sigma algebra of all 0 and 1 events
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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 at 16:00
Shaun
8,190113577
asked Nov 26 at 15:58
gregor
1496
add a comment |
up vote
0
down vote
favorite
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 at 16:00
Shaun
8,190113577
asked Nov 26 at 15:58
gregor
1496
Maybe you could split the union into two parts, zero measure and full measure events.
– Calculon
Nov 26 at 16:04
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.
– Keen-ameteur
Nov 26 at 16:30
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?
– gregor
Nov 26 at 18:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 16:00
Shaun
8,190113577
asked Nov 26 at 15:58
gregor
1496
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 at 16:00
Shaun
8,190113577
asked Nov 26 at 15:58
gregor
1496
edited Nov 26 at 16:00
Shaun
8,190113577
asked Nov 26 at 15:58
gregor
1496
edited Nov 26 at 16:00
Shaun
8,190113577
edited Nov 26 at 16:00
Shaun
8,190113577
edited Nov 26 at 16:00
Shaun
8,190113577
8,190113577
asked Nov 26 at 15:58
gregor
1496
asked Nov 26 at 15:58
gregor
1496
asked Nov 26 at 15:58
gregor
1496
1496
Maybe you could split the union into two parts, zero measure and full measure events.
– Calculon
Nov 26 at 16:04
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.
– Keen-ameteur
Nov 26 at 16:30
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?
– gregor
Nov 26 at 18:23
add a comment |
Maybe you could split the union into two parts, zero measure and full measure events.
– Calculon
Nov 26 at 16:04
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.
– Keen-ameteur
Nov 26 at 16:30
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?
– gregor
Nov 26 at 18:23
Maybe you could split the union into two parts, zero measure and full measure events.
– Calculon
Nov 26 at 16:04
Maybe you could split the union into two parts, zero measure and full measure events.
– Calculon
Nov 26 at 16:04
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.
– Keen-ameteur
Nov 26 at 16:30
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.
– Keen-ameteur
Nov 26 at 16:30
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?
– gregor
Nov 26 at 18:23
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?
– gregor
Nov 26 at 18:23
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Maybe you could split the union into two parts, zero measure and full measure events.
– Calculon
Nov 26 at 16:04
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.
– Keen-ameteur
Nov 26 at 16:30
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?
– gregor
Nov 26 at 18:23