb{R}$, $emptyset$}. If B is a Borel set, and f is a measurable function, then:
$f(A) subseteq B$
$f(A^c) subseteq B^c$
b) I'm really not sure how to approach this question. Also, if my understanding for part a is incorrect then I will be on the wrong track here too. Any suggestions for how to approach this? Also, is my understanding of part A correct?
Thank you for any and all help. Again, studying for quals, not doing homework!
real-analysis measure-theory radon-nikodym
real-analysis measure-theory radon-nikodym
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
asked Dec 27 '18 at 15:21
Math LadyMath Lady
1196
1196
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
a) It is correct that $N={varnothing, A,A^{complement},X}$.
So a function $g:Xtomathbb R$ is $N$-measurable if $g^{-1}(B)in{varnothing, A,A^{complement},X}$ for every Borel set $Bsubseteqmathbb R$.
This implies that $g$ can take at most $2$ values (do you see why?).
We find that $g$ is $N$-measurable that if it can be written as $amathbf1_A+bmathbf1_{A^{complement}}$ or - equivalently - as $c+dmathbf1_A$.
b) The Radon-Nikodym derivative of $nu$ with respect to $mu|N$ must be an $N$-measurable function $g$ so must have the form $amathsf1_A+bmathsf1_{A^{complement}}$ as proved above.
Next to that it must satisfy the condition:$$int_Eg;d(mu|N)=nu(E)=int_E f;dmutext{ for every }Ein N={varnothing,A,A^{complement},X}$$
Substituting for $g=amathsf1_A+bmathsf1_{A^{complement}}$ we find $4$ equalities that must satisfied:
$0=0$ for $E=varnothing$
$amu(A)=int_A f;dmu$ for $E=A$
$bmu(A^{complement})=int_{A^{complement}} f;dmu$ for $E=A^{complement}$
$amu(A)+bmu(A^{complement})=int_Xf;dmu$ for $E=X$
The first is irrelevant and the fourth is a consequence of the second and third.
Since $0<mu(A)<mu(X)<infty$ we know that $mu(A)$ and $mu(A^{complement})$ are both positive and finite.
So the second and third enable us to find $a,b$ hence also $g=amathsf1_A+bmathsf1_{A^{complement}}$.
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
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add a comment |
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1 Answer
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$begingroup$
a) It is correct that $N={varnothing, A,A^{complement},X}$.
So a function $g:Xtomathbb R$ is $N$-measurable if $g^{-1}(B)in{varnothing, A,A^{complement},X}$ for every Borel set $Bsubseteqmathbb R$.
This implies that $g$ can take at most $2$ values (do you see why?).
We find that $g$ is $N$-measurable that if it can be written as $amathbf1_A+bmathbf1_{A^{complement}}$ or - equivalently - as $c+dmathbf1_A$.
b) The Radon-Nikodym derivative of $nu$ with respect to $mu|N$ must be an $N$-measurable function $g$ so must have the form $amathsf1_A+bmathsf1_{A^{complement}}$ as proved above.
Next to that it must satisfy the condition:$$int_Eg;d(mu|N)=nu(E)=int_E f;dmutext{ for every }Ein N={varnothing,A,A^{complement},X}$$
Substituting for $g=amathsf1_A+bmathsf1_{A^{complement}}$ we find $4$ equalities that must satisfied:
$0=0$ for $E=varnothing$
$amu(A)=int_A f;dmu$ for $E=A$
$bmu(A^{complement})=int_{A^{complement}} f;dmu$ for $E=A^{complement}$
$amu(A)+bmu(A^{complement})=int_Xf;dmu$ for $E=X$
The first is irrelevant and the fourth is a consequence of the second and third.
Since $0<mu(A)<mu(X)<infty$ we know that $mu(A)$ and $mu(A^{complement})$ are both positive and finite.
So the second and third enable us to find $a,b$ hence also $g=amathsf1_A+bmathsf1_{A^{complement}}$.
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
a) It is correct that $N={varnothing, A,A^{complement},X}$.
So a function $g:Xtomathbb R$ is $N$-measurable if $g^{-1}(B)in{varnothing, A,A^{complement},X}$ for every Borel set $Bsubseteqmathbb R$.
This implies that $g$ can take at most $2$ values (do you see why?).
We find that $g$ is $N$-measurable that if it can be written as $amathbf1_A+bmathbf1_{A^{complement}}$ or - equivalently - as $c+dmathbf1_A$.
b) The Radon-Nikodym derivative of $nu$ with respect to $mu|N$ must be an $N$-measurable function $g$ so must have the form $amathsf1_A+bmathsf1_{A^{complement}}$ as proved above.
Next to that it must satisfy the condition:$$int_Eg;d(mu|N)=nu(E)=int_E f;dmutext{ for every }Ein N={varnothing,A,A^{complement},X}$$
Substituting for $g=amathsf1_A+bmathsf1_{A^{complement}}$ we find $4$ equalities that must satisfied:
$0=0$ for $E=varnothing$
$amu(A)=int_A f;dmu$ for $E=A$
$bmu(A^{complement})=int_{A^{complement}} f;dmu$ for $E=A^{complement}$
$amu(A)+bmu(A^{complement})=int_Xf;dmu$ for $E=X$
The first is irrelevant and the fourth is a consequence of the second and third.
Since $0<mu(A)<mu(X)<infty$ we know that $mu(A)$ and $mu(A^{complement})$ are both positive and finite.
So the second and third enable us to find $a,b$ hence also $g=amathsf1_A+bmathsf1_{A^{complement}}$.
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
a) It is correct that $N={varnothing, A,A^{complement},X}$.
So a function $g:Xtomathbb R$ is $N$-measurable if $g^{-1}(B)in{varnothing, A,A^{complement},X}$ for every Borel set $Bsubseteqmathbb R$.
This implies that $g$ can take at most $2$ values (do you see why?).
We find that $g$ is $N$-measurable that if it can be written as $amathbf1_A+bmathbf1_{A^{complement}}$ or - equivalently - as $c+dmathbf1_A$.
b) The Radon-Nikodym derivative of $nu$ with respect to $mu|N$ must be an $N$-measurable function $g$ so must have the form $amathsf1_A+bmathsf1_{A^{complement}}$ as proved above.
Next to that it must satisfy the condition:$$int_Eg;d(mu|N)=nu(E)=int_E f;dmutext{ for every }Ein N={varnothing,A,A^{complement},X}$$
Substituting for $g=amathsf1_A+bmathsf1_{A^{complement}}$ we find $4$ equalities that must satisfied:
$0=0$ for $E=varnothing$
$amu(A)=int_A f;dmu$ for $E=A$
$bmu(A^{complement})=int_{A^{complement}} f;dmu$ for $E=A^{complement}$
$amu(A)+bmu(A^{complement})=int_Xf;dmu$ for $E=X$
The first is irrelevant and the fourth is a consequence of the second and third.
Since $0<mu(A)<mu(X)<infty$ we know that $mu(A)$ and $mu(A^{complement})$ are both positive and finite.
So the second and third enable us to find $a,b$ hence also $g=amathsf1_A+bmathsf1_{A^{complement}}$.
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
$endgroup$
a) It is correct that $N={varnothing, A,A^{complement},X}$.
So a function $g:Xtomathbb R$ is $N$-measurable if $g^{-1}(B)in{varnothing, A,A^{complement},X}$ for every Borel set $Bsubseteqmathbb R$.
This implies that $g$ can take at most $2$ values (do you see why?).
We find that $g$ is $N$-measurable that if it can be written as $amathbf1_A+bmathbf1_{A^{complement}}$ or - equivalently - as $c+dmathbf1_A$.
b) The Radon-Nikodym derivative of $nu$ with respect to $mu|N$ must be an $N$-measurable function $g$ so must have the form $amathsf1_A+bmathsf1_{A^{complement}}$ as proved above.
Next to that it must satisfy the condition:$$int_Eg;d(mu|N)=nu(E)=int_E f;dmutext{ for every }Ein N={varnothing,A,A^{complement},X}$$
Substituting for $g=amathsf1_A+bmathsf1_{A^{complement}}$ we find $4$ equalities that must satisfied:
$0=0$ for $E=varnothing$
$amu(A)=int_A f;dmu$ for $E=A$
$bmu(A^{complement})=int_{A^{complement}} f;dmu$ for $E=A^{complement}$
$amu(A)+bmu(A^{complement})=int_Xf;dmu$ for $E=X$
The first is irrelevant and the fourth is a consequence of the second and third.
Since $0<mu(A)<mu(X)<infty$ we know that $mu(A)$ and $mu(A^{complement})$ are both positive and finite.
So the second and third enable us to find $a,b$ hence also $g=amathsf1_A+bmathsf1_{A^{complement}}$.
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
edited Dec 27 '18 at 16:31
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
answered Dec 27 '18 at 16:01
drhabdrhab
101k545136
101k545136
add a comment |
add a comment |
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