Mixing
sigma algebra of all 0 and 1 events
$begingroup$
Let $mathcal{F}$ be a sigma-algebra over a set $Omega$. Now let
$$
mathcal{G} := {E in mathcal{F} : mathbb{P}(E) = 0 ~text{or}~1}.
$$
I have to show that $mathcal{G}$ is a sigma algebra.
The only difficult is to show that if $E_{1},E_{2},cdots in mathcal{G}$ then $bigcup_{k geq 1}E_{k} in mathcal{G}$.
My first attempt was to use $mathbb{P}left(bigcup_{k geq 1}E_{k}right)$ $leq$ $sum_{kgeq 1}mathbb{P}(E_{k})$. Now I'm stuck.
Does anyone have any idea?
Thanks!
probability measure-theory
edited Nov 26 '18 at 16:00
Shaun
9,083113683
asked Nov 26 '18 at 15:58
gregorgregor
1617
$endgroup$
add a comment |
$begingroup$
Let $mathcal{F}$ be a sigma-algebra over a set $Omega$. Now let
$$
mathcal{G} := {E in mathcal{F} : mathbb{P}(E) = 0 ~text{or}~1}.
$$
I have to show that $mathcal{G}$ is a sigma algebra.
The only difficult is to show that if $E_{1},E_{2},cdots in mathcal{G}$ then $bigcup_{k geq 1}E_{k} in mathcal{G}$.
My first attempt was to use $mathbb{P}left(bigcup_{k geq 1}E_{k}right)$ $leq$ $sum_{kgeq 1}mathbb{P}(E_{k})$. Now I'm stuck.
Does anyone have any idea?
Thanks!
probability measure-theory
edited Nov 26 '18 at 16:00
Shaun
9,083113683
asked Nov 26 '18 at 15:58
gregorgregor
1617
$endgroup$
$begingroup$
Maybe you could split the union into two parts, zero measure and full measure events.
$endgroup$
– Calculon
Nov 26 '18 at 16:04
$begingroup$
It depends on whether $mathbb{P}(E_{k_0})=1$ for some $k_0$. If not it is simple, otherwise try looking at the countable intersection of all $1$ measured sets in ${ E_k} $ and everything else.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 16:30
$begingroup$
Lets say we have $mathbb{P}(E_{k_{0}}) = 1$ for some $k_{0}$. Then we have $$ 1 = mathbb{P}(E_{k_{0}}) leq mathbb{P}left(bigcup_{k geq 1}E_{k}right). $$ So $bigcup_{k geq 1}E_{k} in mathcal{G}$. If all are zero events it is clear. Similarly if all are $1$ events. Does this prove, that $mathcal{G}$ is a sigma algebra?
$endgroup$
– gregor
Nov 26 '18 at 18:23
add a comment |
$begingroup$
Let $mathcal{F}$ be a sigma-algebra over a set $Omega$. Now let
$$
mathcal{G} := {E in mathcal{F} : mathbb{P}(E) = 0 ~text{or}~1}.
$$
I have to show that $mathcal{G}$ is a sigma algebra.
The only difficult is to show that if $E_{1},E_{2},cdots in mathcal{G}$ then $bigcup_{k geq 1}E_{k} in mathcal{G}$.
My first attempt was to use $mathbb{P}left(bigcup_{k geq 1}E_{k}right)$ $leq$ $sum_{kgeq 1}mathbb{P}(E_{k})$. Now I'm stuck.
Does anyone have any idea?
Thanks!
probability measure-theory
edited Nov 26 '18 at 16:00
Shaun
9,083113683
asked Nov 26 '18 at 15:58
gregorgregor
1617
$endgroup$
Let $mathcal{F}$ be a sigma-algebra over a set $Omega$. Now let
$$
mathcal{G} := {E in mathcal{F} : mathbb{P}(E) = 0 ~text{or}~1}.
$$
I have to show that $mathcal{G}$ is a sigma algebra.
The only difficult is to show that if $E_{1},E_{2},cdots in mathcal{G}$ then $bigcup_{k geq 1}E_{k} in mathcal{G}$.
My first attempt was to use $mathbb{P}left(bigcup_{k geq 1}E_{k}right)$ $leq$ $sum_{kgeq 1}mathbb{P}(E_{k})$. Now I'm stuck.
Does anyone have any idea?
Thanks!
probability measure-theory
probability measure-theory
edited Nov 26 '18 at 16:00
Shaun
9,083113683
asked Nov 26 '18 at 15:58
gregorgregor
1617
edited Nov 26 '18 at 16:00
Shaun
9,083113683
asked Nov 26 '18 at 15:58
gregorgregor
1617
edited Nov 26 '18 at 16:00
Shaun
9,083113683
edited Nov 26 '18 at 16:00
Shaun
9,083113683
edited Nov 26 '18 at 16:00
Shaun
9,083113683
9,083113683
asked Nov 26 '18 at 15:58
gregorgregor
1617
asked Nov 26 '18 at 15:58
gregorgregor
1617
asked Nov 26 '18 at 15:58
gregorgregor
1617
1617
$begingroup$
Maybe you could split the union into two parts, zero measure and full measure events.
$endgroup$
– Calculon
Nov 26 '18 at 16:04
$begingroup$
It depends on whether $mathbb{P}(E_{k_0})=1$ for some $k_0$. If not it is simple, otherwise try looking at the countable intersection of all $1$ measured sets in ${ E_k} $ and everything else.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 16:30
$begingroup$
Lets say we have $mathbb{P}(E_{k_{0}}) = 1$ for some $k_{0}$. Then we have $$ 1 = mathbb{P}(E_{k_{0}}) leq mathbb{P}left(bigcup_{k geq 1}E_{k}right). $$ So $bigcup_{k geq 1}E_{k} in mathcal{G}$. If all are zero events it is clear. Similarly if all are $1$ events. Does this prove, that $mathcal{G}$ is a sigma algebra?
$endgroup$
– gregor
Nov 26 '18 at 18:23
add a comment |
$begingroup$
Maybe you could split the union into two parts, zero measure and full measure events.
$endgroup$
– Calculon
Nov 26 '18 at 16:04
$begingroup$
It depends on whether $mathbb{P}(E_{k_0})=1$ for some $k_0$. If not it is simple, otherwise try looking at the countable intersection of all $1$ measured sets in ${ E_k} $ and everything else.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 16:30
$begingroup$
Lets say we have $mathbb{P}(E_{k_{0}}) = 1$ for some $k_{0}$. Then we have $$ 1 = mathbb{P}(E_{k_{0}}) leq mathbb{P}left(bigcup_{k geq 1}E_{k}right). $$ So $bigcup_{k geq 1}E_{k} in mathcal{G}$. If all are zero events it is clear. Similarly if all are $1$ events. Does this prove, that $mathcal{G}$ is a sigma algebra?
$endgroup$
– gregor
Nov 26 '18 at 18:23
$begingroup$
Maybe you could split the union into two parts, zero measure and full measure events.
$endgroup$
– Calculon
Nov 26 '18 at 16:04
$begingroup$
Maybe you could split the union into two parts, zero measure and full measure events.
$endgroup$
– Calculon
Nov 26 '18 at 16:04
$begingroup$
It depends on whether $mathbb{P}(E_{k_0})=1$ for some $k_0$. If not it is simple, otherwise try looking at the countable intersection of all $1$ measured sets in ${ E_k} $ and everything else.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 16:30
$begingroup$
It depends on whether $mathbb{P}(E_{k_0})=1$ for some $k_0$. If not it is simple, otherwise try looking at the countable intersection of all $1$ measured sets in ${ E_k} $ and everything else.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 16:30
$begingroup$
Lets say we have $mathbb{P}(E_{k_{0}}) = 1$ for some $k_{0}$. Then we have $$ 1 = mathbb{P}(E_{k_{0}}) leq mathbb{P}left(bigcup_{k geq 1}E_{k}right). $$ So $bigcup_{k geq 1}E_{k} in mathcal{G}$. If all are zero events it is clear. Similarly if all are $1$ events. Does this prove, that $mathcal{G}$ is a sigma algebra?
$endgroup$
– gregor
Nov 26 '18 at 18:23
$begingroup$
Lets say we have $mathbb{P}(E_{k_{0}}) = 1$ for some $k_{0}$. Then we have $$ 1 = mathbb{P}(E_{k_{0}}) leq mathbb{P}left(bigcup_{k geq 1}E_{k}right). $$ So $bigcup_{k geq 1}E_{k} in mathcal{G}$. If all are zero events it is clear. Similarly if all are $1$ events. Does this prove, that $mathcal{G}$ is a sigma algebra?
$endgroup$
– gregor
Nov 26 '18 at 18:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$renewcommand{Pr}{mathbb{P}}$
$renewcommand{c}[1]{left(#1right)}$
Assume without loss of generality that you have $E_1,E_2,dots,F_1,F_2,dotsinmathcal{G}$ s.t. $mathbb{P}c{E_i}=1$ and $Prc{F_i}=0$ and prove that: $bigcup E_i, bigcup F_i in mathcal{G}$.
And all you have left to prove now is that if $E,Finmathcal{G}$ s.t. $Prc{E}=1$ and $Prc{F}=0$ then $Ecup Finmathcal{G}$.
edited Jan 12 at 14:09
amWhy
1
answered Jan 12 at 13:39
roy999roy999
111
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$renewcommand{Pr}{mathbb{P}}$
$renewcommand{c}[1]{left(#1right)}$
Assume without loss of generality that you have $E_1,E_2,dots,F_1,F_2,dotsinmathcal{G}$ s.t. $mathbb{P}c{E_i}=1$ and $Prc{F_i}=0$ and prove that: $bigcup E_i, bigcup F_i in mathcal{G}$.
And all you have left to prove now is that if $E,Finmathcal{G}$ s.t. $Prc{E}=1$ and $Prc{F}=0$ then $Ecup Finmathcal{G}$.
edited Jan 12 at 14:09
amWhy
1
answered Jan 12 at 13:39
roy999roy999
111
$endgroup$
add a comment |
$begingroup$
$renewcommand{Pr}{mathbb{P}}$
$renewcommand{c}[1]{left(#1right)}$
Assume without loss of generality that you have $E_1,E_2,dots,F_1,F_2,dotsinmathcal{G}$ s.t. $mathbb{P}c{E_i}=1$ and $Prc{F_i}=0$ and prove that: $bigcup E_i, bigcup F_i in mathcal{G}$.
And all you have left to prove now is that if $E,Finmathcal{G}$ s.t. $Prc{E}=1$ and $Prc{F}=0$ then $Ecup Finmathcal{G}$.
edited Jan 12 at 14:09
amWhy
1
answered Jan 12 at 13:39
roy999roy999
111
$endgroup$
add a comment |
$begingroup$
$renewcommand{Pr}{mathbb{P}}$
$renewcommand{c}[1]{left(#1right)}$
Assume without loss of generality that you have $E_1,E_2,dots,F_1,F_2,dotsinmathcal{G}$ s.t. $mathbb{P}c{E_i}=1$ and $Prc{F_i}=0$ and prove that: $bigcup E_i, bigcup F_i in mathcal{G}$.
And all you have left to prove now is that if $E,Finmathcal{G}$ s.t. $Prc{E}=1$ and $Prc{F}=0$ then $Ecup Finmathcal{G}$.
edited Jan 12 at 14:09
amWhy
1
answered Jan 12 at 13:39
roy999roy999
111
$endgroup$
$renewcommand{Pr}{mathbb{P}}$
$renewcommand{c}[1]{left(#1right)}$
Assume without loss of generality that you have $E_1,E_2,dots,F_1,F_2,dotsinmathcal{G}$ s.t. $mathbb{P}c{E_i}=1$ and $Prc{F_i}=0$ and prove that: $bigcup E_i, bigcup F_i in mathcal{G}$.
And all you have left to prove now is that if $E,Finmathcal{G}$ s.t. $Prc{E}=1$ and $Prc{F}=0$ then $Ecup Finmathcal{G}$.
edited Jan 12 at 14:09
amWhy
1
answered Jan 12 at 13:39
roy999roy999
111
edited Jan 12 at 14:09
amWhy
1
edited Jan 12 at 14:09
amWhy
1
edited Jan 12 at 14:09
amWhy
1
1
answered Jan 12 at 13:39
roy999roy999
111
answered Jan 12 at 13:39
roy999roy999
111
answered Jan 12 at 13:39
roy999roy999
111
111
add a comment |
add a comment |
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$begingroup$
Maybe you could split the union into two parts, zero measure and full measure events.
$endgroup$
– Calculon
Nov 26 '18 at 16:04
$begingroup$
It depends on whether $mathbb{P}(E_{k_0})=1$ for some $k_0$. If not it is simple, otherwise try looking at the countable intersection of all $1$ measured sets in ${ E_k} $ and everything else.
$endgroup$
– Keen-ameteur
Nov 26 '18 at 16:30
$begingroup$
Lets say we have $mathbb{P}(E_{k_{0}}) = 1$ for some $k_{0}$. Then we have $$ 1 = mathbb{P}(E_{k_{0}}) leq mathbb{P}left(bigcup_{k geq 1}E_{k}right). $$ So $bigcup_{k geq 1}E_{k} in mathcal{G}$. If all are zero events it is clear. Similarly if all are $1$ events. Does this prove, that $mathcal{G}$ is a sigma algebra?
$endgroup$
– gregor
Nov 26 '18 at 18:23