Classifying measurability of functions for $\mathcal{L}(\mathbb{R})$

Question

In a recent exam we constructed a whole lot of measure-theory-related counterexamples, such as a continuous function which isn't measurable for the Lebesgue-$\sigma$-algebra $\mathcal{L}(\mathbb{R})$ over $\mathbb{R}$. Although I understand why this is possible, it got me thinking: Let $f:(\mathbb{R},\mathcal{L}(\mathbb{R}))\to (\mathbb{R},\mathcal{L}(\mathbb{R}))$ be some function. What are sufficient conditions to guarantee measurability of $f$? Can we even classify such functions with a simple-to-verify condition? And do we sometimes need this stronger measurability compared with the weaker measurability if the range is equipped with the Borel-$\sigma$-algebra $\mathfrak{B}(\mathbb{R})$, or is this more of a curious artifact? Thanks for any clarifications.

Answer 1

1) By sard’s theorem, Diffeomorphism preserves zero measure sets. So one of the sufficient condition I can think of is diffeomorphism to get measurability in stronger sense.

2) Only advantage I can see with this stronger measurability is composition of stronger measurable functions is again stronger measurable.

As you know contionous functions are not stronger measurable. We stick to weaker definition of measurability because we want give definition of integration for larger class of functions (at least larger than class of Riemann integrable functions).