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Do Vitali sets really explain measure-theoretic probability?

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3

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I’ve read several answers about how how crazy sets like Vitali’s set are (one of the reasons) why we need $sigma$-algebra and measure theory for probability spaces. However, Vitali’s set is only eliminated by requiring non-trivial translation-invariant measures. So either (a) Vitali’s set wasn’t a problem in the first place or (b) probability spaces need translation-invariance.

No definition of probability spaces I’ve seen mentions translation invariance, leaving us with (a): Vitali’s set is not a problem for probabilists. Then:

1. Why do people mention Vitali's set while explaining the measure-theoretic formulation of probability?


2. Why do, in fact, we need measure theory for probability?


3. Are there any other paradoxes arising from uncountable sets to worry about that do apply to regular measures and probability?

probability-theory

share|cite|improve this question

edited Mar 12 at 23:12

Yatharth Agarwal

asked Mar 12 at 21:40

Yatharth AgarwalYatharth Agarwal

542418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Non-measurable set is not the only reason that we define a measure on $sigma$-algebra instead of the whole power set.
  $endgroup$
  – BigbearZzz
  Mar 12 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BigbearZzz Some of the other reasons I’ve seen are $sigma$-algebra as “information” which is fine. But still, what’s even the relevance of people mentioning the Vitali set if can’t be solved for all probability spaces anyway?
  $endgroup$
  – Yatharth Agarwal
  Mar 12 at 21:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Vitali set serves as an example of a Lebesgue non-measurable set, perhaps the first example of why the Lebesgue outer measure is not a measure on the power set of $Bbb R$. You can say that this has more to do with real analysis than probability theory.
  $endgroup$
  – BigbearZzz
  Mar 12 at 22:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should look into the construction of a probability measure via Caratheodory extension theorem. The presence/absence of Vitali set and translation invariant is not relevant at all, but still the theorem doesn't, in general, let you define a probability measure on the whole power set of your space anyway. This is another reason why $sigma$-algebra is preferred.
  $endgroup$
  – BigbearZzz
  Mar 12 at 22:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BigbearZzz That makes a ton of sense: These Vitali set examples are not actually relevant for probability theory; people just tend to mention them because they’re familiar with analysis, or they’re just trying to present an analogy of how things go wrong with uncountable sets! Does that sound fair?
  $endgroup$
  – Yatharth Agarwal
  Mar 12 at 22:16
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

I’ve read several answers about how how crazy sets like Vitali’s set are (one of the reasons) why we need $sigma$-algebra and measure theory for probability spaces. However, Vitali’s set is only eliminated by requiring non-trivial translation-invariant measures. So either (a) Vitali’s set wasn’t a problem in the first place or (b) probability spaces need translation-invariance.

No definition of probability spaces I’ve seen mentions translation invariance, leaving us with (a): Vitali’s set is not a problem for probabilists. Then:

1. Why do people mention Vitali's set while explaining the measure-theoretic formulation of probability?


2. Why do, in fact, we need measure theory for probability?


3. Are there any other paradoxes arising from uncountable sets to worry about that do apply to regular measures and probability?

probability-theory

share|cite|improve this question

edited Mar 12 at 23:12

Yatharth Agarwal

asked Mar 12 at 21:40

Yatharth AgarwalYatharth Agarwal

542418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Non-measurable set is not the only reason that we define a measure on $sigma$-algebra instead of the whole power set.
  $endgroup$
  – BigbearZzz
  Mar 12 at 21:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BigbearZzz Some of the other reasons I’ve seen are $sigma$-algebra as “information” which is fine. But still, what’s even the relevance of people mentioning the Vitali set if can’t be solved for all probability spaces anyway?
  $endgroup$
  – Yatharth Agarwal
  Mar 12 at 21:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Vitali set serves as an example of a Lebesgue non-measurable set, perhaps the first example of why the Lebesgue outer measure is not a measure on the power set of $Bbb R$. You can say that this has more to do with real analysis than probability theory.
  $endgroup$
  – BigbearZzz
  Mar 12 at 22:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should look into the construction of a probability measure via Caratheodory extension theorem. The presence/absence of Vitali set and translation invariant is not relevant at all, but still the theorem doesn't, in general, let you define a probability measure on the whole power set of your space anyway. This is another reason why $sigma$-algebra is preferred.
  $endgroup$
  – BigbearZzz
  Mar 12 at 22:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BigbearZzz That makes a ton of sense: These Vitali set examples are not actually relevant for probability theory; people just tend to mention them because they’re familiar with analysis, or they’re just trying to present an analogy of how things go wrong with uncountable sets! Does that sound fair?
  $endgroup$
  – Yatharth Agarwal
  Mar 12 at 22:16
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I’ve read several answers about how how crazy sets like Vitali’s set are (one of the reasons) why we need $sigma$-algebra and measure theory for probability spaces. However, Vitali’s set is only eliminated by requiring non-trivial translation-invariant measures. So either (a) Vitali’s set wasn’t a problem in the first place or (b) probability spaces need translation-invariance.

No definition of probability spaces I’ve seen mentions translation invariance, leaving us with (a): Vitali’s set is not a problem for probabilists. Then:

1. Why do people mention Vitali's set while explaining the measure-theoretic formulation of probability?


2. Why do, in fact, we need measure theory for probability?


3. Are there any other paradoxes arising from uncountable sets to worry about that do apply to regular measures and probability?

probability-theory

share|cite|improve this question

edited Mar 12 at 23:12

Yatharth Agarwal

asked Mar 12 at 21:40

Yatharth AgarwalYatharth Agarwal

542418

$endgroup$

share|cite|improve this question

edited Mar 12 at 23:12

Yatharth Agarwal

asked Mar 12 at 21:40

Yatharth AgarwalYatharth Agarwal

542418

share|cite|improve this question

edited Mar 12 at 23:12

Yatharth Agarwal

asked Mar 12 at 21:40

Yatharth AgarwalYatharth Agarwal

542418

asked Mar 12 at 21:40

Yatharth AgarwalYatharth Agarwal

542418

asked Mar 12 at 21:40

Yatharth AgarwalYatharth Agarwal

542418