The question in the title has been answered by this MOF post. One reason is that Lebesgue measurable function is bad since may not be Lebesgue measurable even if function and random variable are both Lebesgue measurable. A good explanation is from Exercise 2.9.d of real analysis book by [Folland 1999], which will be given below.
Before we start, we define the following notations: Let and be Borel and Lebesgue -algebra of real number set , respectively. Since is the completion of , all subset of zero Lebesgue measure is again zero measure and belongs to . Therfore, any subset of Cantor set is Lebesgue measurable set. A function is said to be Lebesgue measurable if .
[Problem] (Ex.2.9.d of [Fol99]). There exists a Lebesgue measurable function and a continuous function (of course Lebesgue measurable) such that is not a Lebesgue measurable.
Proof: Let be the Cantor function and be . According to Ex2.9a-c of [Fol99], there exists a subset of Cantor set satisfying
Note that (But ). Consider an indicator function . By definition, is Lebesgue measurable function and
Now satisfies the requirement, since its inverse image of is , i.e