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$sigma(C_1) subset sigma(C_2) iff C_1 subset C_2$?

$mathcalC_1 subseteq mathcalC_2 implies sigma( mathcalC_1) subseteq sigma( mathcalC_2) $If $D_1subseteq D_2$ then the generated sigma-algebras are such that $sigma(D_1)subseteq sigma(D_2)$About the smallest sigma field under certain conditions.Two ways to generate the same $sigma$-algebrasProving equality of sigma-algebras$sigma$-field generate by one point setsWhy does $sigma (X_t) subset sigma (X)$ hold?Atoms in measurable spaces$sigma$-algebra generated by a subsetWhy is $mathcal M(mathcal C) subset sigma (mathcal C)$?Why the sigma algebra generated by a function of a random variable is a subset of the sigma algebra generated by the random variable itself?Hint on simple problem regarding countably generated sigma algebras

1

$begingroup$

Let $(Omega, mathcalF)$ be a measurable space and $C_1, C_2 subset mathcalP(Omega)$. Then, we know that $C_1 subset C_2 implies sigma(C_1) subset sigma(C_2)$, where
$$
sigma(A)
:= bigcap mathcalB: mathcalB text is a sigmatext-algebra containing A
$$
is the generated $sigma$-algebra of $A subset mathcalP(Omega)$.
Does the converse also hold?

Since in our lecture, we only formulated one direction, I would be surprised if both directions worked, but haven't been able to come up with a counterexample.

Any help is greatly appreciated.

probability-theory measure-theory examples-counterexamples

share|cite|improve this question

edited Mar 29 at 21:45

Viktor Glombik

asked Mar 29 at 21:11

Viktor GlombikViktor Glombik

1,3072628

$endgroup$

add a comment | 

1

$begingroup$

Let $(Omega, mathcalF)$ be a measurable space and $C_1, C_2 subset mathcalP(Omega)$. Then, we know that $C_1 subset C_2 implies sigma(C_1) subset sigma(C_2)$, where
$$
sigma(A)
:= bigcap mathcalB: mathcalB text is a sigmatext-algebra containing A
$$
is the generated $sigma$-algebra of $A subset mathcalP(Omega)$.
Does the converse also hold?

Since in our lecture, we only formulated one direction, I would be surprised if both directions worked, but haven't been able to come up with a counterexample.

Any help is greatly appreciated.

probability-theory measure-theory examples-counterexamples

share|cite|improve this question

edited Mar 29 at 21:45

Viktor Glombik

asked Mar 29 at 21:11

Viktor GlombikViktor Glombik

1,3072628

$endgroup$

add a comment | 

$begingroup$

Let $(Omega, mathcalF)$ be a measurable space and $C_1, C_2 subset mathcalP(Omega)$. Then, we know that $C_1 subset C_2 implies sigma(C_1) subset sigma(C_2)$, where
$$
sigma(A)
:= bigcap mathcalB: mathcalB text is a sigmatext-algebra containing A
$$
is the generated $sigma$-algebra of $A subset mathcalP(Omega)$.
Does the converse also hold?

Since in our lecture, we only formulated one direction, I would be surprised if both directions worked, but haven't been able to come up with a counterexample.

Any help is greatly appreciated.

probability-theory measure-theory examples-counterexamples

share|cite|improve this question

edited Mar 29 at 21:45

Viktor Glombik

asked Mar 29 at 21:11

Viktor GlombikViktor Glombik

1,3072628

$endgroup$

share|cite|improve this question

edited Mar 29 at 21:45

Viktor Glombik

asked Mar 29 at 21:11

Viktor GlombikViktor Glombik

1,3072628

share|cite|improve this question

edited Mar 29 at 21:45

Viktor Glombik

asked Mar 29 at 21:11

Viktor GlombikViktor Glombik

1,3072628

asked Mar 29 at 21:11

Viktor GlombikViktor Glombik

1,3072628

asked Mar 29 at 21:11

Viktor GlombikViktor Glombik

1,3072628

2 Answers
2

active

oldest

votes

1

$begingroup$

I assume here your "$subset$" means "$subseteq$" (containment) and not "$subsetneq$" (proper containment).

Assuming this, then it's not true. Suppose $S$ is an element of $sigma(C_2)$ which is not actually already in $C_2$, and define $C_1 = C_2cup S$.

Then $sigma(C_1)=sigma(C_2)$, so $ sigma(C_1)subset sigma(C_2)$, but $C_1notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:20

MPWMPW

31.1k12157

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the following a good example? $Omega := 0,1,2$, $C_1 := 1, 2, 0 $ and $C_2 := 1,2, 0,2, 0,1$? Then, clearly, $C_1 notsubset C_2$, but $sigma(C_1) = sigma(C_2) = mathcalP(Omega)$.
  $endgroup$
  – Viktor Glombik
  Mar 29 at 21:54
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Take $Omega := mathbbR, C_1 := A subset Omega: A text open $ and $C_2 := A subset Omega: A text closed $.
Then, $sigma(C_1) = sigma(C_2) = mathcalB$, where $mathcalB$ is the Borel-$sigma$-Algebra on $mathbbR$, and therefore, $sigma(C_1) subset sigma(C_2)$ but $C_1 notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:44

Viktor GlombikViktor Glombik

1,3072628

$endgroup$

add a comment | 

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1

$begingroup$

I assume here your "$subset$" means "$subseteq$" (containment) and not "$subsetneq$" (proper containment).

Assuming this, then it's not true. Suppose $S$ is an element of $sigma(C_2)$ which is not actually already in $C_2$, and define $C_1 = C_2cup S$.

Then $sigma(C_1)=sigma(C_2)$, so $ sigma(C_1)subset sigma(C_2)$, but $C_1notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:20

MPWMPW

31.1k12157

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the following a good example? $Omega := 0,1,2$, $C_1 := 1, 2, 0 $ and $C_2 := 1,2, 0,2, 0,1$? Then, clearly, $C_1 notsubset C_2$, but $sigma(C_1) = sigma(C_2) = mathcalP(Omega)$.
  $endgroup$
  – Viktor Glombik
  Mar 29 at 21:54
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I assume here your "$subset$" means "$subseteq$" (containment) and not "$subsetneq$" (proper containment).

Assuming this, then it's not true. Suppose $S$ is an element of $sigma(C_2)$ which is not actually already in $C_2$, and define $C_1 = C_2cup S$.

Then $sigma(C_1)=sigma(C_2)$, so $ sigma(C_1)subset sigma(C_2)$, but $C_1notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:20

MPWMPW

31.1k12157

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the following a good example? $Omega := 0,1,2$, $C_1 := 1, 2, 0 $ and $C_2 := 1,2, 0,2, 0,1$? Then, clearly, $C_1 notsubset C_2$, but $sigma(C_1) = sigma(C_2) = mathcalP(Omega)$.
  $endgroup$
  – Viktor Glombik
  Mar 29 at 21:54
  
  

  
  
  
  

add a comment | 

$begingroup$

I assume here your "$subset$" means "$subseteq$" (containment) and not "$subsetneq$" (proper containment).

Assuming this, then it's not true. Suppose $S$ is an element of $sigma(C_2)$ which is not actually already in $C_2$, and define $C_1 = C_2cup S$.

Then $sigma(C_1)=sigma(C_2)$, so $ sigma(C_1)subset sigma(C_2)$, but $C_1notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:20

MPWMPW

31.1k12157

$endgroup$

share|cite|improve this answer

answered Mar 29 at 21:20

MPWMPW

31.1k12157

answered Mar 29 at 21:20

MPWMPW

31.1k12157

answered Mar 29 at 21:20

MPWMPW

31.1k12157

1

$begingroup$

Take $Omega := mathbbR, C_1 := A subset Omega: A text open $ and $C_2 := A subset Omega: A text closed $.
Then, $sigma(C_1) = sigma(C_2) = mathcalB$, where $mathcalB$ is the Borel-$sigma$-Algebra on $mathbbR$, and therefore, $sigma(C_1) subset sigma(C_2)$ but $C_1 notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:44

Viktor GlombikViktor Glombik

1,3072628

$endgroup$

add a comment | 

1

$begingroup$

Take $Omega := mathbbR, C_1 := A subset Omega: A text open $ and $C_2 := A subset Omega: A text closed $.
Then, $sigma(C_1) = sigma(C_2) = mathcalB$, where $mathcalB$ is the Borel-$sigma$-Algebra on $mathbbR$, and therefore, $sigma(C_1) subset sigma(C_2)$ but $C_1 notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:44

Viktor GlombikViktor Glombik

1,3072628

$endgroup$

add a comment | 

$begingroup$

Take $Omega := mathbbR, C_1 := A subset Omega: A text open $ and $C_2 := A subset Omega: A text closed $.
Then, $sigma(C_1) = sigma(C_2) = mathcalB$, where $mathcalB$ is the Borel-$sigma$-Algebra on $mathbbR$, and therefore, $sigma(C_1) subset sigma(C_2)$ but $C_1 notsubset C_2$.

share|cite|improve this answer

answered Mar 29 at 21:44

Viktor GlombikViktor Glombik

1,3072628

$endgroup$

share|cite|improve this answer

answered Mar 29 at 21:44

Viktor GlombikViktor Glombik

1,3072628

answered Mar 29 at 21:44

Viktor GlombikViktor Glombik

1,3072628

answered Mar 29 at 21:44

Viktor GlombikViktor Glombik

1,3072628