Is this measure on extension of $sigma$-algebra well defined?
$begingroup$
Let $(X, xi, mu)$ be a measure space and define:
$$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:
$$barmu(A) := mu(B)$$
Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:
$$A ,triangle, B_1 subseteq C_1$$
$$A ,triangle, B_2 subseteq C_2$$
but I'm stuck showing $mu(B_1) = mu(B_2)$.
measure-theory
asked Jan 7 at 14:52
user7802048user7802048
382211
$endgroup$
add a comment |
$begingroup$
Let $(X, xi, mu)$ be a measure space and define:
$$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:
$$barmu(A) := mu(B)$$
Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:
$$A ,triangle, B_1 subseteq C_1$$
$$A ,triangle, B_2 subseteq C_2$$
but I'm stuck showing $mu(B_1) = mu(B_2)$.
measure-theory
asked Jan 7 at 14:52
user7802048user7802048
382211
$endgroup$
add a comment |
$begingroup$
Let $(X, xi, mu)$ be a measure space and define:
$$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:
$$barmu(A) := mu(B)$$
Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:
$$A ,triangle, B_1 subseteq C_1$$
$$A ,triangle, B_2 subseteq C_2$$
but I'm stuck showing $mu(B_1) = mu(B_2)$.
measure-theory
asked Jan 7 at 14:52
user7802048user7802048
382211
$endgroup$
Let $(X, xi, mu)$ be a measure space and define:
$$xi_mu := {A in mathcal{P} : exists B, C in xi: mu(C) = 0 text{ and } A ,triangle, Bsubseteq C}$$
I already showed that $xi_mu$ is a $sigma$-algebra and $xi subseteq xi_mu$. For $A in xi_mu$ with $B, C in xi$ as in the definition we define:
$$barmu(A) := mu(B)$$
Now I'll need to prove that this is even well defined. Let's consider two $B_1, B_2, C_1, C_2$ with $mu(C_1), mu(C_2) = 0$ and:
$$A ,triangle, B_1 subseteq C_1$$
$$A ,triangle, B_2 subseteq C_2$$
but I'm stuck showing $mu(B_1) = mu(B_2)$.
measure-theory
measure-theory
asked Jan 7 at 14:52
user7802048user7802048
382211
asked Jan 7 at 14:52
user7802048user7802048
382211
asked Jan 7 at 14:52
user7802048user7802048
382211
asked Jan 7 at 14:52
user7802048user7802048
382211
asked Jan 7 at 14:52
user7802048user7802048
382211
382211
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint: First prove that
$$
B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
$$
answered Jan 7 at 15:11
SongSong
13.6k633
$endgroup$
$begingroup$
Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
$endgroup$
– user7802048
Jan 7 at 15:48
$begingroup$
Sure :) I'm glad it helped.
$endgroup$
– Song
Jan 7 at 15:49
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: First prove that
$$
B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
$$
answered Jan 7 at 15:11
SongSong
13.6k633
$endgroup$
$begingroup$
Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
$endgroup$
– user7802048
Jan 7 at 15:48
$begingroup$
Sure :) I'm glad it helped.
$endgroup$
– Song
Jan 7 at 15:49
add a comment |
$begingroup$
Hint: First prove that
$$
B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
$$
answered Jan 7 at 15:11
SongSong
13.6k633
$endgroup$
$begingroup$
Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
$endgroup$
– user7802048
Jan 7 at 15:48
$begingroup$
Sure :) I'm glad it helped.
$endgroup$
– Song
Jan 7 at 15:49
add a comment |
$begingroup$
Hint: First prove that
$$
B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
$$
answered Jan 7 at 15:11
SongSong
13.6k633
$endgroup$
Hint: First prove that
$$
B_1triangle B_2subset (Atriangle B_1)cup(Atriangle B_2)subset C_1cup C_2.
$$
answered Jan 7 at 15:11
SongSong
13.6k633
answered Jan 7 at 15:11
SongSong
13.6k633
answered Jan 7 at 15:11
SongSong
13.6k633
answered Jan 7 at 15:11
SongSong
13.6k633
13.6k633
$begingroup$
Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
$endgroup$
– user7802048
Jan 7 at 15:48
$begingroup$
Sure :) I'm glad it helped.
$endgroup$
– Song
Jan 7 at 15:49
add a comment |
$begingroup$
Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
$endgroup$
– user7802048
Jan 7 at 15:48
$begingroup$
Sure :) I'm glad it helped.
$endgroup$
– Song
Jan 7 at 15:49
$begingroup$
Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
$endgroup$
– user7802048
Jan 7 at 15:48
$begingroup$
Perfect. I concluded $mu(B_1) = mu(B_1 cup B_2) - mu(B_2 setminus B_1) = mu(B_1 cup B_2) - mu(B_1 setminus B_2) = mu(B_2)$. Your hint proves the second equality, right?
$endgroup$
– user7802048
Jan 7 at 15:48
$begingroup$
Sure :) I'm glad it helped.
$endgroup$
– Song
Jan 7 at 15:49
$begingroup$
Sure :) I'm glad it helped.
$endgroup$
– Song
Jan 7 at 15:49
add a comment |
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