My proof is as follow (S is the event space in the probability space of an experiment):If A ⊂ B ∈ S then A c ∩ B ∈ S so this means that A c ∩ B contains B but not A. This implies that B ∖ A ∈ SI'm not sure if this prove is correct and if the rigor is strong enough.

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My proof is as follow (S is the event space in the probability space of an experiment):If   A  ⊂  B  ∈  S then       A    c    ∩  B  ∈  S so this means that       A    c    ∩  B contains B but not A. This implies that   B  ∖  A  ∈  SI&#039;m not sure if this prove is correct and if the rigor is strong enough.

Jayvion Tyler · Accepted Answer

You don&#039;t even need that   A  ⊆  B. As you mentioned, it is sufficient to see that  B  ∖  A  =  B  ∩      A    ∁  and to recall that the sigma-algebra   S is closed under taking complements and intersections.

slijmigrd · Accepted Answer

Let   C be the set of elementary events (the largest element of the sigma algebra   S). As with each set in   S also its complement is in   S, we have   C  ∖  B  ∈  S. As with any two sets, also its union is in   S, we have  B  ∖  A  =  (  A  ∪  (  C  ∖  B  )      )    c    ∈  S  .[edit: the complement       A    c   of any set   A is the same as   C  ∖  A in the example]

My proof is as follow (S is the event space in the probability space of an experiment): If A

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