I really could use a hand on this questionLet f : R ↦ f ( R ) := E, a continuous function. Let μ e be a probability on ( E , B ( E ) ).Show that there exists a measure ν on R such that : ∀ B ∈ B ( E ) , μ e ( B ) = ν ( f − 1 ( B ) )I really can't think of a single way to approache this. The obvious choice would seem to be ν ( B ) = μ e ( f ( B ) ), but this doesn't seem to work. Another idea would be to construct an outer measure with one of the methods available but I don't know how I would go about itI thank you all in advance

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I really could use a hand on this questionLet   f  :      R    ↦  f  (      R    )  :=  E, a continuous function. Let       μ    e   be a probability on   (  E  ,  B  (  E  )  ).Show that there exists a measure   ν on       R   such that :  ∀  B  ∈  B  (  E  )  ,      μ    e    (  B  )  =  ν  (      f          −      1        (  B  )  )I really can&#039;t think of a single way to approache this. The obvious choice would seem to be   ν  (  B  )  =      μ    e    (  f  (  B  )  ), but this doesn&#039;t seem to work. Another idea would be to construct an outer measure with one of the methods available but I don&#039;t know how I would go about itI thank you all in advance

persstemc1 · Accepted Answer

Hint:  E is an interval (possibly a single point) since   f is continuous. Let       A    =  {      f          −      1        (  B  )  :  B  ∈      B    (  E  )  }. This is a   σ-algebra, define   ν  (      f          −      1        (  B  )  )  =  μ  (  B  ).  ν is well defined since   f  :      R    →  E is surjective; thus, if       f          −      1        (  B  )  =      f          −      1        (      B    ′    ), then   B  =  f  (      f          −      1        (  B  )  )  =  f  (      f          −      1        (      B    ′    )  )  =      B    ′  .If   {      f          −      1        (      B    n    )  :  n  ∈      N    } are disjoint, then   {      B    n    :  n  ∈      N    } are disjoint since   f is surjective. Thus, if the   {      B    n    :  n  ∈      N    }  ⊂      B    (  E  ),  ν            (            ⋃    n        f          −      1        (      B    n    )            )        =  ν            (            f          −      1        (      ⋃    n        B    n    )            )        =  μ  (      ⋃    n        B    n    )  =      ∑    n    μ  (      B    n    )  =      ∑    n    ν  (      f          −      1        (      B    n    )  )

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