We want to prove the following implication : A ∈ F S ⟹ A ∩ { S ≤ T } ∈ F T Since A ∈ F S , we have A ∈ F ∞ and A ∩ { S ≤ t } ∈ F t .We will then show that A ∩ { S ≤ T } ∈ F T Meaning : A ∩ { S ≤ T } ∩ { T ≤ t } ∈ F t We have : A ∩ { S ≤ T } ∩ { T ≤ t } = A ∩ { S ≤ t } ∩ { T ≤ t } ∩ { S ∧ t ≤ T ∧ t }The ''chosen'' set { S ∧ t ≤ T ∧ t }, according to the book, guarantees that the stopping time S will be less or equal to T, and at the same time, the chosen set is in fact F t -measurable, which helps a lot in the proofIt's not very clear to me where it did come from.I understand that { S ≤ T } ≠ { S ≤ t } ∩ { T ≤ t } , ∀ t ∈ R + but the exact choice of { S ∧ t ≤ T ∧ t } is still not obvious.

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We want to prove the following implication :  A  ∈            F        S      ⟹    A  ∩  {  S  ≤  T  }  ∈            F        T  Since   A  ∈            F        S  , we have   A  ∈            F        ∞   and   A  ∩  {  S  ≤  t  }  ∈            F        t  .We will then show that  A  ∩  {  S  ≤  T  }  ∈            F        T  Meaning :  A  ∩  {  S  ≤  T  }  ∩  {  T  ≤  t  }  ∈            F        t  We have :  A  ∩  {  S  ≤  T  }  ∩  {  T  ≤  t  }  =  A  ∩  {  S  ≤  t  }  ∩  {  T  ≤  t  }  ∩  {  S  ∧  t  ≤  T  ∧  t  }The &#039;&#039;chosen&#039;&#039; set   {  S  ∧  t  ≤  T  ∧  t  }, according to the book, guarantees that the stopping time   S will be less or equal to   T, and at the same time, the chosen set is in fact             F        t  -measurable, which helps a lot in the proofIt&#039;s not very clear to me where it did come from.I understand that  {  S  ≤  T  }  ≠  {  S  ≤  t  }  ∩  {  T  ≤  t  }  ,    ∀  t  ∈            R        +  but the exact choice of   {  S  ∧  t  ≤  T  ∧  t  } is still not obvious.

oedfeuonbk203 · Accepted Answer

(  T  ∧  t  ≤  u  )  =  (  T  ≤  u  ∧  t  )  ∪  (  T  &amp;gt;  t  ,  t  ≤  u  ). Note that   (  T  &amp;gt;  t  ,  t  ≤  u  ) is empty if   u  &amp;lt;  t and it is   (  T  &amp;gt;  t  ) if   u  ≥  t. Conclude that   T  ∧  t is meaurable on   (  Ω  ,            F        t    ). And so is   S  ∧  t by the same argument. It follows that   T  ∧  t  −  S  ∧  t is also measurable on this sapce and   (  T  ∧  t  −  S  ∧  t  ≥  0  ) is measurable on   (  Ω  ,            F        t    ).

We want to prove the following implication : A ∈ <mrow class="MJ

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