I am trying to find an example that shows the image of a measurable function is not measurable.Here is what I found: Let X={0,1} with σ-algebra { ∅ , X }. Let f : X → X and f(x)=0. Then f(X)={0} is not measurable.But how do we know that f is measurable in the first place? Obviously, we have f − 1 ( { } ) = { }. However, how can we say that f − 1 ( X ) = f − 1 ( { 0 , 1 } ) = X = { 0 , 1 }? I mean f is not surjective; it is impossible that f(x)=1.

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I am trying to find an example that shows the image of a measurable function is not measurable.Here is what I found: Let X={0,1} with   σ-algebra   {  ∅  ,  X  }. Let   f  :  X  →  X and f(x)=0. Then f(X)={0} is not measurable.But how do we know that f is measurable in the first place? Obviously, we have       f          −      1        (  {  }  )  =  {  }. However, how can we say that       f          −      1        (  X  )  =      f          −      1        (  {  0  ,  1  }  )  =  X  =  {  0  ,  1  }? I mean f is not surjective; it is impossible that f(x)=1.

Dwayne James · Accepted Answer

Remember that       f          −      1        (  S  ) just means &#039;&#039;all the points   x  ∈  X such that   f  (  x  )  ∈  S&#039;&#039;. You know that   f  (  0  )  ∈  X and   f  (  1  )  ∈  X, so   0  ,  1  ∈      f          −      1        (  X  ). It&#039;s true that       f          −      1        (  {  1  }  )  =  ∅, but that isn&#039;t important.

I am trying to find an example that shows the image of a measurable function is not measurable. He

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