Let X ( ω ) be a continuous random variable defined on a probability space ( Ω , F , P ).How does the measurability condition of this random variable (i.e. ∀ r ∈ R , { ω : X ( ω ) ≤ r } ∈ F ) cohere with the σ-additivity of the σ-algebra F ?I understand that a σ-algebra requires that the union of countable many elements of F must also be contained in F . But if we consider a continuous random variable, then the events that the σ-algebra F must contain are uncountable, since r can be any value in R and X ( ω ) is a continuous r.v.What am I missing?

Question

Let   X  (  ω  ) be a continuous random variable defined on a probability space   (  Ω  ,      F    ,      P    ).How does the measurability condition of this random variable (i.e.   ∀  r  ∈      R    ,  {  ω  :  X  (  ω  )  ≤  r  }  ∈      F  ) cohere with the   σ-additivity of the   σ-algebra       F  ?I understand that a   σ-algebra requires that the union of countable many elements of       F   must also be contained in       F  . But if we consider a continuous random variable, then the events that the   σ-algebra       F   must contain are uncountable, since   r can be any value in       R   and   X  (  ω  ) is a continuous r.v.What am I missing?

diamondogsaz · Accepted Answer

A   σ-algebra is closed under countable unions. This does not mean that its elements cannot be uncountable, or that it&#039;s impossible that some (or even all) uncountable unions of elements in the   σ-algebra are in the   σ-algebra.For example, if   Ω is an arbitrary nonempty set (possibly an uncountable one, like       R   or             R                      R            ), both   {  ∅  ,  Ω  } and       P    (  Ω  ) (the whole power set) are perfectly good   σ-algebras of subsets of   Ω, and both of them are closed under arbitrary (even uncountable) unions and have uncountable elements, if   Ω is not countable.In fact, unless   Ω is countable, there is always an uncountable elements in any   σ-algebra: namely,   Ω itself.Further, if       F   is a   σ-algebra of subsets of an uncountable set   Ω, and       F   contains singletons, then there is always at least one uncountable disjoint union of elements of       F   which is in       F  , namely,   Ω  =      ⋃          ω      ∈      Ω        {  ω  }.