Is composition of measurable functions measurable?

We know that if f:ER is a Lebesgue-measurable function and g:RR is a continuous function, then gf is Lebesgue-measurable. Can one replace the continuous function g by a Lebesgue-measurable function without affecting the validity of the previous result?

Answer

Here is the standard example:

Let f:[0,1][0,1] be the Cantor–Lebesgue function. This is a monotonic and continuous function, and the image f(C) of the Cantor set C is all of [0,1]. Define g(x)=x+f(x). Then g:[0,1][0,2] is a strictly monotonic and continuous map, so its inverse h=g1 is continuous, too.

Observe that g(C) measure one in [0,2]: this is because f is constant on every interval in the complement of C, so g maps such an interval to an interval of the same length. It follows that there is a non-Lebesgue measurable subset A of g(C) (Vitali’s theorem: a subset of R is a Lebesgue null set if and only if all its subsets are Lebesgue measurable).

Put B=g1(A)C. Then B is a Lebesgue measurable set as a subset of the Lebesgue null set C, so the characteristic function 1B of B is Lebesgue measurable.

The function k=1Bh is the composition of the Lebesgue measurable function 1B and and the continuous function h, but k is not Lebesgue measurable, since k1(1)=(1Bh)1(1)=h1(B)=g(B)=A.

Attribution
Source : Link , Question Author : user54843 , Answer Author : Mirjam

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