Mixing
Let $(X, {cal E}, mu)$ be a measure space and let $B,C in {cal E}$. Show $mu_B + mu_C leq mu$.
$begingroup$
Let $(X, {cal E}, mu)$ be a measure space and let $B,C in {cal E}$. Define $mu_B:{cal E} to [0,infty]$ by $mu_B (E) = mu(Ecap B)$. (I have already proved that this does indeed define a new measure).
Find a necessary and sufficient condition on sets $B,C in mathcal{E}$ such that $mu_B + mu_C leq mu$.
Any ideas?
measure-theory
asked Jan 31 at 6:01
johnny133253johnny133253
470111
$endgroup$
add a comment |
$begingroup$
Let $(X, {cal E}, mu)$ be a measure space and let $B,C in {cal E}$. Define $mu_B:{cal E} to [0,infty]$ by $mu_B (E) = mu(Ecap B)$. (I have already proved that this does indeed define a new measure).
Find a necessary and sufficient condition on sets $B,C in mathcal{E}$ such that $mu_B + mu_C leq mu$.
Any ideas?
measure-theory
asked Jan 31 at 6:01
johnny133253johnny133253
470111
$endgroup$
$begingroup$
If $mathrm{B}$ and $mathrm{C}$ are disjoint then $mu_{mathrm{B} cup mathrm{C}} = mu_{mathrm{B}} + mu_{mathrm{C}}.$ This is a sufficient condition. Likely, it is also necessary but that needs more thought.
$endgroup$
– Will M.
Jan 31 at 6:04
$begingroup$
Thanks, I did find this condition but struggled to find a proof that shows it is necessary.
$endgroup$
– johnny133253
Jan 31 at 6:05
add a comment |
$begingroup$
Let $(X, {cal E}, mu)$ be a measure space and let $B,C in {cal E}$. Define $mu_B:{cal E} to [0,infty]$ by $mu_B (E) = mu(Ecap B)$. (I have already proved that this does indeed define a new measure).
Find a necessary and sufficient condition on sets $B,C in mathcal{E}$ such that $mu_B + mu_C leq mu$.
Any ideas?
measure-theory
asked Jan 31 at 6:01
johnny133253johnny133253
470111
$endgroup$
Let $(X, {cal E}, mu)$ be a measure space and let $B,C in {cal E}$. Define $mu_B:{cal E} to [0,infty]$ by $mu_B (E) = mu(Ecap B)$. (I have already proved that this does indeed define a new measure).
Find a necessary and sufficient condition on sets $B,C in mathcal{E}$ such that $mu_B + mu_C leq mu$.
Any ideas?
measure-theory
measure-theory
asked Jan 31 at 6:01
johnny133253johnny133253
470111
asked Jan 31 at 6:01
johnny133253johnny133253
470111
asked Jan 31 at 6:01
johnny133253johnny133253
470111
asked Jan 31 at 6:01
johnny133253johnny133253
470111
asked Jan 31 at 6:01
johnny133253johnny133253
470111
470111
$begingroup$
If $mathrm{B}$ and $mathrm{C}$ are disjoint then $mu_{mathrm{B} cup mathrm{C}} = mu_{mathrm{B}} + mu_{mathrm{C}}.$ This is a sufficient condition. Likely, it is also necessary but that needs more thought.
$endgroup$
– Will M.
Jan 31 at 6:04
$begingroup$
Thanks, I did find this condition but struggled to find a proof that shows it is necessary.
$endgroup$
– johnny133253
Jan 31 at 6:05
add a comment |
$begingroup$
If $mathrm{B}$ and $mathrm{C}$ are disjoint then $mu_{mathrm{B} cup mathrm{C}} = mu_{mathrm{B}} + mu_{mathrm{C}}.$ This is a sufficient condition. Likely, it is also necessary but that needs more thought.
$endgroup$
– Will M.
Jan 31 at 6:04
$begingroup$
Thanks, I did find this condition but struggled to find a proof that shows it is necessary.
$endgroup$
– johnny133253
Jan 31 at 6:05
$begingroup$
If $mathrm{B}$ and $mathrm{C}$ are disjoint then $mu_{mathrm{B} cup mathrm{C}} = mu_{mathrm{B}} + mu_{mathrm{C}}.$ This is a sufficient condition. Likely, it is also necessary but that needs more thought.
$endgroup$
– Will M.
Jan 31 at 6:04
$begingroup$
If $mathrm{B}$ and $mathrm{C}$ are disjoint then $mu_{mathrm{B} cup mathrm{C}} = mu_{mathrm{B}} + mu_{mathrm{C}}.$ This is a sufficient condition. Likely, it is also necessary but that needs more thought.
$endgroup$
– Will M.
Jan 31 at 6:04
$begingroup$
Thanks, I did find this condition but struggled to find a proof that shows it is necessary.
$endgroup$
– johnny133253
Jan 31 at 6:05
$begingroup$
Thanks, I did find this condition but struggled to find a proof that shows it is necessary.
$endgroup$
– johnny133253
Jan 31 at 6:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Claim. For $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ it is a necessary and sufficient condition that $mu(mathrm{B} cap mathrm{C}) = 0.$
Proof. If $mu(mathrm{B} cap mathrm{C}) = 0$ then $mu_{mathrm{B}} = mu_{mathrm{B} setminus mathrm{C}},$ and similarly interchanging the rôles of $mathrm{B}$ and $mathrm{C}.$ Hence, $mu_{mathrm{B}} + mu_{mathrm{C}} = mu_{(mathrm{B} setminus mathrm{C}) cup (mathrm{C} setminus mathrm{B})} leq mu_{mathrm{X}} = mu;$ the condition is therefore sufficient. It is also necessary for we can evaluate the inequality $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ in $mathrm{B} cap mathrm{C}$ to reach $0 leq 2 mu(mathrm{B} cap mathrm{C}) leq mu(mathrm{B} cap mathrm{C}),$ hence $mu(mathrm{B} cap mathrm{C}) = 0.$ Q.E.D.
answered Jan 31 at 6:11
Will M.Will M.
2,890315
$endgroup$
$begingroup$
How did you show $mu_B = mu_{Bsetminus C}$?
$endgroup$
– johnny133253
Jan 31 at 7:16
$begingroup$
Because $mu_mathrm{B}(mathrm{E}) = mu(mathrm{E} cap (mathrm{B} setminus mathrm{C})) + mu(mathrm{E} cap mathrm{B} cap mathrm{C}).$
$endgroup$
– Will M.
Jan 31 at 7:43
add a comment |
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1 Answer
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$begingroup$
Claim. For $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ it is a necessary and sufficient condition that $mu(mathrm{B} cap mathrm{C}) = 0.$
Proof. If $mu(mathrm{B} cap mathrm{C}) = 0$ then $mu_{mathrm{B}} = mu_{mathrm{B} setminus mathrm{C}},$ and similarly interchanging the rôles of $mathrm{B}$ and $mathrm{C}.$ Hence, $mu_{mathrm{B}} + mu_{mathrm{C}} = mu_{(mathrm{B} setminus mathrm{C}) cup (mathrm{C} setminus mathrm{B})} leq mu_{mathrm{X}} = mu;$ the condition is therefore sufficient. It is also necessary for we can evaluate the inequality $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ in $mathrm{B} cap mathrm{C}$ to reach $0 leq 2 mu(mathrm{B} cap mathrm{C}) leq mu(mathrm{B} cap mathrm{C}),$ hence $mu(mathrm{B} cap mathrm{C}) = 0.$ Q.E.D.
answered Jan 31 at 6:11
Will M.Will M.
2,890315
$endgroup$
$begingroup$
How did you show $mu_B = mu_{Bsetminus C}$?
$endgroup$
– johnny133253
Jan 31 at 7:16
$begingroup$
Because $mu_mathrm{B}(mathrm{E}) = mu(mathrm{E} cap (mathrm{B} setminus mathrm{C})) + mu(mathrm{E} cap mathrm{B} cap mathrm{C}).$
$endgroup$
– Will M.
Jan 31 at 7:43
add a comment |
$begingroup$
Claim. For $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ it is a necessary and sufficient condition that $mu(mathrm{B} cap mathrm{C}) = 0.$
Proof. If $mu(mathrm{B} cap mathrm{C}) = 0$ then $mu_{mathrm{B}} = mu_{mathrm{B} setminus mathrm{C}},$ and similarly interchanging the rôles of $mathrm{B}$ and $mathrm{C}.$ Hence, $mu_{mathrm{B}} + mu_{mathrm{C}} = mu_{(mathrm{B} setminus mathrm{C}) cup (mathrm{C} setminus mathrm{B})} leq mu_{mathrm{X}} = mu;$ the condition is therefore sufficient. It is also necessary for we can evaluate the inequality $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ in $mathrm{B} cap mathrm{C}$ to reach $0 leq 2 mu(mathrm{B} cap mathrm{C}) leq mu(mathrm{B} cap mathrm{C}),$ hence $mu(mathrm{B} cap mathrm{C}) = 0.$ Q.E.D.
answered Jan 31 at 6:11
Will M.Will M.
2,890315
$endgroup$
$begingroup$
How did you show $mu_B = mu_{Bsetminus C}$?
$endgroup$
– johnny133253
Jan 31 at 7:16
$begingroup$
Because $mu_mathrm{B}(mathrm{E}) = mu(mathrm{E} cap (mathrm{B} setminus mathrm{C})) + mu(mathrm{E} cap mathrm{B} cap mathrm{C}).$
$endgroup$
– Will M.
Jan 31 at 7:43
add a comment |
$begingroup$
Claim. For $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ it is a necessary and sufficient condition that $mu(mathrm{B} cap mathrm{C}) = 0.$
Proof. If $mu(mathrm{B} cap mathrm{C}) = 0$ then $mu_{mathrm{B}} = mu_{mathrm{B} setminus mathrm{C}},$ and similarly interchanging the rôles of $mathrm{B}$ and $mathrm{C}.$ Hence, $mu_{mathrm{B}} + mu_{mathrm{C}} = mu_{(mathrm{B} setminus mathrm{C}) cup (mathrm{C} setminus mathrm{B})} leq mu_{mathrm{X}} = mu;$ the condition is therefore sufficient. It is also necessary for we can evaluate the inequality $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ in $mathrm{B} cap mathrm{C}$ to reach $0 leq 2 mu(mathrm{B} cap mathrm{C}) leq mu(mathrm{B} cap mathrm{C}),$ hence $mu(mathrm{B} cap mathrm{C}) = 0.$ Q.E.D.
answered Jan 31 at 6:11
Will M.Will M.
2,890315
$endgroup$
Claim. For $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ it is a necessary and sufficient condition that $mu(mathrm{B} cap mathrm{C}) = 0.$
Proof. If $mu(mathrm{B} cap mathrm{C}) = 0$ then $mu_{mathrm{B}} = mu_{mathrm{B} setminus mathrm{C}},$ and similarly interchanging the rôles of $mathrm{B}$ and $mathrm{C}.$ Hence, $mu_{mathrm{B}} + mu_{mathrm{C}} = mu_{(mathrm{B} setminus mathrm{C}) cup (mathrm{C} setminus mathrm{B})} leq mu_{mathrm{X}} = mu;$ the condition is therefore sufficient. It is also necessary for we can evaluate the inequality $mu_{mathrm{B}} + mu_{mathrm{C}} leq mu$ in $mathrm{B} cap mathrm{C}$ to reach $0 leq 2 mu(mathrm{B} cap mathrm{C}) leq mu(mathrm{B} cap mathrm{C}),$ hence $mu(mathrm{B} cap mathrm{C}) = 0.$ Q.E.D.
answered Jan 31 at 6:11
Will M.Will M.
2,890315
answered Jan 31 at 6:11
Will M.Will M.
2,890315
answered Jan 31 at 6:11
Will M.Will M.
2,890315
answered Jan 31 at 6:11
Will M.Will M.
2,890315
2,890315
$begingroup$
How did you show $mu_B = mu_{Bsetminus C}$?
$endgroup$
– johnny133253
Jan 31 at 7:16
$begingroup$
Because $mu_mathrm{B}(mathrm{E}) = mu(mathrm{E} cap (mathrm{B} setminus mathrm{C})) + mu(mathrm{E} cap mathrm{B} cap mathrm{C}).$
$endgroup$
– Will M.
Jan 31 at 7:43
add a comment |
$begingroup$
How did you show $mu_B = mu_{Bsetminus C}$?
$endgroup$
– johnny133253
Jan 31 at 7:16
$begingroup$
Because $mu_mathrm{B}(mathrm{E}) = mu(mathrm{E} cap (mathrm{B} setminus mathrm{C})) + mu(mathrm{E} cap mathrm{B} cap mathrm{C}).$
$endgroup$
– Will M.
Jan 31 at 7:43
$begingroup$
How did you show $mu_B = mu_{Bsetminus C}$?
$endgroup$
– johnny133253
Jan 31 at 7:16
$begingroup$
How did you show $mu_B = mu_{Bsetminus C}$?
$endgroup$
– johnny133253
Jan 31 at 7:16
$begingroup$
Because $mu_mathrm{B}(mathrm{E}) = mu(mathrm{E} cap (mathrm{B} setminus mathrm{C})) + mu(mathrm{E} cap mathrm{B} cap mathrm{C}).$
$endgroup$
– Will M.
Jan 31 at 7:43
$begingroup$
Because $mu_mathrm{B}(mathrm{E}) = mu(mathrm{E} cap (mathrm{B} setminus mathrm{C})) + mu(mathrm{E} cap mathrm{B} cap mathrm{C}).$
$endgroup$
– Will M.
Jan 31 at 7:43
add a comment |
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$begingroup$
If $mathrm{B}$ and $mathrm{C}$ are disjoint then $mu_{mathrm{B} cup mathrm{C}} = mu_{mathrm{B}} + mu_{mathrm{C}}.$ This is a sufficient condition. Likely, it is also necessary but that needs more thought.
$endgroup$
– Will M.
Jan 31 at 6:04
$begingroup$
Thanks, I did find this condition but struggled to find a proof that shows it is necessary.
$endgroup$
– johnny133253
Jan 31 at 6:05