Talk:Riemann–Lebesgue lemma

which "special cases" in proof?

The first sentence of the proof is confusing, and it is not clear which "special cases" it refers to, and which if any of them is the "first one". Perhaps it could be replaced with something like the following:

"The proof can be organized into steps, proving increasingly general special cases; the 4th step extends the result to the original formulation."

but personally I think that it might be better to remove that sentence altogether --AmitAronovitch (talk) 18:35, 21 May 2010 (UTC)

The statement of the Theorem seems to be flawed. f is assumed to be a measurable function from R to C. But the proof deals with an interval [a,b]. Should we say f:[a,b] to C? or should the proof omit [a,b]? —Preceding unsigned comment added by 69.142.45.52 (talk) 01:22, 29 April 2011 (UTC)

Also the proof is only for f on R, while the statement of the theorem has f on R^d — Preceding unsigned comment added by Bzhao2017 (talk • contribs) 04:10, 2 December 2019 (UTC)

Abstract measure spaces

The proof in the research gate article is wrong, the reference was removed. — Preceding unsigned comment added by 81.243.243.97 (talk) 21:10, 21 November 2018 (UTC)

Does this proof work?

The lemma is first proof for step functions but is applied to g a simple function. It is not explain how we can jump from step function to simple function and it seems unclear (and difficult) to me. — Preceding unsigned comment added by 2001:861:3DC3:44B0:A07:DFC0:8480:1E9 (talk) 12:59, 12 March 2020 (UTC)

Deficient proof

The theorem says the domain of $f$ is $\mathbb {R} ^{d}$ but the proof works only for $\mathbb {R} ^{1}.$ Michael Hardy (talk) 21:50, 2 August 2020 (UTC)

Does this hold for complex z?

This proof only seems to work for real z. In fact, I think that along the imaginary axis the theorem does not hold.

Perhaps this should be stated explicitly — Preceding unsigned comment added by 83.130.90.163 (talk) 13:16, 26 May 2021 (UTC)