Do subset sigma algebras exist in general?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let $X$ be a set and $M$ be a sigma algebra of $X$. If $U subset X$, is it true that the set $M_U = E in M$ is a sigma algebra? I'm getting the countable union part, but not compliments:
$$(Ecap U)^C = E^C cup U^C$$
I'm not sure what to do with $U^C$
real-analysis measure-theory
asked Sep 3 at 14:27
yoshi
878716
add a comment |Â
up vote
0
down vote
favorite
Let $X$ be a set and $M$ be a sigma algebra of $X$. If $U subset X$, is it true that the set $M_U = E in M$ is a sigma algebra? I'm getting the countable union part, but not compliments:
$$(Ecap U)^C = E^C cup U^C$$
I'm not sure what to do with $U^C$
real-analysis measure-theory
asked Sep 3 at 14:27
yoshi
878716
2
You need the complement relative to $U$, which is $U setminus E$. $M_U$ is a $sigma$-algebra on $U$, not on $X$.
– Daniel Fischer♦
Sep 3 at 14:34
1
"..and $M$ be it's sigma algebra.." A set $X$ does not have a (canonical) $sigma$-algebra.
– drhab
Sep 3 at 14:42
^ ya, sorry, fixed
– yoshi
Sep 3 at 14:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a set and $M$ be a sigma algebra of $X$. If $U subset X$, is it true that the set $M_U = E in M$ is a sigma algebra? I'm getting the countable union part, but not compliments:
$$(Ecap U)^C = E^C cup U^C$$
I'm not sure what to do with $U^C$
real-analysis measure-theory
asked Sep 3 at 14:27
yoshi
878716
Let $X$ be a set and $M$ be a sigma algebra of $X$. If $U subset X$, is it true that the set $M_U = E in M$ is a sigma algebra? I'm getting the countable union part, but not compliments:
$$(Ecap U)^C = E^C cup U^C$$
I'm not sure what to do with $U^C$
real-analysis measure-theory
real-analysis measure-theory
asked Sep 3 at 14:27
yoshi
878716
asked Sep 3 at 14:27
yoshi
878716
edited Sep 3 at 14:43
asked Sep 3 at 14:27
yoshi
878716
asked Sep 3 at 14:27
yoshi
878716
asked Sep 3 at 14:27
yoshi
878716
878716
2
You need the complement relative to $U$, which is $U setminus E$. $M_U$ is a $sigma$-algebra on $U$, not on $X$.
– Daniel Fischer♦
Sep 3 at 14:34
1
"..and $M$ be it's sigma algebra.." A set $X$ does not have a (canonical) $sigma$-algebra.
– drhab
Sep 3 at 14:42
^ ya, sorry, fixed
– yoshi
Sep 3 at 14:43
add a comment |Â
2
You need the complement relative to $U$, which is $U setminus E$. $M_U$ is a $sigma$-algebra on $U$, not on $X$.
– Daniel Fischer♦
Sep 3 at 14:34
1
"..and $M$ be it's sigma algebra.." A set $X$ does not have a (canonical) $sigma$-algebra.
– drhab
Sep 3 at 14:42
^ ya, sorry, fixed
– yoshi
Sep 3 at 14:43
2
2
You need the complement relative to $U$, which is $U setminus E$. $M_U$ is a $sigma$-algebra on $U$, not on $X$.
– Daniel Fischer♦
Sep 3 at 14:34
You need the complement relative to $U$, which is $U setminus E$. $M_U$ is a $sigma$-algebra on $U$, not on $X$.
– Daniel Fischer♦
Sep 3 at 14:34
1
1
"..and $M$ be it's sigma algebra.." A set $X$ does not have a (canonical) $sigma$-algebra.
– drhab
Sep 3 at 14:42
"..and $M$ be it's sigma algebra.." A set $X$ does not have a (canonical) $sigma$-algebra.
– drhab
Sep 3 at 14:42
^ ya, sorry, fixed
– yoshi
Sep 3 at 14:43
^ ya, sorry, fixed
– yoshi
Sep 3 at 14:43
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Well, you want to show that $mathcalM_U$ is an $U$ $sigma-$ algebra, thus you have to show that $$U cap (E cap U)^c in mathcalM_U$$
But $$U cap (E cap U)^c = U cap (E^c cup U^c) = U cap E^c in mathcalM_U$$
answered Sep 3 at 14:34
LucaMac
1,44414
add a comment |Â
up vote
1
down vote
If $X$ is a set, $mathcal M$ a $sigma$-algebra on it and $f:Yto X$ is a function then the collection $f^-1(mathcal M)$ can be shown to be a $sigma$-algebra on $Y$.
Observe that in that situation $f^-1(E^complement)=f^-1(E)^complement$.
You can apply that here by taking $Y=U$ and $f:Uto X$ the inclusion.
That will lead to $f^-1(mathcal M)=mathcal M_U$ here.
The complement of $Esubseteq U$ in universe $U$ is $E^complementcap U$.
This because you are focusing on a collection of subsets of $U$.
answered Sep 3 at 14:35
drhab
89k541122
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Well, you want to show that $mathcalM_U$ is an $U$ $sigma-$ algebra, thus you have to show that $$U cap (E cap U)^c in mathcalM_U$$
But $$U cap (E cap U)^c = U cap (E^c cup U^c) = U cap E^c in mathcalM_U$$
answered Sep 3 at 14:34
LucaMac
1,44414
add a comment |Â
up vote
2
down vote
accepted
Well, you want to show that $mathcalM_U$ is an $U$ $sigma-$ algebra, thus you have to show that $$U cap (E cap U)^c in mathcalM_U$$
But $$U cap (E cap U)^c = U cap (E^c cup U^c) = U cap E^c in mathcalM_U$$
answered Sep 3 at 14:34
LucaMac
1,44414
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Well, you want to show that $mathcalM_U$ is an $U$ $sigma-$ algebra, thus you have to show that $$U cap (E cap U)^c in mathcalM_U$$
But $$U cap (E cap U)^c = U cap (E^c cup U^c) = U cap E^c in mathcalM_U$$
answered Sep 3 at 14:34
LucaMac
1,44414
Well, you want to show that $mathcalM_U$ is an $U$ $sigma-$ algebra, thus you have to show that $$U cap (E cap U)^c in mathcalM_U$$
But $$U cap (E cap U)^c = U cap (E^c cup U^c) = U cap E^c in mathcalM_U$$
answered Sep 3 at 14:34
LucaMac
1,44414
answered Sep 3 at 14:34
LucaMac
1,44414
answered Sep 3 at 14:34
LucaMac
1,44414
answered Sep 3 at 14:34
LucaMac
1,44414
1,44414
add a comment |Â
add a comment |Â
up vote
1
down vote
If $X$ is a set, $mathcal M$ a $sigma$-algebra on it and $f:Yto X$ is a function then the collection $f^-1(mathcal M)$ can be shown to be a $sigma$-algebra on $Y$.
Observe that in that situation $f^-1(E^complement)=f^-1(E)^complement$.
You can apply that here by taking $Y=U$ and $f:Uto X$ the inclusion.
That will lead to $f^-1(mathcal M)=mathcal M_U$ here.
The complement of $Esubseteq U$ in universe $U$ is $E^complementcap U$.
This because you are focusing on a collection of subsets of $U$.
answered Sep 3 at 14:35
drhab
89k541122
add a comment |Â
up vote
1
down vote
If $X$ is a set, $mathcal M$ a $sigma$-algebra on it and $f:Yto X$ is a function then the collection $f^-1(mathcal M)$ can be shown to be a $sigma$-algebra on $Y$.
Observe that in that situation $f^-1(E^complement)=f^-1(E)^complement$.
You can apply that here by taking $Y=U$ and $f:Uto X$ the inclusion.
That will lead to $f^-1(mathcal M)=mathcal M_U$ here.
The complement of $Esubseteq U$ in universe $U$ is $E^complementcap U$.
This because you are focusing on a collection of subsets of $U$.
answered Sep 3 at 14:35
drhab
89k541122
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $X$ is a set, $mathcal M$ a $sigma$-algebra on it and $f:Yto X$ is a function then the collection $f^-1(mathcal M)$ can be shown to be a $sigma$-algebra on $Y$.
Observe that in that situation $f^-1(E^complement)=f^-1(E)^complement$.
You can apply that here by taking $Y=U$ and $f:Uto X$ the inclusion.
That will lead to $f^-1(mathcal M)=mathcal M_U$ here.
The complement of $Esubseteq U$ in universe $U$ is $E^complementcap U$.
This because you are focusing on a collection of subsets of $U$.
answered Sep 3 at 14:35
drhab
89k541122
If $X$ is a set, $mathcal M$ a $sigma$-algebra on it and $f:Yto X$ is a function then the collection $f^-1(mathcal M)$ can be shown to be a $sigma$-algebra on $Y$.
Observe that in that situation $f^-1(E^complement)=f^-1(E)^complement$.
You can apply that here by taking $Y=U$ and $f:Uto X$ the inclusion.
That will lead to $f^-1(mathcal M)=mathcal M_U$ here.
The complement of $Esubseteq U$ in universe $U$ is $E^complementcap U$.
This because you are focusing on a collection of subsets of $U$.
answered Sep 3 at 14:35
drhab
89k541122
answered Sep 3 at 14:35
drhab
89k541122
answered Sep 3 at 14:35
drhab
89k541122
answered Sep 3 at 14:35
drhab
89k541122
89k541122
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2903920%2fdo-subset-sigma-algebras-exist-in-general%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
You need the complement relative to $U$, which is $U setminus E$. $M_U$ is a $sigma$-algebra on $U$, not on $X$.
– Daniel Fischer♦
Sep 3 at 14:34
1
"..and $M$ be it's sigma algebra.." A set $X$ does not have a (canonical) $sigma$-algebra.
– drhab
Sep 3 at 14:42
^ ya, sorry, fixed
– yoshi
Sep 3 at 14:43