Fix some probability space ( Ω , F , P ), and let A ⊂ F be a sub- σ-algebra. Suppose that g , h : Ω → R are two P -integrable and A -measurable functions.I need help understanding why { h &gt; g } ∈ A ? I understand that since g and h both are A -measurable, it holds that for all B ∈ B ( R ): h − 1 ( B ) ∈ A , g − 1 ( B ) ∈ A .However, I can't seem to make the connection as to why { h &gt; g } ∈ A also holds true.

Question

Fix some probability space   (  Ω  ,      F    ,      P    ), and let       A    ⊂      F   be a sub-  σ-algebra. Suppose that   g  ,  h  :  Ω  →      R   are two       P  -integrable and       A  -measurable functions.I need help understanding why   {  h  &amp;gt;  g  }  ∈      A  ? I understand that since   g and   h both are       A  -measurable, it holds that for all   B  ∈      B    (      R    ):      h          −      1        (  B  )  ∈      A    ,        g          −      1        (  B  )  ∈      A    .However, I can&#039;t seem to make the connection as to why   {  h  &amp;gt;  g  }  ∈      A   also holds true.

Keegan Barry · Accepted Answer

{  h  &amp;gt;  g  }  =  {  h  −  g  &amp;gt;  0  }  ∈      A    because if   h and   g are       A  -measurable, then the difference   h  −  g is also       A  -measurable.To see this, note that for all   t  ∈      R  ,   h  −  g  &amp;lt;  t    ⟺    h  &amp;lt;  t  +  g    ⟺    h  &amp;lt;  q  &amp;lt;  t  +  g for some rational   q. So  {  h  −  g  &amp;lt;  t  }  =      ⋃          q      ∈              Q              {  h  &amp;lt;  q  &amp;lt;  t  +  g  }  =      ⋃          q      ∈              Q              {  h  &amp;lt;  q  }  ∩  {  g  &amp;gt;  q  −  t  }Hence   {  h  −  g  &amp;lt;  t  }  ∈      A   as a countable union of elements of       A  , and therefore   h  −  g is       A  -measurable.

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