The space of square Lebesgue integrable functions is said to be a Hilbert space. Why is if the integral is Riemann then this is not a Hilbert space? In other words, why not the space of Riemann square integrable functions not a Hilbert space? An example or two would be great. Thanks.
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It would be incomplete. See this. – David Mitra Feb 15 '14 at 17:05
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A $L^2$-Riemann Cauchy sequence of $L^2$-Riemann functions can have a nonintegrable limit (see the example of David Mitra).
In general, The Riemann integral behaves much worse that the Lebesge integral. Example: pick a enumeration of rationals in $[0,1]$, $r_1,r_2,\cdots$. The sequence $$f_n=\chi_{\{r_1,\cdots,r_n\}}$$ of Riemann integrable functions has a nonintegrable pointwise limit.
Martín-Blas Pérez Pinilla
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Could you please explain how that happens and how that will not happen with Lebesgue integral? – triomphe Feb 15 '14 at 16:41
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2The limit of your sequence in $L_2$, using Riemann integration, is the zero function. – David Mitra Feb 15 '14 at 17:06
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True, but my point was about the inadequacy of Riemann integral in general. I will edit. – Martín-Blas Pérez Pinilla Feb 15 '14 at 17:45