Extension of measures by adding one extra set
Suppose that $mu$ is a measure on a measurable set $(X,mathcal{B})$.
And suppose that $Asubseteq X$ is not in $mathcal{B}$.
Is there a measure $mu'$ on the $sigma$-algebra generated by $mathcal{B}cup {A}$ that coincide with $mu$ on $mathcal{B}$ and is such that $mu'(A)=0$ ?
measure-theory
asked Nov 19 '18 at 1:33
Phil-W
28417
add a comment |
Suppose that $mu$ is a measure on a measurable set $(X,mathcal{B})$.
And suppose that $Asubseteq X$ is not in $mathcal{B}$.
Is there a measure $mu'$ on the $sigma$-algebra generated by $mathcal{B}cup {A}$ that coincide with $mu$ on $mathcal{B}$ and is such that $mu'(A)=0$ ?
measure-theory
asked Nov 19 '18 at 1:33
Phil-W
28417
See here: math.stackexchange.com/questions/1181869/…
– Robson
Nov 19 '18 at 1:46
1
Thanks for the link, but it seems the question there doesn't require the value zero on $A$. The answer there also doesn't show why the "K" must exist.
– Phil-W
Nov 19 '18 at 1:55
It is possible if $inf{mu(B):Asubset B, Binmathcal{B}}=0$.
– d.k.o.
Nov 19 '18 at 1:57
add a comment |
Suppose that $mu$ is a measure on a measurable set $(X,mathcal{B})$.
And suppose that $Asubseteq X$ is not in $mathcal{B}$.
Is there a measure $mu'$ on the $sigma$-algebra generated by $mathcal{B}cup {A}$ that coincide with $mu$ on $mathcal{B}$ and is such that $mu'(A)=0$ ?
measure-theory
asked Nov 19 '18 at 1:33
Phil-W
28417
Suppose that $mu$ is a measure on a measurable set $(X,mathcal{B})$.
And suppose that $Asubseteq X$ is not in $mathcal{B}$.
Is there a measure $mu'$ on the $sigma$-algebra generated by $mathcal{B}cup {A}$ that coincide with $mu$ on $mathcal{B}$ and is such that $mu'(A)=0$ ?
measure-theory
measure-theory
asked Nov 19 '18 at 1:33
Phil-W
28417
asked Nov 19 '18 at 1:33
Phil-W
28417
asked Nov 19 '18 at 1:33
Phil-W
28417
asked Nov 19 '18 at 1:33
Phil-W
28417
asked Nov 19 '18 at 1:33
Phil-W
28417
28417
See here: math.stackexchange.com/questions/1181869/…
– Robson
Nov 19 '18 at 1:46
1
Thanks for the link, but it seems the question there doesn't require the value zero on $A$. The answer there also doesn't show why the "K" must exist.
– Phil-W
Nov 19 '18 at 1:55
It is possible if $inf{mu(B):Asubset B, Binmathcal{B}}=0$.
– d.k.o.
Nov 19 '18 at 1:57
add a comment |
See here: math.stackexchange.com/questions/1181869/…
– Robson
Nov 19 '18 at 1:46
1
Thanks for the link, but it seems the question there doesn't require the value zero on $A$. The answer there also doesn't show why the "K" must exist.
– Phil-W
Nov 19 '18 at 1:55
It is possible if $inf{mu(B):Asubset B, Binmathcal{B}}=0$.
– d.k.o.
Nov 19 '18 at 1:57
See here: math.stackexchange.com/questions/1181869/…
– Robson
Nov 19 '18 at 1:46
See here: math.stackexchange.com/questions/1181869/…
– Robson
Nov 19 '18 at 1:46
1
1
Thanks for the link, but it seems the question there doesn't require the value zero on $A$. The answer there also doesn't show why the "K" must exist.
– Phil-W
Nov 19 '18 at 1:55
Thanks for the link, but it seems the question there doesn't require the value zero on $A$. The answer there also doesn't show why the "K" must exist.
– Phil-W
Nov 19 '18 at 1:55
It is possible if $inf{mu(B):Asubset B, Binmathcal{B}}=0$.
– d.k.o.
Nov 19 '18 at 1:57
It is possible if $inf{mu(B):Asubset B, Binmathcal{B}}=0$.
– d.k.o.
Nov 19 '18 at 1:57
add a comment |
1 Answer
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Borrowing from the above link:
Let $a := sup{mu(B) : B subseteq A, B in mathcal{B}}$. There exists $K$ satisfying $K subseteq A$ and $K in mathcal{B}$ and $mu(K) = a$,
(namely, $K := bigcup_{B in mathcal{B} : B subseteq A} B)$.
Note that the new $sigma$-algebra can be expressed as ${(B_1 cap A) cup (B_2 cap A^c) : B_1, B_2 in mathcal{B}}$.
We define $mu'((B_1 cap A) cup (B_2 cap A^c)) := mu(B_1 cap K) + mu(B_2 cap K^c)$. One can check that $mu'$ is a measure, that $mu'(B) = mu(B)$ for $B in mathcal{B}$, and that $mu'(A) = mu(K) = a$. So if $a$ (as defined above) is zero (which I believe is equivalent to the condition in d.k.o.'s comment above), then this construction gives you what you want.
answered Nov 19 '18 at 3:58
angryavian
39k23180
add a comment |
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Borrowing from the above link:
Let $a := sup{mu(B) : B subseteq A, B in mathcal{B}}$. There exists $K$ satisfying $K subseteq A$ and $K in mathcal{B}$ and $mu(K) = a$,
(namely, $K := bigcup_{B in mathcal{B} : B subseteq A} B)$.
Note that the new $sigma$-algebra can be expressed as ${(B_1 cap A) cup (B_2 cap A^c) : B_1, B_2 in mathcal{B}}$.
We define $mu'((B_1 cap A) cup (B_2 cap A^c)) := mu(B_1 cap K) + mu(B_2 cap K^c)$. One can check that $mu'$ is a measure, that $mu'(B) = mu(B)$ for $B in mathcal{B}$, and that $mu'(A) = mu(K) = a$. So if $a$ (as defined above) is zero (which I believe is equivalent to the condition in d.k.o.'s comment above), then this construction gives you what you want.
answered Nov 19 '18 at 3:58
angryavian
39k23180
add a comment |
Borrowing from the above link:
Let $a := sup{mu(B) : B subseteq A, B in mathcal{B}}$. There exists $K$ satisfying $K subseteq A$ and $K in mathcal{B}$ and $mu(K) = a$,
(namely, $K := bigcup_{B in mathcal{B} : B subseteq A} B)$.
Note that the new $sigma$-algebra can be expressed as ${(B_1 cap A) cup (B_2 cap A^c) : B_1, B_2 in mathcal{B}}$.
We define $mu'((B_1 cap A) cup (B_2 cap A^c)) := mu(B_1 cap K) + mu(B_2 cap K^c)$. One can check that $mu'$ is a measure, that $mu'(B) = mu(B)$ for $B in mathcal{B}$, and that $mu'(A) = mu(K) = a$. So if $a$ (as defined above) is zero (which I believe is equivalent to the condition in d.k.o.'s comment above), then this construction gives you what you want.
answered Nov 19 '18 at 3:58
angryavian
39k23180
add a comment |
Borrowing from the above link:
Let $a := sup{mu(B) : B subseteq A, B in mathcal{B}}$. There exists $K$ satisfying $K subseteq A$ and $K in mathcal{B}$ and $mu(K) = a$,
(namely, $K := bigcup_{B in mathcal{B} : B subseteq A} B)$.
Note that the new $sigma$-algebra can be expressed as ${(B_1 cap A) cup (B_2 cap A^c) : B_1, B_2 in mathcal{B}}$.
We define $mu'((B_1 cap A) cup (B_2 cap A^c)) := mu(B_1 cap K) + mu(B_2 cap K^c)$. One can check that $mu'$ is a measure, that $mu'(B) = mu(B)$ for $B in mathcal{B}$, and that $mu'(A) = mu(K) = a$. So if $a$ (as defined above) is zero (which I believe is equivalent to the condition in d.k.o.'s comment above), then this construction gives you what you want.
answered Nov 19 '18 at 3:58
angryavian
39k23180
Borrowing from the above link:
Let $a := sup{mu(B) : B subseteq A, B in mathcal{B}}$. There exists $K$ satisfying $K subseteq A$ and $K in mathcal{B}$ and $mu(K) = a$,
(namely, $K := bigcup_{B in mathcal{B} : B subseteq A} B)$.
Note that the new $sigma$-algebra can be expressed as ${(B_1 cap A) cup (B_2 cap A^c) : B_1, B_2 in mathcal{B}}$.
We define $mu'((B_1 cap A) cup (B_2 cap A^c)) := mu(B_1 cap K) + mu(B_2 cap K^c)$. One can check that $mu'$ is a measure, that $mu'(B) = mu(B)$ for $B in mathcal{B}$, and that $mu'(A) = mu(K) = a$. So if $a$ (as defined above) is zero (which I believe is equivalent to the condition in d.k.o.'s comment above), then this construction gives you what you want.
answered Nov 19 '18 at 3:58
angryavian
39k23180
answered Nov 19 '18 at 3:58
angryavian
39k23180
answered Nov 19 '18 at 3:58
angryavian
39k23180
answered Nov 19 '18 at 3:58
angryavian
39k23180
39k23180
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See here: math.stackexchange.com/questions/1181869/…
– Robson
Nov 19 '18 at 1:46
1
Thanks for the link, but it seems the question there doesn't require the value zero on $A$. The answer there also doesn't show why the "K" must exist.
– Phil-W
Nov 19 '18 at 1:55
It is possible if $inf{mu(B):Asubset B, Binmathcal{B}}=0$.
– d.k.o.
Nov 19 '18 at 1:57