Completion of a $sigma$-algebra is a $sigma$-algebra
$begingroup$
Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
begin{align*}
mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
Prove that $mathcal{G}$ is a $sigma$-algebra.
I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?
I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.
probability-theory measure-theory elementary-set-theory
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
begin{align*}
mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
Prove that $mathcal{G}$ is a $sigma$-algebra.
I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?
I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.
probability-theory measure-theory elementary-set-theory
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
begin{align*}
mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
Prove that $mathcal{G}$ is a $sigma$-algebra.
I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?
I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.
probability-theory measure-theory elementary-set-theory
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
$endgroup$
Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
begin{align*}
mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
Prove that $mathcal{G}$ is a $sigma$-algebra.
I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?
I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.
probability-theory measure-theory elementary-set-theory
probability-theory measure-theory elementary-set-theory
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
edited Jan 22 at 12:35
Davide Giraudo
128k17156268
128k17156268
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
asked Jan 18 at 17:42
J. DoeJ. Doe
16613
16613
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,
$$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$
As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that
$$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$
here we have used the general fact that
$$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that
$$G^c = tilde{A} cup tilde{N} tag{2}$$
for
$$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$
As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.
answered Jan 21 at 18:51
sazsaz
82.7k863132
$endgroup$
$begingroup$
Why is $(1)$ true? @saz
$endgroup$
– J. Doe
Jan 22 at 10:26
1
$begingroup$
@J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
$endgroup$
– saz
Jan 23 at 7:10
add a comment |
$begingroup$
Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.
Then
$$
begin{aligned}
A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
end{aligned}
$$
so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
$$
0le
Bbb P(A^c-(A^ccap F^c))
=
Bbb P(A^ccap F^{cc})
=
Bbb P(A^ccap F)
le
Bbb P(F)
=0
.
$$
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
$endgroup$
$begingroup$
I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:07
1
$begingroup$
$G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
$endgroup$
– dan_fulea
Jan 18 at 18:11
$begingroup$
Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:23
$begingroup$
$G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
$endgroup$
– dan_fulea
Jan 18 at 18:27
$begingroup$
We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
$endgroup$
– J. Doe
Jan 19 at 9:59
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,
$$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$
As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that
$$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$
here we have used the general fact that
$$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that
$$G^c = tilde{A} cup tilde{N} tag{2}$$
for
$$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$
As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.
answered Jan 21 at 18:51
sazsaz
82.7k863132
$endgroup$
$begingroup$
Why is $(1)$ true? @saz
$endgroup$
– J. Doe
Jan 22 at 10:26
1
$begingroup$
@J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
$endgroup$
– saz
Jan 23 at 7:10
add a comment |
$begingroup$
Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,
$$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$
As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that
$$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$
here we have used the general fact that
$$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that
$$G^c = tilde{A} cup tilde{N} tag{2}$$
for
$$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$
As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.
answered Jan 21 at 18:51
sazsaz
82.7k863132
$endgroup$
$begingroup$
Why is $(1)$ true? @saz
$endgroup$
– J. Doe
Jan 22 at 10:26
1
$begingroup$
@J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
$endgroup$
– saz
Jan 23 at 7:10
add a comment |
$begingroup$
Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,
$$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$
As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that
$$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$
here we have used the general fact that
$$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that
$$G^c = tilde{A} cup tilde{N} tag{2}$$
for
$$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$
As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.
answered Jan 21 at 18:51
sazsaz
82.7k863132
$endgroup$
Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,
$$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$
As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that
$$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$
here we have used the general fact that
$$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that
$$G^c = tilde{A} cup tilde{N} tag{2}$$
for
$$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$
As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.
answered Jan 21 at 18:51
sazsaz
82.7k863132
edited Jan 23 at 7:53
answered Jan 21 at 18:51
sazsaz
82.7k863132
answered Jan 21 at 18:51
sazsaz
82.7k863132
answered Jan 21 at 18:51
sazsaz
82.7k863132
82.7k863132
$begingroup$
Why is $(1)$ true? @saz
$endgroup$
– J. Doe
Jan 22 at 10:26
1
$begingroup$
@J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
$endgroup$
– saz
Jan 23 at 7:10
add a comment |
$begingroup$
Why is $(1)$ true? @saz
$endgroup$
– J. Doe
Jan 22 at 10:26
1
$begingroup$
@J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
$endgroup$
– saz
Jan 23 at 7:10
$begingroup$
Why is $(1)$ true? @saz
$endgroup$
– J. Doe
Jan 22 at 10:26
$begingroup$
Why is $(1)$ true? @saz
$endgroup$
– J. Doe
Jan 22 at 10:26
1
1
$begingroup$
@J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
$endgroup$
– saz
Jan 23 at 7:10
$begingroup$
@J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
$endgroup$
– saz
Jan 23 at 7:10
add a comment |
$begingroup$
Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.
Then
$$
begin{aligned}
A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
end{aligned}
$$
so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
$$
0le
Bbb P(A^c-(A^ccap F^c))
=
Bbb P(A^ccap F^{cc})
=
Bbb P(A^ccap F)
le
Bbb P(F)
=0
.
$$
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
$endgroup$
$begingroup$
I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:07
1
$begingroup$
$G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
$endgroup$
– dan_fulea
Jan 18 at 18:11
$begingroup$
Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:23
$begingroup$
$G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
$endgroup$
– dan_fulea
Jan 18 at 18:27
$begingroup$
We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
$endgroup$
– J. Doe
Jan 19 at 9:59
add a comment |
$begingroup$
Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.
Then
$$
begin{aligned}
A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
end{aligned}
$$
so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
$$
0le
Bbb P(A^c-(A^ccap F^c))
=
Bbb P(A^ccap F^{cc})
=
Bbb P(A^ccap F)
le
Bbb P(F)
=0
.
$$
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
$endgroup$
$begingroup$
I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:07
1
$begingroup$
$G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
$endgroup$
– dan_fulea
Jan 18 at 18:11
$begingroup$
Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:23
$begingroup$
$G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
$endgroup$
– dan_fulea
Jan 18 at 18:27
$begingroup$
We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
$endgroup$
– J. Doe
Jan 19 at 9:59
add a comment |
$begingroup$
Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.
Then
$$
begin{aligned}
A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
end{aligned}
$$
so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
$$
0le
Bbb P(A^c-(A^ccap F^c))
=
Bbb P(A^ccap F^{cc})
=
Bbb P(A^ccap F)
le
Bbb P(F)
=0
.
$$
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
$endgroup$
Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.
Then
$$
begin{aligned}
A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
end{aligned}
$$
so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
$$
0le
Bbb P(A^c-(A^ccap F^c))
=
Bbb P(A^ccap F^{cc})
=
Bbb P(A^ccap F)
le
Bbb P(F)
=0
.
$$
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
answered Jan 18 at 17:54
dan_fuleadan_fulea
7,1981513
7,1981513
$begingroup$
I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:07
1
$begingroup$
$G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
$endgroup$
– dan_fulea
Jan 18 at 18:11
$begingroup$
Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:23
$begingroup$
$G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
$endgroup$
– dan_fulea
Jan 18 at 18:27
$begingroup$
We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
$endgroup$
– J. Doe
Jan 19 at 9:59
add a comment |
$begingroup$
I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:07
1
$begingroup$
$G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
$endgroup$
– dan_fulea
Jan 18 at 18:11
$begingroup$
Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:23
$begingroup$
$G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
$endgroup$
– dan_fulea
Jan 18 at 18:27
$begingroup$
We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
$endgroup$
– J. Doe
Jan 19 at 9:59
$begingroup$
I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:07
$begingroup$
I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:07
1
1
$begingroup$
$G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
$endgroup$
– dan_fulea
Jan 18 at 18:11
$begingroup$
$G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
$endgroup$
– dan_fulea
Jan 18 at 18:11
$begingroup$
Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:23
$begingroup$
Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
$endgroup$
– J. Doe
Jan 18 at 18:23
$begingroup$
$G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
$endgroup$
– dan_fulea
Jan 18 at 18:27
$begingroup$
$G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
$endgroup$
– dan_fulea
Jan 18 at 18:27
$begingroup$
We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
$endgroup$
– J. Doe
Jan 19 at 9:59
$begingroup$
We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
$endgroup$
– J. Doe
Jan 19 at 9:59
add a comment |
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