Having had no (proper) answer to [*this question*](https://mathoverflow.net/questions/57813/stability-of-convex-sets-w-r-t-integration-over-0-1), I formulate the remaining case as a new question as follows. With $I=[0,1]$, let $E$ be a separable (real) Banach space, and let $\gamma:I\to E$ be continuous. *Do there* then *exist* sequences $\boldsymbol c,\boldsymbol t\in I^{\ \mathbb N\_0}$, $\boldsymbol c(i)=c\_i$ and $\boldsymbol t(i)=t\_i$, with



> 
> (1) $\quad\mathbb R\ \text{-}\ \lim\_{\ k\to\infty\ }\sum\_{i=0}^kc\_i=1 \quad$ and


> (2) $\quad
> E\ \text{-}\ \lim\_{\ k\to\infty\ }\big(k^{-1}\sum\_{i=0}^{k-1}\gamma(k^{-1}i)\big) =
> E\ \text{-}\ \lim\_{\ k\to\infty\ }\sum\_{i=0}^k(c\_i\gamma(t\_i)) \quad$ ?




Either a (sketch of a) proof of the positive case or a counterexample is welcome. Countable or σ−convexity has also been considered in [*this question*](https://mathoverflow.net/questions/56161/infinite-convex-combinations-in-a-banach-space).



Integral in a σ−convex set.
Counterexample. $E = L^2[0,1]$ and $\gamma \colon [0,1] \to E$ defined by $\gamma(x) = 1\_{[0,x]}$, the characteristic function of interval $[0,x]$. Then $\gamma$ is continuous, in fact $\|\gamma(x) - \gamma(y)\| = \sqrt{|y-x|}$.
Now suppose $c\_i$ and $t\_i$ are as given. Let $$u :=
E\ \text{-}\ \lim\_{\ k\to\infty\ }\big(k^{-1}\sum\_{i=0}^{k-1}\gamma(k^{-1}i)\big)
$$ and
$$ v :=
E\ \text{-}\ \lim\_{\ k\to\infty\ }\sum\_{i=0}^k(c\_i\gamma(t\_i))
$$
Then $u(t) = 1-t$ for $t \in [0,1]$. Note $u$ is continous. But (assuming the $t\_i$ are distinct) $v$ has a jump of size $c\_i$ at $t\_i$ for all $i$, so $v$ is certainly not continuous. Not even equal a.e. to a continuous function.


