Regarding dominated convergence theorem in Folland

[psie]

TL;DR

I am stuck at the very first sentence in the proof of Folland's version of the dominated convergence theorem. The wording confuses me and I'm not sure if he assumes the measure to be complete and the limiting function to be measurable.

The Dominated Convergence Theorem. Let $\{f_n\}$ be a sequence in $L^1$ such that (a) $f_n\to f$ a.e., and (b) there exists a nonnegative $g\in L^1$ such that $|f_n|\leq g$ a.e. for all $n$. Then $f\in L^1$ and $\int f=\lim\int f_n$.

Proof. $f$ is measurable (perhaps after redefinition on a null set) by Prop. 2.11 and 2.12, and since $|f|\leq g$ a.e., we have $f\in L^1$. ...

That's the first sentence in the proof. Prior to this Folland mentions the spaces $L^1(\overline{\mu})$ and $L^1(\mu)$ and how "we can (and shall) identify these spaces." (here $\overline{\mu}$ is the completion of $\mu$). The propositions mentioned in the proof read as follows:

Proposition. 2.11. The following implications are valid iff the measure $\mu$ 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 $(X, \mathcal M,m)$ be a measure space and let $(X,\overline{\mathcal M} ,\overline \mu)$ be its completion. If $f$ is an $\overline{\mathcal M}$-measurable function on $X$, there is an $\mathcal M$-measurable function $g$ such that $f=g$ $\overline{\mu}$-almost everywhere.

I'm really confused by Folland's first sentence in the proof of the dominated convergence theorem. My interpretation of Folland's theorem and first sentence is that he assumes $\{f_n\}\subset L^1(\overline{\mu})$, so by Prop. 2.11b $f$ is measurable. Now by Prop. 2.12, i.e. by redefinition on a null set, we can find $f_0\in L^1(\mu)$ such that $f=f_0$ a.e. Does this make sense to you? My interpretation assumes the measure to be complete; I don't see otherwise why he'd refer to Proposition. 2.11.

Grateful for any thoughts or comments.

 

[psie]

Here's an alternative interpretation, which I think is the correct one.

The measure $\mu$ is not assumed to be complete. As for the measurability of $f$; we know that $f_n$ converge to $f$ $\mu$-almost everywhere. Consider temporarily the completion of your $\sigma$-algebra and measure. The $f_n$'s are measurable with respect to that completion, and still converge $\overline{\mu}$-a.e to $f$. This is because if a function is measurable with respect to some $\sigma$-algebra, then also with respect to any larger $\sigma$-algebra (also, $\{x:f_n(x)\not\to f(x)\}$ is still a $\overline{\mu}$-null set). By Proposition 2.11, $f$ is measurable with respect to the complete $\sigma$-algebra. By Proposition 2.12, $f$ is equal $\overline{\mu}$-a.e. to a function that's measurable with respect to the incomplete $\sigma$-algebra (and we agree to denote that measurable function by $f$, which is what Folland means by redefining $f$ on a null set).

 

[fresh_42]

I'm not sure I understand your concerns. 2.11(b) makes $f$ measurable and $|f|<g$ guarantees $\int |f|<\int g<\infty .$

 

[psie]

I'm not sure I understand your concerns. 2.11(b) makes $f$ measurable and $|f|<g$ guarantees $\int |f|<\int g<\infty .$

2.11b makes $f$ measurable provided the measure is complete, but Folland does not state the measure is complete.

 

[FactChecker]

FYI, my Firefox browser is not displaying the Latex overline. My Chrome and Microsoft Edge browsers display it properly. Readers might want to not use Firefox for this thread.

 

[martinbn]

2.12 says that in questions about $L^1$ it doesn't matter if you work with the given measure or its completion. So assume the measure is complete.