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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."

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Being measurable does imply measurability in the completion. There are at least two reasons why these this proposition is important.

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 $L^{1}(\bar{\mu})$.

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