Why do probabilists take random variables to be Borel (and not Lebesgue) measurable?

The question in the title has been answered by this MOF post. One reason is that Lebesgue measurable function is bad since {f(X)} may not be Lebesgue measurable even if function {f} and random variable {X} 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 {\mathcal B} and {\mathcal L} be Borel and Lebesgue {\sigma}-algebra of real number set {\mathbb R}, respectively. Since {\mathcal L} is the completion of {\mathcal B}, all subset of zero Lebesgue measure is again zero measure and belongs to {\mathcal L}. Therfore, any subset of Cantor set is Lebesgue measurable set. A function {f: \mathbb R \mapsto \mathbb R} is said to be Lebesgue measurable if {f^{-1}(B) \in \mathcal L}.

[Problem] (Ex.2.9.d of [Fol99]). There exists a Lebesgue measurable function {h} and a continuous function {g} (of course Lebesgue measurable) such that {h\circ g} is not a Lebesgue measurable.

Proof: Let {f:[0,1] \mapsto [0,1]} be the Cantor function and {g:[0,1] \mapsto [0,2]} be {g(x) = f(x) + x}. According to Ex2.9a-c of [Fol99], there exists a subset {A} of Cantor set satisfying

\displaystyle g(A) \notin \mathcal L.

Note that {A\in \mathcal L} (But {A\notin \mathcal B}). Consider an indicator function {h(x) = I(x\in A)}. By definition, {h} is Lebesgue measurable function and

\displaystyle h^{-1} (1) = A \in \mathcal L.

Now {h\circ g} satisfies the requirement, since its inverse image of {\{1\}} is {g^{-1}(A) \notin \mathcal L}, i.e

\displaystyle (h\circ g)^{-1}(1) = g^{-1}\circ h^{-1}(1) = g^{-1}(A) \notin \mathcal L.

\Box

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