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Why can you write the Borel sets in n dimensions as the sigma-field generated by the product of half open intervals?

1

$begingroup$

My professor defines $mathcalB^2 equiv sigmaA times E : A,E in mathcalB $ where $mathcalB$ is the Borel sets on the real line.

He then says an equivalent definition is that $mathcalB^2 equiv sigma(a_1, b_2] times (a_2, b_2] : a_1 leq b_1, a_2 leq b_2$.

I am not seeing how these two are equivalent. I understand that the Borel sets on the real line are defined as $sigma(a, b] : a_1 leq b_1$.

measure-theory

share|cite|improve this question

edited Jul 24 '13 at 12:44

NewNameStat

asked Jul 24 '13 at 4:30

NewNameStatNewNameStat

193211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The title is (wrong and) not what the question is asking.
  $endgroup$
  – Did
  Jul 24 '13 at 8:26
  

  
  
  
  

add a comment | 

1

$begingroup$

My professor defines $mathcalB^2 equiv sigmaA times E : A,E in mathcalB $ where $mathcalB$ is the Borel sets on the real line.

He then says an equivalent definition is that $mathcalB^2 equiv sigma(a_1, b_2] times (a_2, b_2] : a_1 leq b_1, a_2 leq b_2$.

I am not seeing how these two are equivalent. I understand that the Borel sets on the real line are defined as $sigma(a, b] : a_1 leq b_1$.

measure-theory

share|cite|improve this question

edited Jul 24 '13 at 12:44

NewNameStat

asked Jul 24 '13 at 4:30

NewNameStatNewNameStat

193211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The title is (wrong and) not what the question is asking.
  $endgroup$
  – Did
  Jul 24 '13 at 8:26
  

  
  
  
  

add a comment | 

$begingroup$

My professor defines $mathcalB^2 equiv sigmaA times E : A,E in mathcalB $ where $mathcalB$ is the Borel sets on the real line.

He then says an equivalent definition is that $mathcalB^2 equiv sigma(a_1, b_2] times (a_2, b_2] : a_1 leq b_1, a_2 leq b_2$.

I am not seeing how these two are equivalent. I understand that the Borel sets on the real line are defined as $sigma(a, b] : a_1 leq b_1$.

measure-theory

share|cite|improve this question

edited Jul 24 '13 at 12:44

NewNameStat

asked Jul 24 '13 at 4:30

NewNameStatNewNameStat

193211

$endgroup$

share|cite|improve this question

edited Jul 24 '13 at 12:44

NewNameStat

asked Jul 24 '13 at 4:30

NewNameStatNewNameStat

193211

share|cite|improve this question

edited Jul 24 '13 at 12:44

NewNameStat

asked Jul 24 '13 at 4:30

NewNameStatNewNameStat

193211

asked Jul 24 '13 at 4:30

NewNameStatNewNameStat

193211

asked Jul 24 '13 at 4:30

NewNameStatNewNameStat

193211

1 Answer
1

active

oldest

votes

0

$begingroup$

Let $mathcalB^2 = sigmaA times E : A,E in mathcalB$, as you defined. Let $mathcalB_1^2$ be the $sigma$-algebra generated by products of half-open intervals. We want to show that $mathcalB^2 = mathcalB_1^2$.

Clearly, $mathcalB_1^2 subseteq mathcalB^2$. Now, to show the reverse inclusion, let $A,E in mathcalB$. Note that

$$ A times E = (A times mathbbR) cap (mathbbR times E). $$

Clearly, $A times mathbbR, mathbbR times E$ are in $mathcalB_1^2$ so that $A times E in mathcalB_1^2$ as well. This means that the generator of $mathcalB^2$ is contained in $mathcalB_1^2$; this implies that $mathcalB_1^2 supseteq mathcalB^2$.

share|cite|improve this answer

answered Jul 24 '13 at 8:24

Alex StrifeAlex Strife

556314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is it clear that $A times mathbbR$ is in $mathcalB_1^2$? It is clear to me that $A times mathbbR in sigma(a_1,b_1]:a_1,b_1 in mathbbR times sigma(a_2,b_2]:a_2,b_2 in mathbbR$.
  $endgroup$
  – NewNameStat
  Jul 24 '13 at 12:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, sorry that wasn't clear; I didn't realize it at first as well. Anyway, please check this out instead: math.stackexchange.com/questions/195841/…
  $endgroup$
  – Alex Strife
  Jul 24 '13 at 14:56
  

  
  
  
  

add a comment | 

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0

$begingroup$

Let $mathcalB^2 = sigmaA times E : A,E in mathcalB$, as you defined. Let $mathcalB_1^2$ be the $sigma$-algebra generated by products of half-open intervals. We want to show that $mathcalB^2 = mathcalB_1^2$.

Clearly, $mathcalB_1^2 subseteq mathcalB^2$. Now, to show the reverse inclusion, let $A,E in mathcalB$. Note that

$$ A times E = (A times mathbbR) cap (mathbbR times E). $$

Clearly, $A times mathbbR, mathbbR times E$ are in $mathcalB_1^2$ so that $A times E in mathcalB_1^2$ as well. This means that the generator of $mathcalB^2$ is contained in $mathcalB_1^2$; this implies that $mathcalB_1^2 supseteq mathcalB^2$.

share|cite|improve this answer

answered Jul 24 '13 at 8:24

Alex StrifeAlex Strife

556314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is it clear that $A times mathbbR$ is in $mathcalB_1^2$? It is clear to me that $A times mathbbR in sigma(a_1,b_1]:a_1,b_1 in mathbbR times sigma(a_2,b_2]:a_2,b_2 in mathbbR$.
  $endgroup$
  – NewNameStat
  Jul 24 '13 at 12:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, sorry that wasn't clear; I didn't realize it at first as well. Anyway, please check this out instead: math.stackexchange.com/questions/195841/…
  $endgroup$
  – Alex Strife
  Jul 24 '13 at 14:56
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $mathcalB^2 = sigmaA times E : A,E in mathcalB$, as you defined. Let $mathcalB_1^2$ be the $sigma$-algebra generated by products of half-open intervals. We want to show that $mathcalB^2 = mathcalB_1^2$.

Clearly, $mathcalB_1^2 subseteq mathcalB^2$. Now, to show the reverse inclusion, let $A,E in mathcalB$. Note that

$$ A times E = (A times mathbbR) cap (mathbbR times E). $$

Clearly, $A times mathbbR, mathbbR times E$ are in $mathcalB_1^2$ so that $A times E in mathcalB_1^2$ as well. This means that the generator of $mathcalB^2$ is contained in $mathcalB_1^2$; this implies that $mathcalB_1^2 supseteq mathcalB^2$.

share|cite|improve this answer

answered Jul 24 '13 at 8:24

Alex StrifeAlex Strife

556314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is it clear that $A times mathbbR$ is in $mathcalB_1^2$? It is clear to me that $A times mathbbR in sigma(a_1,b_1]:a_1,b_1 in mathbbR times sigma(a_2,b_2]:a_2,b_2 in mathbbR$.
  $endgroup$
  – NewNameStat
  Jul 24 '13 at 12:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, sorry that wasn't clear; I didn't realize it at first as well. Anyway, please check this out instead: math.stackexchange.com/questions/195841/…
  $endgroup$
  – Alex Strife
  Jul 24 '13 at 14:56
  

  
  
  
  

add a comment | 

$begingroup$

Let $mathcalB^2 = sigmaA times E : A,E in mathcalB$, as you defined. Let $mathcalB_1^2$ be the $sigma$-algebra generated by products of half-open intervals. We want to show that $mathcalB^2 = mathcalB_1^2$.

Clearly, $mathcalB_1^2 subseteq mathcalB^2$. Now, to show the reverse inclusion, let $A,E in mathcalB$. Note that

$$ A times E = (A times mathbbR) cap (mathbbR times E). $$

Clearly, $A times mathbbR, mathbbR times E$ are in $mathcalB_1^2$ so that $A times E in mathcalB_1^2$ as well. This means that the generator of $mathcalB^2$ is contained in $mathcalB_1^2$; this implies that $mathcalB_1^2 supseteq mathcalB^2$.

share|cite|improve this answer

answered Jul 24 '13 at 8:24

Alex StrifeAlex Strife

556314

$endgroup$

share|cite|improve this answer

answered Jul 24 '13 at 8:24

Alex StrifeAlex Strife

556314

answered Jul 24 '13 at 8:24

Alex StrifeAlex Strife

556314

answered Jul 24 '13 at 8:24

Alex StrifeAlex Strife

556314