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.


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."


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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