Given a probability space ( Ω , F , P ) and a random variable X defined on it, s.t. X ≥ 0, I want to show that the mapping ( ω , t ) ↦ 1 { X ( ω ) ≥ t } is measurable on the measure space ( Ω × [ 0 , ∞ ) , F × B ( [ 0 , ∞ ) ) , P ⊗ λ ), for λ denoting the Lebesgue measure.My thoughts were to simply use the definition and look at the preimages of this mapping. One would then have, e.g. 1 { X ( ω ) ≥ t } − 1 ( { 0 } ) = { ( ω , t ) : X ( ω ) &lt; t } = { ( ω , t ) : ω ∈ X − 1 ( [ 0 , t ) ) } .Since X is a random variable, it is measurable, so X − 1 ( [ 0 , t ) ) ∈ F . However, I cannot see at the moment why { ( ω , t ) : ω ∈ X − 1 ( [ 0 , t ) ) } belongs to the product σ-algebra, since one coordinate depends on the other. I'm grateful for any hints.

Question

Given a probability space   (  Ω  ,      F    ,  P  ) and a random variable   X defined on it, s.t.   X  ≥  0, I want to show that the mapping  (  ω  ,  t  )  ↦            1              {      X      (      ω      )      ≥      t      }      is measurable on the measure space   (  Ω  ×  [  0  ,  ∞  )  ,      F    ×      B    (  [  0  ,  ∞  )  )  ,  P  ⊗  λ  ), for   λ denoting the Lebesgue measure.My thoughts were to simply use the definition and look at the preimages of this mapping. One would then have, e.g.            1              {      X      (      ω      )      ≥      t      }              −      1        (  {  0  }  )  =  {  (  ω  ,  t  )  :  X  (  ω  )  &amp;lt;  t  }  =  {  (  ω  ,  t  )  :  ω  ∈      X          −      1        (  [  0  ,  t  )  )  }  .Since   X is a random variable, it is measurable, so       X          −      1        (  [  0  ,  t  )  )  ∈      F  . However, I cannot see at the moment why   {  (  ω  ,  t  )  :  ω  ∈      X          −      1        (  [  0  ,  t  )  )  } belongs to the product   σ-algebra, since one coordinate depends on the other. I&#039;m grateful for any hints.

Ryan Fitzgerald · Accepted Answer

If   a  &amp;lt;  b then there exists   q  ∈      Q   such that   a  &amp;lt;  r  &amp;lt;  b. Thus  {  (  ω  ,  t  )  :  X  (  ω  )  &amp;lt;  t  }  =      ⋃          r      ∈              Q              {  (  ω  ,  t  )  :  X  (  ω  )  &amp;lt;  r  &amp;lt;  t  }where  {  (  ω  ,  t  )  :  X  (  ω  )  &amp;lt;  r  &amp;lt;  t  }  =  {  (  ω  ,  t  )  :  X  (  ω  )  &amp;lt;  r  }  ∩  {  (  ω  ,  t  )  :  r  &amp;lt;  t  }Futher  {  (  ω  ,  t  )  :  X  (  ω  )  &amp;lt;  r  }  =  {      X          −      1        (  (  ∞  ,  r  )  )  }  ×  [  0  ,  ∞  )  ∈      F    ×      B    (  [  0  ,  ∞  )  )  {  (  ω  ,  t  )  :  r  &amp;lt;  t  }  =  Ω  ×  (  r  ,  +  ∞  )  ∈      F    ×      B    (  [  0  ,  ∞  )  )q.e.d.

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