Motivation: The composition of 2 Lebesgue measurable functions need not be measurable. This problem can be dealt with in a case by case basis( like with convolutions). Or as Big Rudin does, apply lüsins theorem to get borel measurable functions that differ only by a set of measure zero from the original 2 functions. The composition Of these functions will differ from the composition of the original by a set of measutev 0, and will also be borel and hence Lebesgue measureable.
This begs the question ' can two functions differ by a set of measure 0 with one measurable and the other not measurable, even if the measure is complete?'
Suppose $f, g: \mathbb R^d \to \mathbb R$ agree a.e. and $f$ is measurable. Then $\{f>\lambda\}$ and $\{g>\lambda\}$ differ by a set of measure zero, so there exist null sets $E, F$ with $\{g > \lambda\} \cap F = \emptyset$ such that $$\{f > \lambda\} \cup E = \{g>\lambda\} \cup F.$$
Since the union of measurable sets is measurable, $\{f > \lambda\} \cup E$ is measurable, so $\{g>\lambda\} \cup F$ is measurable. Then $\{g>\lambda\} \cup F \setminus F = \{g>\lambda\}$ is measurable, so $g$ is measurable.