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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
$begingroup$
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
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
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
asked Mar 29 at 21:11
Viktor GlombikViktor Glombik
1,3072628
$endgroup$
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
probability-theory measure-theory examples-counterexamples
asked Mar 29 at 21:11
Viktor GlombikViktor Glombik
1,3072628
asked Mar 29 at 21:11
Viktor GlombikViktor Glombik
1,3072628
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
1,3072628
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $C_1notsubset C_2$.
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 |
$begingroup$
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 $C_1 notsubset C_2$.
answered Mar 29 at 21:44
Viktor GlombikViktor Glombik
1,3072628
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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oldest
votes
$begingroup$
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 $C_1notsubset C_2$.
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 $C_1notsubset C_2$.
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 $C_1notsubset C_2$.
answered Mar 29 at 21:20
MPWMPW
31.1k12157
$endgroup$
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 $C_1notsubset C_2$.
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
answered Mar 29 at 21:20
MPWMPW
31.1k12157
31.1k12157
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
$begingroup$
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 $C_1 notsubset C_2$.
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 $C_1 notsubset C_2$.
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 $C_1 notsubset C_2$.
answered Mar 29 at 21:44
Viktor GlombikViktor Glombik
1,3072628
$endgroup$
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 $C_1 notsubset C_2$.
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
answered Mar 29 at 21:44
Viktor GlombikViktor Glombik
1,3072628
1,3072628
add a comment |
add a comment |
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