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.
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.