Regarding dominated convergence theorem in Folland

In summary, the dominated convergence theorem in Folland addresses the conditions under which the limit of a sequence of measurable functions can be interchanged with the integral. It states that if a sequence of functions converges pointwise to a function and is dominated by an integrable function, then the integral of the limit equals the limit of the integrals. This theorem is essential in analysis as it provides a powerful tool for evaluating limits and integrals, particularly in probability and statistical applications.
  • #1
psie
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TL;DR Summary
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 be a sequence in such that (a) a.e., and (b) there exists a nonnegative such that a.e. for all . Then and .

Proof. is measurable (perhaps after redefinition on a null set) by Prop. 2.11 and 2.12, and since a.e., we have . ...

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

Proposition. 2.11. The following implications are valid iff the measure is complete:
a) If is measurable and -a.e., then is measurable.
b) If is measurable for and -a.e., then is measurable.

Proposition. 2.12. Let be a measure space and let be its completion. If is an -measurable function on , there is an -measurable function such that -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 , so by Prop. 2.11b is measurable. Now by Prop. 2.12, i.e. by redefinition on a null set, we can find 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.
 
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  • #2
Here's an alternative interpretation, which I think is the correct one.

The measure is not assumed to be complete. As for the measurability of ; we know that converge to -almost everywhere. Consider temporarily the completion of your -algebra and measure. The 's are measurable with respect to that completion, and still converge -a.e to . This is because if a function is measurable with respect to some -algebra, then also with respect to any larger -algebra (also, is still a -null set). By Proposition 2.11, is measurable with respect to the complete -algebra. By Proposition 2.12, is equal -a.e. to a function that's measurable with respect to the incomplete -algebra (and we agree to denote that measurable function by , which is what Folland means by redefining on a null set).
 
  • #3
I'm not sure I understand your concerns. 2.11(b) makes measurable and guarantees
 
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  • #4
fresh_42 said:
I'm not sure I understand your concerns. 2.11(b) makes measurable and guarantees
2.11b makes measurable provided the measure is complete, but Folland does not state the measure is complete.
 
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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.
 
  • #6
2.12 says that in questions about it doesn't matter if you work with the given measure or its completion. So assume the measure is complete.
 
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