Mixing
Sigma-algebra generated by integer multiples
$begingroup$
I came across this question and I'd like to check if my answer is correct. The problem is the following:
Consider the collection $mathcal{A}$ of subsets $A_1,A_2,...$ of $mathbb{Z}$
such that
$$A_i = {ni ,|,n text{ is an integer}}.$$
Determine the sigma-algebra generated by $mathcal{A}$, denoted by $sigma(mathcal{A})$.
My answer: Since the sets $A_i$ are symmetric around ${0}$, I first noticed that, for any $kinmathbb{N}$, I can write
$$
{-k,k}=A_kbackslashbigcup_{i=1}^infty A_{i+k}insigma(mathcal{A}).
$$
Note also that, when $k=0$, $A_0$ is not defined, but we can write
$$
{0}=bigcap_{i=1}^infty A_iinsigma(mathcal{A}).
$$
Thus any complement, countable union or intersection of sets in $mathcal{A}$ can be written as a union of sets of the form ${-k,k}$ and ${0}$, which are all in $sigma(mathcal{A})$. Hence, $sigma(mathcal{A})$ consists of all subsets $B$ in $mathbb{Z}$ such that for each interger $kin B$, we have $-kin B$.
Is this enough?
measure-theory
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
$endgroup$
add a comment |
$begingroup$
I came across this question and I'd like to check if my answer is correct. The problem is the following:
Consider the collection $mathcal{A}$ of subsets $A_1,A_2,...$ of $mathbb{Z}$
such that
$$A_i = {ni ,|,n text{ is an integer}}.$$
Determine the sigma-algebra generated by $mathcal{A}$, denoted by $sigma(mathcal{A})$.
My answer: Since the sets $A_i$ are symmetric around ${0}$, I first noticed that, for any $kinmathbb{N}$, I can write
$$
{-k,k}=A_kbackslashbigcup_{i=1}^infty A_{i+k}insigma(mathcal{A}).
$$
Note also that, when $k=0$, $A_0$ is not defined, but we can write
$$
{0}=bigcap_{i=1}^infty A_iinsigma(mathcal{A}).
$$
Thus any complement, countable union or intersection of sets in $mathcal{A}$ can be written as a union of sets of the form ${-k,k}$ and ${0}$, which are all in $sigma(mathcal{A})$. Hence, $sigma(mathcal{A})$ consists of all subsets $B$ in $mathbb{Z}$ such that for each interger $kin B$, we have $-kin B$.
Is this enough?
measure-theory
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
$endgroup$
2
$begingroup$
You showed that all those sets are in $sigma(mathcal{A})$. You only have left to show that the sets you describe actually form a $sigma$-algebra!
$endgroup$
– csprun
Jan 24 at 1:13
add a comment |
$begingroup$
I came across this question and I'd like to check if my answer is correct. The problem is the following:
Consider the collection $mathcal{A}$ of subsets $A_1,A_2,...$ of $mathbb{Z}$
such that
$$A_i = {ni ,|,n text{ is an integer}}.$$
Determine the sigma-algebra generated by $mathcal{A}$, denoted by $sigma(mathcal{A})$.
My answer: Since the sets $A_i$ are symmetric around ${0}$, I first noticed that, for any $kinmathbb{N}$, I can write
$$
{-k,k}=A_kbackslashbigcup_{i=1}^infty A_{i+k}insigma(mathcal{A}).
$$
Note also that, when $k=0$, $A_0$ is not defined, but we can write
$$
{0}=bigcap_{i=1}^infty A_iinsigma(mathcal{A}).
$$
Thus any complement, countable union or intersection of sets in $mathcal{A}$ can be written as a union of sets of the form ${-k,k}$ and ${0}$, which are all in $sigma(mathcal{A})$. Hence, $sigma(mathcal{A})$ consists of all subsets $B$ in $mathbb{Z}$ such that for each interger $kin B$, we have $-kin B$.
Is this enough?
measure-theory
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
$endgroup$
I came across this question and I'd like to check if my answer is correct. The problem is the following:
Consider the collection $mathcal{A}$ of subsets $A_1,A_2,...$ of $mathbb{Z}$
such that
$$A_i = {ni ,|,n text{ is an integer}}.$$
Determine the sigma-algebra generated by $mathcal{A}$, denoted by $sigma(mathcal{A})$.
My answer: Since the sets $A_i$ are symmetric around ${0}$, I first noticed that, for any $kinmathbb{N}$, I can write
$$
{-k,k}=A_kbackslashbigcup_{i=1}^infty A_{i+k}insigma(mathcal{A}).
$$
Note also that, when $k=0$, $A_0$ is not defined, but we can write
$$
{0}=bigcap_{i=1}^infty A_iinsigma(mathcal{A}).
$$
Thus any complement, countable union or intersection of sets in $mathcal{A}$ can be written as a union of sets of the form ${-k,k}$ and ${0}$, which are all in $sigma(mathcal{A})$. Hence, $sigma(mathcal{A})$ consists of all subsets $B$ in $mathbb{Z}$ such that for each interger $kin B$, we have $-kin B$.
Is this enough?
measure-theory
measure-theory
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
edited Jan 24 at 1:02
sam wolfe
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
asked Jan 23 at 17:59
sam wolfesam wolfe
791525
791525
2
$begingroup$
You showed that all those sets are in $sigma(mathcal{A})$. You only have left to show that the sets you describe actually form a $sigma$-algebra!
$endgroup$
– csprun
Jan 24 at 1:13
add a comment |
2
$begingroup$
You showed that all those sets are in $sigma(mathcal{A})$. You only have left to show that the sets you describe actually form a $sigma$-algebra!
$endgroup$
– csprun
Jan 24 at 1:13
2
2
$begingroup$
You showed that all those sets are in $sigma(mathcal{A})$. You only have left to show that the sets you describe actually form a $sigma$-algebra!
$endgroup$
– csprun
Jan 24 at 1:13
$begingroup$
You showed that all those sets are in $sigma(mathcal{A})$. You only have left to show that the sets you describe actually form a $sigma$-algebra!
$endgroup$
– csprun
Jan 24 at 1:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To make your argument precise consider ${B in sigma (mathcal A): -B=B}$. ($-B$ stands for ${-b: bin B}$). Verify that this is a sigma algebra and that it contains $mathcal A$. Conclusion: $-B=B$ holds for every set $B$ in $sigma (mathcal A)$. Conversely, $-B=B$ implies $B=cup_{k in A} {-k,k} cup {0}$ where $A={n in B: n>0}$. Thus, $B$ is a countable union of sets in $sigma (mathcal A)$ proving that $B in sigma (mathcal A)$.
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
$endgroup$
$begingroup$
Thank you. What about the argument regarding the 'smallest $sigma$-algebra', what can be said?
$endgroup$
– sam wolfe
Jan 24 at 20:30
add a comment |
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1 Answer
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$begingroup$
To make your argument precise consider ${B in sigma (mathcal A): -B=B}$. ($-B$ stands for ${-b: bin B}$). Verify that this is a sigma algebra and that it contains $mathcal A$. Conclusion: $-B=B$ holds for every set $B$ in $sigma (mathcal A)$. Conversely, $-B=B$ implies $B=cup_{k in A} {-k,k} cup {0}$ where $A={n in B: n>0}$. Thus, $B$ is a countable union of sets in $sigma (mathcal A)$ proving that $B in sigma (mathcal A)$.
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
$endgroup$
$begingroup$
Thank you. What about the argument regarding the 'smallest $sigma$-algebra', what can be said?
$endgroup$
– sam wolfe
Jan 24 at 20:30
add a comment |
$begingroup$
To make your argument precise consider ${B in sigma (mathcal A): -B=B}$. ($-B$ stands for ${-b: bin B}$). Verify that this is a sigma algebra and that it contains $mathcal A$. Conclusion: $-B=B$ holds for every set $B$ in $sigma (mathcal A)$. Conversely, $-B=B$ implies $B=cup_{k in A} {-k,k} cup {0}$ where $A={n in B: n>0}$. Thus, $B$ is a countable union of sets in $sigma (mathcal A)$ proving that $B in sigma (mathcal A)$.
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
$endgroup$
$begingroup$
Thank you. What about the argument regarding the 'smallest $sigma$-algebra', what can be said?
$endgroup$
– sam wolfe
Jan 24 at 20:30
add a comment |
$begingroup$
To make your argument precise consider ${B in sigma (mathcal A): -B=B}$. ($-B$ stands for ${-b: bin B}$). Verify that this is a sigma algebra and that it contains $mathcal A$. Conclusion: $-B=B$ holds for every set $B$ in $sigma (mathcal A)$. Conversely, $-B=B$ implies $B=cup_{k in A} {-k,k} cup {0}$ where $A={n in B: n>0}$. Thus, $B$ is a countable union of sets in $sigma (mathcal A)$ proving that $B in sigma (mathcal A)$.
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
$endgroup$
To make your argument precise consider ${B in sigma (mathcal A): -B=B}$. ($-B$ stands for ${-b: bin B}$). Verify that this is a sigma algebra and that it contains $mathcal A$. Conclusion: $-B=B$ holds for every set $B$ in $sigma (mathcal A)$. Conversely, $-B=B$ implies $B=cup_{k in A} {-k,k} cup {0}$ where $A={n in B: n>0}$. Thus, $B$ is a countable union of sets in $sigma (mathcal A)$ proving that $B in sigma (mathcal A)$.
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
answered Jan 24 at 6:24
Kavi Rama MurthyKavi Rama Murthy
67.2k53067
67.2k53067
$begingroup$
Thank you. What about the argument regarding the 'smallest $sigma$-algebra', what can be said?
$endgroup$
– sam wolfe
Jan 24 at 20:30
add a comment |
$begingroup$
Thank you. What about the argument regarding the 'smallest $sigma$-algebra', what can be said?
$endgroup$
– sam wolfe
Jan 24 at 20:30
$begingroup$
Thank you. What about the argument regarding the 'smallest $sigma$-algebra', what can be said?
$endgroup$
– sam wolfe
Jan 24 at 20:30
$begingroup$
Thank you. What about the argument regarding the 'smallest $sigma$-algebra', what can be said?
$endgroup$
– sam wolfe
Jan 24 at 20:30
add a comment |
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2
$begingroup$
You showed that all those sets are in $sigma(mathcal{A})$. You only have left to show that the sets you describe actually form a $sigma$-algebra!
$endgroup$
– csprun
Jan 24 at 1:13