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The set of bounded and Riemann integrable maps, with the norm $\|\cdot\|_{\infty}$ are a Banach space? To be more clear of what am I talking about, the set is $$\mathcal{RI}:= \{\phi: A\rightarrow B: \phi \text{ is bounded and Riemann integrable}\}$$ where $A$ is a box in $\mathbb{R}^d$

Anyway142
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    What needs to be shown? What did you try? – Hagen von Eitzen Jun 26 '21 at 20:18
  • @HagenvonEitzen I am studying how the space $(RI,|\cdot|_{\infty})$ works and, so, I want to know if it is a Banach space or not. It is clear to me that is a normed vector space, however, I am struggling to show if it is (or not) a complete space. EDIT: Starting from a Cauchy sequence, I could not show it is convergent, however, I couldnt found a counterexample either. – Anyway142 Jun 26 '21 at 20:37
  • Hint: if $(f_n)$ is Cauchy in your space, then define a limit candidate $f$ to be the pointwise limit of $f_n(x)$ for each $x$ in $A$ (why does this exist?). Show that $\Vert f_n-f\Vert\to0$ so that $f_n\to f$ and that $f$ lies in your space as well. – csch2 Jun 26 '21 at 21:37
  • @csch2 I tried that way and, honestly, I think it was very useful, thanks! Could you just give me a tip to show that $f \in RI$? Showing that $f$ is bounded is clear, but I concluded that is Riemann integrable seeing that the set of discontinous points of $f_n$ equals to the set of discontinous points of $f$ and, since the first one has measure zero, so do the second. Is it right? – Anyway142 Jun 26 '21 at 22:12
  • Easier I believe is to use the fact that uniform convergence preserves Riemann integrability. Since $\Vert f_n-f\Vert_\infty\to0$, $f_n\to f$ uniformly and thus $f$ is Riemann integrable. – csch2 Jun 27 '21 at 00:10

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