# measureability of a function and its relation to the completion of the measure

I am trying to understand the need for proposition 2.12 in Folland.

Proposition 2.11: The following implications are valid iff the measure is complete:

a. If $$f$$ is measurable and $$f=g$$ $$\mu$$-a.e., then $$g$$ is measurable

b. If $$f_n$$ is measurable for $$n\in \mathbb{N}$$ and $$f_n\to f$$ $$\mu$$-a.e., then $$f$$ is measurable

Proposition 2.12: Let $$(X,\mathcal{M},\mu)$$ be a measure space and let $$(X,\overline{\mathcal{M}},\overline{\mu})$$ be its completion. If $$f$$ is $$\overline{\mathcal{M}}$$ measurable on $$X$$, there is a $$\mathcal{M}$$-measurable function $$g$$ such that $$f=g$$ $$\bar{\mu}$$-almost everywhere.

Questions

Let $$(X,\mathcal{M},\mu)$$ be a measure space. Let $$f:X\to \mathbb{R}$$. If $$f$$ is measurable then, for every Borel set $$B$$, $$f^{-1}(B)$$ is measurable. But doesn't this imply that $$f$$ is $$\overline{\mu}$$ measureable?

If that were true, then if $$f=g$$ $$\mu$$-a.e. we have that $$f=g$$ $$\bar{\mu}$$-a.e.. Thus $$g$$ would be $$\bar{\mu}$$ measurable (prop 2.11).

Is it possible to find a measure $$v$$, measurable function $$f$$, such that $$f=g$$ a.e. but $$g$$ is not measurable?

I am trying to understand why Folland says "one is unlikely to commit any serious blunders by forgetting to worry about completeness of the measure."

1) If $$f=g$$ a.e. then we can think of $$g$$ as being measurable (as it is measurable in the completion) minus a null set by prop 2.12. This becomes important when $$g=\lim f_n$$ for example.
2) Theorem 2.12 provides a 1 to 1 correspondence between $$L^{1}(\mu)$$ and $$L^{1}(\bar{\mu})$$.