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How can a probability measure satisfy the finite super additivity property?

Semialgebra logical errorfinite additivity&countable additivityShowing countable additivitiy of Lebesgue measureIs $P(bigcup_k=1^infty A_k)=lim_nrightarrowinftyP(bigcup_k=1^n A_k)$ equivalent to countable additivity?A Special Property of Finite Probability Space.Probability measure is countably additive over almost disjoint sets.Show that the function $mathbbQ(A) = sum _ i:omega _ i in A ^,p_i$ is a probability measure on $(Omega, mathcalF)$Proving Countable Additivity From Finite AdditivityFrom additivity of the measure, couldn't we prove the $sigma -$additivity?Countable additivity does not imply continuity at $emptyset$ for non-finite measure

1

$begingroup$

When introducing the Extention Theorem the book said:

"Let $J$ be a semialgebra of subsets of $Omega$. Let P:$J$->[0,1] with P($emptyset$) = 0 and P($Omega$) = 1, satisfying the finite super additive property that.

P($bigcup_i=1^kA_i$)$geqsum_i=1^kP(A_i)$ whenever $A_1,...,A_k epsilon J$, and $bigcup_i=1^kA_iepsilon J $ and the $A_i are disjoint, and also the countable monotonicity property. "

I don't understand how this could be a probability measure. If I have these disjoint events, it shouldn't matter whether I take the probability of the union or the sum of the probability of the individual events. Under what circumstances would P($bigcup_i=1^kA_i$)$>sum_i=1^kP(A_i)$ be true?

probability probability-theory measure-theory

share|cite|improve this question

asked Mar 16 at 10:47

QwertfordQwertford

323212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If only $=$ is satisfied still $geq$ is true.
  $endgroup$
  – user647486
  Mar 16 at 10:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
  $endgroup$
  – GEdgar
  Mar 16 at 11:27
  
  

  
  
  
  

add a comment | 

1

$begingroup$

When introducing the Extention Theorem the book said:

"Let $J$ be a semialgebra of subsets of $Omega$. Let P:$J$->[0,1] with P($emptyset$) = 0 and P($Omega$) = 1, satisfying the finite super additive property that.

P($bigcup_i=1^kA_i$)$geqsum_i=1^kP(A_i)$ whenever $A_1,...,A_k epsilon J$, and $bigcup_i=1^kA_iepsilon J $ and the $A_i are disjoint, and also the countable monotonicity property. "

I don't understand how this could be a probability measure. If I have these disjoint events, it shouldn't matter whether I take the probability of the union or the sum of the probability of the individual events. Under what circumstances would P($bigcup_i=1^kA_i$)$>sum_i=1^kP(A_i)$ be true?

probability probability-theory measure-theory

share|cite|improve this question

asked Mar 16 at 10:47

QwertfordQwertford

323212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If only $=$ is satisfied still $geq$ is true.
  $endgroup$
  – user647486
  Mar 16 at 10:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So far, $P$ is not a probability measure. It is merely a set-function with the stated properties. You are setting up to construct a measure $mu$, say. The strict inequality can hold only for non-measurable sets. If your original set-function $P$ has instances where $>$ holds, then either (1) some of the sets of $J$ will turn out to be non-measurable for $mu$ or (2) for some elements $E in J$ will we will get $mu(E) ne P(E)$.
  $endgroup$
  – GEdgar
  Mar 16 at 11:27
  
  

  
  
  
  

add a comment | 

$begingroup$

When introducing the Extention Theorem the book said:

"Let $J$ be a semialgebra of subsets of $Omega$. Let P:$J$->[0,1] with P($emptyset$) = 0 and P($Omega$) = 1, satisfying the finite super additive property that.

P($bigcup_i=1^kA_i$)$geqsum_i=1^kP(A_i)$ whenever $A_1,...,A_k epsilon J$, and $bigcup_i=1^kA_iepsilon J $ and the $A_i are disjoint, and also the countable monotonicity property. "

I don't understand how this could be a probability measure. If I have these disjoint events, it shouldn't matter whether I take the probability of the union or the sum of the probability of the individual events. Under what circumstances would P($bigcup_i=1^kA_i$)$>sum_i=1^kP(A_i)$ be true?

probability probability-theory measure-theory

share|cite|improve this question

asked Mar 16 at 10:47

QwertfordQwertford

323212

$endgroup$

share|cite|improve this question

asked Mar 16 at 10:47

QwertfordQwertford

323212

share|cite|improve this question

asked Mar 16 at 10:47

QwertfordQwertford

323212

asked Mar 16 at 10:47

QwertfordQwertford

323212

asked Mar 16 at 10:47

QwertfordQwertford

323212