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It seems to me, on $\mathbb{R}$ or $[0, 2\pi]$, the completion of Riemann integrable functions should be $L^1$. However, on p.76 of Stein's Fourier analysis: an introduction, it is also claimed the completion is $L^2$.

I'm a little confused. Should I understand it as, we can equip different norms on the vector space of Riemann integrable functions. Then, if we equip the norm as the Lebesgue integral of absolute value, the completion is $L^1$. But if we equip an inner-product and make it into a pre-Hilbert space, then the completion becomes $L^2$?

Does this mean other spaces can also be the completion of Riemann integrable functions?

Thanks in advance.

user760
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    Related: https://math.stackexchange.com/q/4675674/1104384 Completions are always tied to a metric (here a norm), so of course the completion will change based on the norm you imbued the space of Riemann integrable functions with. – Bruno B Jun 19 '23 at 19:11
  • One or both of these questions deserve actual answers @BrunoB. – Lee Mosher Jun 19 '23 at 19:18
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    Depends on the norm you are considering. Both answers are correct. In fact, the coalition of the space of Riemann integrable functions under the norm $|f|_p=\Big(\int^b_a |f|\big)^{1/p}$, $0\leq p<\infty$ is $\mathcal{L}_p([a,b]$ (identifying a.s. zero functions). The reason is that all continuous functions are Riemann integrable. – Mittens Jun 19 '23 at 19:38
  • @LeeMosher I mean, the one I linked was pretty much just helping OP with hints then saying "yes your solution works", I don't really think it needs an answer (though I won't prevent anyone from making one!), but sure I can turn my comment here into an answer I guess. Don't really know what else to add though, is the thing. – Bruno B Jun 19 '23 at 19:55
  • Well, small answers are sometimes really good. – Lee Mosher Jun 19 '23 at 20:53

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For a given set there is almost never one single completion, as you need to set a metric (or norm, but it's tied to that as you can consider the metric induced by the norm) on your set to speak of completeness and completions, and each metric may provide a different completion.

For instance, due to the density of continuous functions in every $L^p$ for the $L^p$ norm, and the fact that Riemann integrable functions contain continuous functions and are contained in $L^p$, the completion of the Riemann integrable functions for the $L^p$ norm is $L^p$. In fact, there is even a metric, that of uniform convergence, for which the Riemann integrable functions form a complete space themselves, meaning they are their own completion for that metric.

Bruno B
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