$\def\RiemInt{\,\text{-}\,\lower.5mm\hbox{$^{^{\rm Riem}}$}\kern-1mm\int}$ 
This is essentially a reformulation of [this](https://math.stackexchange.com/questions/570972/riemann-integrability-in-not-sequentially-complete-lcs) 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".



Is sequential completeness of LCS strictly stronger that Riemann integrability of curves?
$\def\RiemInt{\,\text{-}\,\lower.5mm\hbox{$^{^{\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".


