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$\def\RiemInt{\,\text{-}\,\lower.5mm\hbox{$^{^{\rm Riem}}$}\kern-1mm\int}$ This is essentially a reformulation of this MSE question which has not received any answers for about three weeks. To formulate the question more precisely, let $E$ be any (real) Hausdorff locally convex space. Say that $c$ is a curve in $E$ iff $c$ is a continuous function $[0,1]\to E$ where the interval $[0,1]$ is equipped with the usual topology induced from $\mathbb R$ . Say that $c$ is Riemann integrable with integral $x$ iff $x$ is in $E$ and for every neighbourhood $V$ of $x$ in $E$ there is an integer $n_0>0$ such that $n^{-1}\sum_{m=1}^{\,n}c\,(n^{-1}m)\in V$ for all $n\in\mathbb Z$ with $n\ge n_0$ . If such an $x$ exists, it is unique, and we write $E\RiemInt_{\,0}^{\,1}c$ for it. The precise questions now are the following two:

Q1. If $E$ is a Hausdorff locally convex space where every curve is Riemann integrable, is $E$ necessarily sequentially complete?

Let $F$ be the space $\ell^{+\infty}(\mathbb N_0)$ of bounded real sequences equipped with (the pointwise addition and scalar multiplication and) the weak topology $\sigma(\ell^{+\infty},\ell^1)$ , and let $S$ be the subset formed by the sequences converging to zero. Let $E=c_0(\mathbb N_0)_\sigma$ be the topological linear subspace of $F$ with underlying set $S$ . Then $F$ is sequentially complete and $E$ is not. Hence every curve in $F$ is Riemann integrable. If $c$ is a curve in $E$ , it is also such in $F$ , and hence $F\RiemInt_{\,0}^{\,1}c$ is in $F$ . Without success, I have been trying to find a curve $c$ in $E$ with $F\RiemInt_{\,0}^{\,1}c\not\in S$ .

Q2. Is every curve in $E$ Riemann integrable?

Of course, if Q2 has answer "yes", then Q1 has answer "no".

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    $\begingroup$ I would use an alternative definition of Riemann integrability: Let $E'$ be the dual of $E$. A curve $c:[0,1]\to E$ is called Riemann integrable if $\ell\circ c$ is Riemann integrable, for every $\ell\in E'$. Note that if $E$ is locally convex, then $E'$ separates points, and hence the integral of $c$, if $c$ integrable, is uniquely defined. $\endgroup$
    – smyrlis
    Commented Dec 8, 2013 at 9:25
  • $\begingroup$ @ smyrlis: Your suggestion corresponds to the Pettis or Gelfand integral which are usually defined so that $\ell\circ c$ is required to be Lebesgue integrable. My choice aims to generalize the definition of Riemann integral of scalar functions via Riemann sums. Riemann integrability, as defined in the question, of all curves in a given LCS is certain completeness property (R) of the space, and now it turned out that it lies between sequential completeness (S) and Mackey completeness (M) so that the implications ${\rm(S)}\Rightarrow{\rm(R)}\Rightarrow{\rm(M)}$ hold, and (cont.) $\endgroup$
    – TaQ
    Commented Dec 8, 2013 at 15:21
  • $\begingroup$ that at least the first one cannot be reversed. Note that (M) is equivalent to both "scalarwise" and "Riemann-sumwise" Riemann integrability of all Lipschitz curves. I suspect that for continuous curves the "scalarwise" and "Riemann-sumwise" definitions of Riemann integrability define distinct completeness conditions, the former lying between (R) and (M). I am now wondering what would possibly be an example of a Mackey complete space where not all (continuous) curves are "Riemann-sumwise" integrable, i.e. showing that the implication ${\rm(R)}\Rightarrow{\rm(M)}$ cannot be reversed. $\endgroup$
    – TaQ
    Commented Dec 8, 2013 at 15:22

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$\def\RiemInt{\,\text{-}\,\lower.5mm\hbox{$^{^{\rm Riem}}$}\kern-1mm\int}$The answer to Q2 is "yes", and hence to Q1 it is "no". It is a simple consequence of the dominated convergence theorem. Indeed, if $c$ is a curve in $E$ , then ${\rm rng\kern1mm}c$ is a bounded set in $E$ , equivalently in $c_0(\mathbb N_0)$ . So all the coordinate functions $c_i:t\mapsto c(t)(i)$ satisfy $|\,c_i(t)\,|\leq M$ for some $M$ in $\mathbb R^+$ for all $i$ in $\mathbb N_0$ and for $0\leq t\leq 1$ . Since $\lim_{\,i\to\infty\,}c_i(t)=0$ for $0\leq t\leq 1$ , it follows that $\lim_{\,i\to\infty\,}\int_{\,0}^{\,1}c_i=0$ , and hence $F\RiemInt_{\,0}^{\,1}c$ is in $S$ .

One could also use Pettis' theorem to conclude that every curve $c$ in $E$ represents a vector in $L^1([0,1],c_0(\mathbb N_0))$ , and hence has Bochner integral over $[0,1]$ in $c_0(\mathbb N_0)$ . Obviously, this Bochner integral coincides with $F\RiemInt_{\,0}^{\,1}c$ which hence belongs to $S$ .

In conclusion, the answer to the question of the title is "yes".

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