Let C b ( R ) be the set of real-valued bounded continuous functions on R . The σ −−algebra generated by C b ( R ) is the smallest σ −−algebra, say σ ( C b ( R ) ), such that all functions in C b ( R ) are measurable with respect to σ ( C b ( R ) ).Do the following equations hold?1. σ ( C b ( R ) ) = σ ( { f − 1 ( A ) : A ∈ B ( R ) , f ∈ C b ( R ) } ) .2. σ ( C b ( R ) ) = B ( R )

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Let       C    b    (      R    ) be the set of real-valued bounded continuous functions on       R  . The   σ  −−algebra generated by       C    b    (      R    ) is the smallest   σ  −−algebra, say   σ  (      C    b    (      R    )  ), such that all functions in       C    b    (      R    ) are measurable with respect to   σ  (      C    b    (      R    )  ).Do the following equations hold?1.   σ  (      C    b    (      R    )  )  =  σ  (  {      f          −      1        (  A  )  :  A  ∈      B    (      R    )  ,  f  ∈      C    b    (      R    )  }  )  .2.   σ  (      C    b    (      R    )  )  =      B    (      R    )

barbesdestyle2k · Accepted Answer

You are right on both counts.More generally, let   {      f    i    :  X  →      X    i        }          i      ∈      I       be a collection of functions that take values in measurable spaces   (      X    i    ,            F        i    ). The smallest   σ-algebra on X such that all       f    i  &#039;s become measurable is precisely  σ  {      f    i          −      1        (      A    i    )  :  i  ∈  I  ,      A    i    ∈            F        i    }  .Your first claim follows from this general fact.Next, let us show that  σ  (      C    b    (      R    )  )  =      B    (      R    )  .Clearly the inclusion   ⊇ holds, because any closed interval of       R   is the Borel-preimage of a function in       C    b    (      R    ) [If the interval is [a,b], consider the function   f  :      R    →      R   defined by f(x)=x if   a  ≤  x  ≤  b and extend it to a continuous bounded function on all of       R   such f does not attain the values [a,b] outside the interval [a,b] (draw a picture if you don&#039;t see that this exists).] Then recall that closed intervals generate the Borel   σ-algebra on       R  .Conversely, clearly all preimages       f          −      1        (  A  ) where   A is Borel and   f  ∈      C    b    (      R    ) are again Borel, because continuous functions are measurable. Hence, the inclusion   ⊆ also holds.

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