Mixing
Proving Countable Additivity From Finite Additivity
Clash Royale CLAN TAG#URR8PPP
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$J$ is a semialgebra of subsets of a sample space $Omega$ and $mathbbP : J to [0,1]$, with $mathbbP (Omega)=1.$
If $D_j$ is a finite collection of disjoint elements of $J$, such that $bigcup_ j leq n D_j in J$, then $$mathbbP left (bigcup_j leq n D_j right)=sum_ j leq nmathbbP(D_j). qquad (1)$$
Further, for any collection $A_n$, where the $A_i$ are all finite unions of elements of $J$, such that $A_n+1 subseteq A_n$ and $bigcap_n A_n=emptyset$, then $$lim_n to infty mathbbP(A_n)=0.$$
Show that $mathbbP$ also satisfies (1) for countable collections of disjoint sets $D_n$.
[Edit: question removed as counterexample was not applicable, as pointed out by Eric in comments.]
I am also wondering if there is any more general significance to the conditions placed on the collection $A_n$?
I have broken the proof into two parts:
Lemma 1: $mathbbP$ is countably subadditive: $$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
Proof:
Note that we assume $$ bigcup_jgeq n D_j in J,$$ for all $ngeq 1$.
If we let $$A_n= Big( bigcup_j D_j Big) setminus Big(bigcup_ j leq n D_j Big),$$ then it is not hard to show that the collection $A_n$ satisfies the given criteria to ensure that $$lim_n to infty mathbbP(A_n)=0.$$
We also have that $$A_n = bigcup_j >n D_j.$$
Putting these together, for any $epsilon>0$ we have:
$$ mathbbP left(bigcup_ j leq n D_j right)+mathbbP(A_n)= mathbbP left( bigcup_j D_j right) leq sum_ j leq n mathbbP (D_j) + epsilon quad (ngeq N),$$
where $N$ is some fixed integer. Thus, we can conclude
$$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
$square$
Lemma 2: $mathbbP$ is countably superadditive: $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
Proof:
Let $A,B in J$ such that $A subseteq B$. Since $J$ is a semialgebra, we can write $$A^c=E_1 dot cup E_2 dot cup cdots dot cup E_k,$$
for disjoint $E_1, E_2, cdots, E_k in J$.
By the finite additivity property of $mathbbP$:
$$mathbbP (B) = mathbbP (A) + mathbbP (B cap E_1) + cdots + mathbbP (B cap E_k) geq mathbbP (A).$$
So, for any positive integer $n$, we have
$$mathbbPleft( bigcup_j D_j right ) geq mathbbP left( bigcup_jleq n D_j right) = sum_j leq n mathbbP (D_j).$$
Since this result holds for all $nin mathbbN$, we conclude that $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
$square$
sequences-and-series limits probability-theory measure-theory proof-verification
asked Jul 21 at 3:28
Moed Pol Bollo
19318
add a comment |Â
up vote
0
down vote
favorite
$J$ is a semialgebra of subsets of a sample space $Omega$ and $mathbbP : J to [0,1]$, with $mathbbP (Omega)=1.$
If $D_j$ is a finite collection of disjoint elements of $J$, such that $bigcup_ j leq n D_j in J$, then $$mathbbP left (bigcup_j leq n D_j right)=sum_ j leq nmathbbP(D_j). qquad (1)$$
Further, for any collection $A_n$, where the $A_i$ are all finite unions of elements of $J$, such that $A_n+1 subseteq A_n$ and $bigcap_n A_n=emptyset$, then $$lim_n to infty mathbbP(A_n)=0.$$
Show that $mathbbP$ also satisfies (1) for countable collections of disjoint sets $D_n$.
[Edit: question removed as counterexample was not applicable, as pointed out by Eric in comments.]
I am also wondering if there is any more general significance to the conditions placed on the collection $A_n$?
I have broken the proof into two parts:
Lemma 1: $mathbbP$ is countably subadditive: $$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
Proof:
Note that we assume $$ bigcup_jgeq n D_j in J,$$ for all $ngeq 1$.
If we let $$A_n= Big( bigcup_j D_j Big) setminus Big(bigcup_ j leq n D_j Big),$$ then it is not hard to show that the collection $A_n$ satisfies the given criteria to ensure that $$lim_n to infty mathbbP(A_n)=0.$$
We also have that $$A_n = bigcup_j >n D_j.$$
Putting these together, for any $epsilon>0$ we have:
$$ mathbbP left(bigcup_ j leq n D_j right)+mathbbP(A_n)= mathbbP left( bigcup_j D_j right) leq sum_ j leq n mathbbP (D_j) + epsilon quad (ngeq N),$$
where $N$ is some fixed integer. Thus, we can conclude
$$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
$square$
Lemma 2: $mathbbP$ is countably superadditive: $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
Proof:
Let $A,B in J$ such that $A subseteq B$. Since $J$ is a semialgebra, we can write $$A^c=E_1 dot cup E_2 dot cup cdots dot cup E_k,$$
for disjoint $E_1, E_2, cdots, E_k in J$.
By the finite additivity property of $mathbbP$:
$$mathbbP (B) = mathbbP (A) + mathbbP (B cap E_1) + cdots + mathbbP (B cap E_k) geq mathbbP (A).$$
So, for any positive integer $n$, we have
$$mathbbPleft( bigcup_j D_j right ) geq mathbbP left( bigcup_jleq n D_j right) = sum_j leq n mathbbP (D_j).$$
Since this result holds for all $nin mathbbN$, we conclude that $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
$square$
sequences-and-series limits probability-theory measure-theory proof-verification
asked Jul 21 at 3:28
Moed Pol Bollo
19318
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$J$ is a semialgebra of subsets of a sample space $Omega$ and $mathbbP : J to [0,1]$, with $mathbbP (Omega)=1.$
If $D_j$ is a finite collection of disjoint elements of $J$, such that $bigcup_ j leq n D_j in J$, then $$mathbbP left (bigcup_j leq n D_j right)=sum_ j leq nmathbbP(D_j). qquad (1)$$
Further, for any collection $A_n$, where the $A_i$ are all finite unions of elements of $J$, such that $A_n+1 subseteq A_n$ and $bigcap_n A_n=emptyset$, then $$lim_n to infty mathbbP(A_n)=0.$$
Show that $mathbbP$ also satisfies (1) for countable collections of disjoint sets $D_n$.
[Edit: question removed as counterexample was not applicable, as pointed out by Eric in comments.]
I am also wondering if there is any more general significance to the conditions placed on the collection $A_n$?
I have broken the proof into two parts:
Lemma 1: $mathbbP$ is countably subadditive: $$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
Proof:
Note that we assume $$ bigcup_jgeq n D_j in J,$$ for all $ngeq 1$.
If we let $$A_n= Big( bigcup_j D_j Big) setminus Big(bigcup_ j leq n D_j Big),$$ then it is not hard to show that the collection $A_n$ satisfies the given criteria to ensure that $$lim_n to infty mathbbP(A_n)=0.$$
We also have that $$A_n = bigcup_j >n D_j.$$
Putting these together, for any $epsilon>0$ we have:
$$ mathbbP left(bigcup_ j leq n D_j right)+mathbbP(A_n)= mathbbP left( bigcup_j D_j right) leq sum_ j leq n mathbbP (D_j) + epsilon quad (ngeq N),$$
where $N$ is some fixed integer. Thus, we can conclude
$$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
$square$
Lemma 2: $mathbbP$ is countably superadditive: $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
Proof:
Let $A,B in J$ such that $A subseteq B$. Since $J$ is a semialgebra, we can write $$A^c=E_1 dot cup E_2 dot cup cdots dot cup E_k,$$
for disjoint $E_1, E_2, cdots, E_k in J$.
By the finite additivity property of $mathbbP$:
$$mathbbP (B) = mathbbP (A) + mathbbP (B cap E_1) + cdots + mathbbP (B cap E_k) geq mathbbP (A).$$
So, for any positive integer $n$, we have
$$mathbbPleft( bigcup_j D_j right ) geq mathbbP left( bigcup_jleq n D_j right) = sum_j leq n mathbbP (D_j).$$
Since this result holds for all $nin mathbbN$, we conclude that $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
$square$
sequences-and-series limits probability-theory measure-theory proof-verification
asked Jul 21 at 3:28
Moed Pol Bollo
19318
$J$ is a semialgebra of subsets of a sample space $Omega$ and $mathbbP : J to [0,1]$, with $mathbbP (Omega)=1.$
If $D_j$ is a finite collection of disjoint elements of $J$, such that $bigcup_ j leq n D_j in J$, then $$mathbbP left (bigcup_j leq n D_j right)=sum_ j leq nmathbbP(D_j). qquad (1)$$
Further, for any collection $A_n$, where the $A_i$ are all finite unions of elements of $J$, such that $A_n+1 subseteq A_n$ and $bigcap_n A_n=emptyset$, then $$lim_n to infty mathbbP(A_n)=0.$$
Show that $mathbbP$ also satisfies (1) for countable collections of disjoint sets $D_n$.
[Edit: question removed as counterexample was not applicable, as pointed out by Eric in comments.]
I am also wondering if there is any more general significance to the conditions placed on the collection $A_n$?
I have broken the proof into two parts:
Lemma 1: $mathbbP$ is countably subadditive: $$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
Proof:
Note that we assume $$ bigcup_jgeq n D_j in J,$$ for all $ngeq 1$.
If we let $$A_n= Big( bigcup_j D_j Big) setminus Big(bigcup_ j leq n D_j Big),$$ then it is not hard to show that the collection $A_n$ satisfies the given criteria to ensure that $$lim_n to infty mathbbP(A_n)=0.$$
We also have that $$A_n = bigcup_j >n D_j.$$
Putting these together, for any $epsilon>0$ we have:
$$ mathbbP left(bigcup_ j leq n D_j right)+mathbbP(A_n)= mathbbP left( bigcup_j D_j right) leq sum_ j leq n mathbbP (D_j) + epsilon quad (ngeq N),$$
where $N$ is some fixed integer. Thus, we can conclude
$$mathbbP left( bigcup_j D_j right) leq sum_j mathbbP (D_j).$$
$square$
Lemma 2: $mathbbP$ is countably superadditive: $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
Proof:
Let $A,B in J$ such that $A subseteq B$. Since $J$ is a semialgebra, we can write $$A^c=E_1 dot cup E_2 dot cup cdots dot cup E_k,$$
for disjoint $E_1, E_2, cdots, E_k in J$.
By the finite additivity property of $mathbbP$:
$$mathbbP (B) = mathbbP (A) + mathbbP (B cap E_1) + cdots + mathbbP (B cap E_k) geq mathbbP (A).$$
So, for any positive integer $n$, we have
$$mathbbPleft( bigcup_j D_j right ) geq mathbbP left( bigcup_jleq n D_j right) = sum_j leq n mathbbP (D_j).$$
Since this result holds for all $nin mathbbN$, we conclude that $$mathbbP left( bigcup_j D_j right) geq sum_j mathbbP (D_j).$$
$square$
sequences-and-series limits probability-theory measure-theory proof-verification
asked Jul 21 at 3:28
Moed Pol Bollo
19318
edited Jul 21 at 7:01
asked Jul 21 at 3:28
Moed Pol Bollo
19318
asked Jul 21 at 3:28
Moed Pol Bollo
19318
asked Jul 21 at 3:28
Moed Pol Bollo
19318
19318
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The statement you have proved is not the same as the statement in the question you linked! In your statement, the $A_n$ are required only to be finite unions of elements of $J$. In the other question, they are required to be elements of $J$. George's counterexample does not satisfy your stronger hypotheses.
Your proof appears to me to be correct.
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply.
– Moed Pol Bollo
Jul 21 at 4:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The statement you have proved is not the same as the statement in the question you linked! In your statement, the $A_n$ are required only to be finite unions of elements of $J$. In the other question, they are required to be elements of $J$. George's counterexample does not satisfy your stronger hypotheses.
Your proof appears to me to be correct.
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply.
– Moed Pol Bollo
Jul 21 at 4:35
add a comment |Â
up vote
1
down vote
accepted
The statement you have proved is not the same as the statement in the question you linked! In your statement, the $A_n$ are required only to be finite unions of elements of $J$. In the other question, they are required to be elements of $J$. George's counterexample does not satisfy your stronger hypotheses.
Your proof appears to me to be correct.
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply.
– Moed Pol Bollo
Jul 21 at 4:35
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The statement you have proved is not the same as the statement in the question you linked! In your statement, the $A_n$ are required only to be finite unions of elements of $J$. In the other question, they are required to be elements of $J$. George's counterexample does not satisfy your stronger hypotheses.
Your proof appears to me to be correct.
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
The statement you have proved is not the same as the statement in the question you linked! In your statement, the $A_n$ are required only to be finite unions of elements of $J$. In the other question, they are required to be elements of $J$. George's counterexample does not satisfy your stronger hypotheses.
Your proof appears to me to be correct.
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
answered Jul 21 at 4:18
Eric Wofsey
162k12189300
162k12189300
Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply.
– Moed Pol Bollo
Jul 21 at 4:35
add a comment |Â
Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply.
– Moed Pol Bollo
Jul 21 at 4:35
Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply.
– Moed Pol Bollo
Jul 21 at 4:35
Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply.
– Moed Pol Bollo
Jul 21 at 4:35
add a comment |Â
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