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Constructing the generated $sigma$-algebra

Increasing limit of sets is in union of familiesField of sets and Sigma algebra of sets$sigma$ -algebra is the smallest collection of sets which…About generated $sigma$-algebras (proof verification).(Countable) partition generated $sigma$-algebraPartition generated $sigma$-algebraDefinition of measurable space - sigma algebraA $lambda$-system $mathcalL$ that is a $pi$-system is automatically a $sigma$-field.Condition on the cardinality of $Omega$ for a countably generated sigma-algebraExample of a $sigma$-algebra in $mathbbR^n$sigma algebra v.s. topology

0

$begingroup$

Consider a collection $mathcalA = A_i_i in I$ where $I$ is some indexing set. I want to construct $sigma(mathcalA)$.

Let $mathcalF_0 = mathcalA$. We now construct inductively: let $mathcalF_i+1$ be the collection of sets which are formed by countable union, intersection, complementation of sets in $mathcalF_i$. This is an increasing sequence of collections.

Let $mathcalF = bigcup_i geq 0 mathcalF_i$. Is $mathcalF = sigma(mathcalA)$?

Clearly $mathcalF$ contains $mathcalA$ and the aim is to show that $mathcalF$ is a $sigma$-algebra. If we show this, then we should be done as $sigma(mathcalA)$ must contain all the sets in our construction.

It is easy to show that $mathcalF$ is closed under complementation and finite intersection, but how can one show it is closed under countable union?

Let $B_1, B_2, dots in mathcalF$. Then $B_i in mathcalF_n_i$ for some $n_i$, Let $B^'_i = bigcup_j=1^i B_j$, $m_j = operatornamemax n_1,dots,n_j$, we see that $B^'_i in mathcalF_m_i$, and we would like to show that $B = bigcup_igeq 0 B^'_i in mathcalF$.

Can we proceed further? If not, what is missing?

measure-theory elementary-set-theory

share|cite|improve this question

asked Mar 29 at 21:12

ArchieR577ArchieR577

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not completely convinced that this is true, as per this. There seems to be some subtlety involved. Can anyone help me out?
  $endgroup$
  – ArchieR577
  Mar 29 at 21:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Seems to be not so simple. There are many books concerning this issue, see e.g. Hewitt-Stromberg "Real and abstract analysis", page 133, the PROOF OF Theorem (10.23), there a construction of $sigma(mathcal A)$ is given using kind of transfinite induction.
  $endgroup$
  – Yu Ding
  Mar 29 at 21:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, looks like there are many more operations that we need to do and this is not sufficient. Is there an easy example of a set in $sigma(mathcalA)$ that is not in $mathcalF$?
  $endgroup$
  – ArchieR577
  Mar 29 at 21:39
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider a collection $mathcalA = A_i_i in I$ where $I$ is some indexing set. I want to construct $sigma(mathcalA)$.

Let $mathcalF_0 = mathcalA$. We now construct inductively: let $mathcalF_i+1$ be the collection of sets which are formed by countable union, intersection, complementation of sets in $mathcalF_i$. This is an increasing sequence of collections.

Let $mathcalF = bigcup_i geq 0 mathcalF_i$. Is $mathcalF = sigma(mathcalA)$?

Clearly $mathcalF$ contains $mathcalA$ and the aim is to show that $mathcalF$ is a $sigma$-algebra. If we show this, then we should be done as $sigma(mathcalA)$ must contain all the sets in our construction.

It is easy to show that $mathcalF$ is closed under complementation and finite intersection, but how can one show it is closed under countable union?

Let $B_1, B_2, dots in mathcalF$. Then $B_i in mathcalF_n_i$ for some $n_i$, Let $B^'_i = bigcup_j=1^i B_j$, $m_j = operatornamemax n_1,dots,n_j$, we see that $B^'_i in mathcalF_m_i$, and we would like to show that $B = bigcup_igeq 0 B^'_i in mathcalF$.

Can we proceed further? If not, what is missing?

measure-theory elementary-set-theory

share|cite|improve this question

asked Mar 29 at 21:12

ArchieR577ArchieR577

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not completely convinced that this is true, as per this. There seems to be some subtlety involved. Can anyone help me out?
  $endgroup$
  – ArchieR577
  Mar 29 at 21:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Seems to be not so simple. There are many books concerning this issue, see e.g. Hewitt-Stromberg "Real and abstract analysis", page 133, the PROOF OF Theorem (10.23), there a construction of $sigma(mathcal A)$ is given using kind of transfinite induction.
  $endgroup$
  – Yu Ding
  Mar 29 at 21:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, looks like there are many more operations that we need to do and this is not sufficient. Is there an easy example of a set in $sigma(mathcalA)$ that is not in $mathcalF$?
  $endgroup$
  – ArchieR577
  Mar 29 at 21:39
  
  

  
  
  
  

add a comment | 

$begingroup$

Consider a collection $mathcalA = A_i_i in I$ where $I$ is some indexing set. I want to construct $sigma(mathcalA)$.

Let $mathcalF_0 = mathcalA$. We now construct inductively: let $mathcalF_i+1$ be the collection of sets which are formed by countable union, intersection, complementation of sets in $mathcalF_i$. This is an increasing sequence of collections.

Let $mathcalF = bigcup_i geq 0 mathcalF_i$. Is $mathcalF = sigma(mathcalA)$?

Clearly $mathcalF$ contains $mathcalA$ and the aim is to show that $mathcalF$ is a $sigma$-algebra. If we show this, then we should be done as $sigma(mathcalA)$ must contain all the sets in our construction.

It is easy to show that $mathcalF$ is closed under complementation and finite intersection, but how can one show it is closed under countable union?

Let $B_1, B_2, dots in mathcalF$. Then $B_i in mathcalF_n_i$ for some $n_i$, Let $B^'_i = bigcup_j=1^i B_j$, $m_j = operatornamemax n_1,dots,n_j$, we see that $B^'_i in mathcalF_m_i$, and we would like to show that $B = bigcup_igeq 0 B^'_i in mathcalF$.

Can we proceed further? If not, what is missing?

measure-theory elementary-set-theory

share|cite|improve this question

asked Mar 29 at 21:12

ArchieR577ArchieR577

114

$endgroup$

share|cite|improve this question

asked Mar 29 at 21:12

ArchieR577ArchieR577

114

share|cite|improve this question

asked Mar 29 at 21:12

ArchieR577ArchieR577

114

asked Mar 29 at 21:12

ArchieR577ArchieR577

114

asked Mar 29 at 21:12

ArchieR577ArchieR577

114