Caratheodory measure problem











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Let $μ$ be a Caratheodory extension of an elementary measure $μ_0$ defined on a semiring $P$. Let $A$ be a measurable set with $μ(A)<∞$. Show there are sets $B,C∈σ(P)$ such that $B⊂A⊂C$ and $μ(Csetminus B)=0$.



I managed to show the existence of $C$
using the identity:



$μ(A)$=$inf{sum_{j=1}^inftymu(R_j):(R_j)_{j=1}^inftysubset P, Asubsetcup_{j=1}^infty R_j}$



Where $μ≡μ_0$ on $P$. I used it to prove that there is a set $C∈σ(P)$ such that $A⊂C$ and $μ(A)=μ(C)$. I believe this is exactly the set $C$ I will need for the exercise. But how to find $B$? The identity I wrote is all about sets that contain $A$, but I don't know anything about the subsets of $A$ itself. I thought about using the fact that Caratheodory measure is complete and about taking the supremum of the measures of subsets of $A$ that are in $σ(P)$, but that didn't help me so far. Any ideas?










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  • whats the definition of semiring?
    – Robson
    Nov 17 at 21:45










  • A semiring $P$ on a set $X$ is subset of $2^X$ which contains the empty set, closed under finite intersections, and the difference between any two sets in $P$ can be written as a finite union of pairwise disjoint sets in $P$.
    – Mark
    Nov 17 at 21:47

















up vote
1
down vote

favorite
1












Let $μ$ be a Caratheodory extension of an elementary measure $μ_0$ defined on a semiring $P$. Let $A$ be a measurable set with $μ(A)<∞$. Show there are sets $B,C∈σ(P)$ such that $B⊂A⊂C$ and $μ(Csetminus B)=0$.



I managed to show the existence of $C$
using the identity:



$μ(A)$=$inf{sum_{j=1}^inftymu(R_j):(R_j)_{j=1}^inftysubset P, Asubsetcup_{j=1}^infty R_j}$



Where $μ≡μ_0$ on $P$. I used it to prove that there is a set $C∈σ(P)$ such that $A⊂C$ and $μ(A)=μ(C)$. I believe this is exactly the set $C$ I will need for the exercise. But how to find $B$? The identity I wrote is all about sets that contain $A$, but I don't know anything about the subsets of $A$ itself. I thought about using the fact that Caratheodory measure is complete and about taking the supremum of the measures of subsets of $A$ that are in $σ(P)$, but that didn't help me so far. Any ideas?










share|cite|improve this question






















  • whats the definition of semiring?
    – Robson
    Nov 17 at 21:45










  • A semiring $P$ on a set $X$ is subset of $2^X$ which contains the empty set, closed under finite intersections, and the difference between any two sets in $P$ can be written as a finite union of pairwise disjoint sets in $P$.
    – Mark
    Nov 17 at 21:47















up vote
1
down vote

favorite
1









up vote
1
down vote

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1





Let $μ$ be a Caratheodory extension of an elementary measure $μ_0$ defined on a semiring $P$. Let $A$ be a measurable set with $μ(A)<∞$. Show there are sets $B,C∈σ(P)$ such that $B⊂A⊂C$ and $μ(Csetminus B)=0$.



I managed to show the existence of $C$
using the identity:



$μ(A)$=$inf{sum_{j=1}^inftymu(R_j):(R_j)_{j=1}^inftysubset P, Asubsetcup_{j=1}^infty R_j}$



Where $μ≡μ_0$ on $P$. I used it to prove that there is a set $C∈σ(P)$ such that $A⊂C$ and $μ(A)=μ(C)$. I believe this is exactly the set $C$ I will need for the exercise. But how to find $B$? The identity I wrote is all about sets that contain $A$, but I don't know anything about the subsets of $A$ itself. I thought about using the fact that Caratheodory measure is complete and about taking the supremum of the measures of subsets of $A$ that are in $σ(P)$, but that didn't help me so far. Any ideas?










share|cite|improve this question













Let $μ$ be a Caratheodory extension of an elementary measure $μ_0$ defined on a semiring $P$. Let $A$ be a measurable set with $μ(A)<∞$. Show there are sets $B,C∈σ(P)$ such that $B⊂A⊂C$ and $μ(Csetminus B)=0$.



I managed to show the existence of $C$
using the identity:



$μ(A)$=$inf{sum_{j=1}^inftymu(R_j):(R_j)_{j=1}^inftysubset P, Asubsetcup_{j=1}^infty R_j}$



Where $μ≡μ_0$ on $P$. I used it to prove that there is a set $C∈σ(P)$ such that $A⊂C$ and $μ(A)=μ(C)$. I believe this is exactly the set $C$ I will need for the exercise. But how to find $B$? The identity I wrote is all about sets that contain $A$, but I don't know anything about the subsets of $A$ itself. I thought about using the fact that Caratheodory measure is complete and about taking the supremum of the measures of subsets of $A$ that are in $σ(P)$, but that didn't help me so far. Any ideas?







measure-theory lebesgue-measure






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asked Nov 17 at 21:36









Mark

5,750415




5,750415












  • whats the definition of semiring?
    – Robson
    Nov 17 at 21:45










  • A semiring $P$ on a set $X$ is subset of $2^X$ which contains the empty set, closed under finite intersections, and the difference between any two sets in $P$ can be written as a finite union of pairwise disjoint sets in $P$.
    – Mark
    Nov 17 at 21:47




















  • whats the definition of semiring?
    – Robson
    Nov 17 at 21:45










  • A semiring $P$ on a set $X$ is subset of $2^X$ which contains the empty set, closed under finite intersections, and the difference between any two sets in $P$ can be written as a finite union of pairwise disjoint sets in $P$.
    – Mark
    Nov 17 at 21:47


















whats the definition of semiring?
– Robson
Nov 17 at 21:45




whats the definition of semiring?
– Robson
Nov 17 at 21:45












A semiring $P$ on a set $X$ is subset of $2^X$ which contains the empty set, closed under finite intersections, and the difference between any two sets in $P$ can be written as a finite union of pairwise disjoint sets in $P$.
– Mark
Nov 17 at 21:47






A semiring $P$ on a set $X$ is subset of $2^X$ which contains the empty set, closed under finite intersections, and the difference between any two sets in $P$ can be written as a finite union of pairwise disjoint sets in $P$.
– Mark
Nov 17 at 21:47












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Let $S$ be the set of all sequences $(R_j)$ in $P$ which satisfies the assertion on the definition of $mu(A)$.



Given a natural $n$, there is a sequence $(R_{j,n})_{jin mathbb{N}}$ in $S$ which satisfy $$mu(bigcup _j R_{j,n} - A)=sum_j mu(R_{j,n})-mu(A) < frac{1}{n}$$



So, what you say about this set: (?) $$C=bigcap _{n=1}^{infty}bigcup_{j=1}^{infty}R_{j,n}$$



You can prove that $C in sigma(P)$ and $mu(Csetminus A)=0$.






share|cite|improve this answer





















  • hmm, now I've seen that this part you already did ...
    – Robson
    Nov 17 at 22:12










  • Thanks for the answer. Yes, I did the same thing to find $C$. But the problem is that $A$ is not necessary in $sigma(P)$. If it is then we take $A=B$ and that's it. But it may not be the case.
    – Mark
    Nov 17 at 22:13











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Let $S$ be the set of all sequences $(R_j)$ in $P$ which satisfies the assertion on the definition of $mu(A)$.



Given a natural $n$, there is a sequence $(R_{j,n})_{jin mathbb{N}}$ in $S$ which satisfy $$mu(bigcup _j R_{j,n} - A)=sum_j mu(R_{j,n})-mu(A) < frac{1}{n}$$



So, what you say about this set: (?) $$C=bigcap _{n=1}^{infty}bigcup_{j=1}^{infty}R_{j,n}$$



You can prove that $C in sigma(P)$ and $mu(Csetminus A)=0$.






share|cite|improve this answer





















  • hmm, now I've seen that this part you already did ...
    – Robson
    Nov 17 at 22:12










  • Thanks for the answer. Yes, I did the same thing to find $C$. But the problem is that $A$ is not necessary in $sigma(P)$. If it is then we take $A=B$ and that's it. But it may not be the case.
    – Mark
    Nov 17 at 22:13















up vote
0
down vote













Let $S$ be the set of all sequences $(R_j)$ in $P$ which satisfies the assertion on the definition of $mu(A)$.



Given a natural $n$, there is a sequence $(R_{j,n})_{jin mathbb{N}}$ in $S$ which satisfy $$mu(bigcup _j R_{j,n} - A)=sum_j mu(R_{j,n})-mu(A) < frac{1}{n}$$



So, what you say about this set: (?) $$C=bigcap _{n=1}^{infty}bigcup_{j=1}^{infty}R_{j,n}$$



You can prove that $C in sigma(P)$ and $mu(Csetminus A)=0$.






share|cite|improve this answer





















  • hmm, now I've seen that this part you already did ...
    – Robson
    Nov 17 at 22:12










  • Thanks for the answer. Yes, I did the same thing to find $C$. But the problem is that $A$ is not necessary in $sigma(P)$. If it is then we take $A=B$ and that's it. But it may not be the case.
    – Mark
    Nov 17 at 22:13













up vote
0
down vote










up vote
0
down vote









Let $S$ be the set of all sequences $(R_j)$ in $P$ which satisfies the assertion on the definition of $mu(A)$.



Given a natural $n$, there is a sequence $(R_{j,n})_{jin mathbb{N}}$ in $S$ which satisfy $$mu(bigcup _j R_{j,n} - A)=sum_j mu(R_{j,n})-mu(A) < frac{1}{n}$$



So, what you say about this set: (?) $$C=bigcap _{n=1}^{infty}bigcup_{j=1}^{infty}R_{j,n}$$



You can prove that $C in sigma(P)$ and $mu(Csetminus A)=0$.






share|cite|improve this answer












Let $S$ be the set of all sequences $(R_j)$ in $P$ which satisfies the assertion on the definition of $mu(A)$.



Given a natural $n$, there is a sequence $(R_{j,n})_{jin mathbb{N}}$ in $S$ which satisfy $$mu(bigcup _j R_{j,n} - A)=sum_j mu(R_{j,n})-mu(A) < frac{1}{n}$$



So, what you say about this set: (?) $$C=bigcap _{n=1}^{infty}bigcup_{j=1}^{infty}R_{j,n}$$



You can prove that $C in sigma(P)$ and $mu(Csetminus A)=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 17 at 22:06









Robson

725221




725221












  • hmm, now I've seen that this part you already did ...
    – Robson
    Nov 17 at 22:12










  • Thanks for the answer. Yes, I did the same thing to find $C$. But the problem is that $A$ is not necessary in $sigma(P)$. If it is then we take $A=B$ and that's it. But it may not be the case.
    – Mark
    Nov 17 at 22:13


















  • hmm, now I've seen that this part you already did ...
    – Robson
    Nov 17 at 22:12










  • Thanks for the answer. Yes, I did the same thing to find $C$. But the problem is that $A$ is not necessary in $sigma(P)$. If it is then we take $A=B$ and that's it. But it may not be the case.
    – Mark
    Nov 17 at 22:13
















hmm, now I've seen that this part you already did ...
– Robson
Nov 17 at 22:12




hmm, now I've seen that this part you already did ...
– Robson
Nov 17 at 22:12












Thanks for the answer. Yes, I did the same thing to find $C$. But the problem is that $A$ is not necessary in $sigma(P)$. If it is then we take $A=B$ and that's it. But it may not be the case.
– Mark
Nov 17 at 22:13




Thanks for the answer. Yes, I did the same thing to find $C$. But the problem is that $A$ is not necessary in $sigma(P)$. If it is then we take $A=B$ and that's it. But it may not be the case.
– Mark
Nov 17 at 22:13


















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