$\def\RiemInt{\,\text{-}\,\lower.5mm\hbox{$^{^{\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".

Lipschitzcurves. 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. – TaQ Dec 8 '13 at 15:22