Two definitions for Riemann integrability of $f:[a,b]\rightarrow X$, where $X$ is a Banach space. Let $X$ be a Banach space and $f:[a,b]\rightarrow X$ be a function. Consider the following two definitions of Riemann integrability:
Definition 1: there exists $x\in X$ and a sequence of partitions $\{P_n\}_{n=1}^\infty$ of $[a,b]$ with mesh tending to $0$, $P_n=\{a=t_0^n<t_1^n<\ldots<t_{r_n}^n=b\}$, such that, for any choice of interior points $s_i^n\in [t_{i-1}^n,t_i^n]$, we have $\lim_{n\rightarrow\infty} \sum_{i=1}^{r_n} f(s_i^n)(t_i^n-t_{i-1}^n)=x$ in $X$.
Definition 2: there exists $x\in X$ such that, for every $\epsilon>0$, there is a $\delta>0$ such that for any partition $P$ with $\|P\|<\delta$ and for any choice of interior points, the corresponding Riemann sum $S(P,f)$ satisfies $\| S(P,f)-x\|_X<\epsilon$.
I have read these two definitions in several notes, but I have never seen a formal proof. Do you know how to prove that Definition 1 implies Definition 2? Obviously, the converse is true.
 A: In the article here (this was linked as a comment at some point, but it seems to have been deleted) it is shown that definition $2$ is equivalent to the following condition (see theorem $5$, condition $4$):
Definition $3$: $\forall \epsilon > 0$ $\exists$ a partition $P_\epsilon$ such that for any two choices of tags on $P_\epsilon$, say giving tagged partitions $P_1$, $P_2$, we have $\|S(P_1,f)-S(P_2,f)\|_X < \epsilon$.
I will show that Definition $1$ $\implies$ Definition $3$.
Let $f$ satisfy definition $1$, and let $x \in X, \{P_n\}_{n=1}^\infty$ be the point and sequence of partitions given by definition $1$.
Suppose $f$ does not satisfy definition $3$, so $\exists \epsilon >0$ such that for every partition $Q$ we can find choices of tags on $Q$ giving tagged partitions $Q_1,Q_2$ such that $\|S(Q_1,f)-S(Q_2,f)\|_X > \epsilon$. In particular, for each $n \in \mathbb{N}$ we may find choices of tags on $P_n$ giving tagged partitions $P_n^1, P_n^2$ such that $\| S(P_n^1,f)-S(P_n^2,f)\|_X > \epsilon$. But by definition $1$, 
$$\lim_{n \to \infty} S(P_n^1,f) = x = \lim_{n \to \infty} S(P_n^2,f)$$
So we must have 
$$\lim_{n \to \infty} \|S(P_n^1,f)-S(P_n^2,f)\|_X =0 $$
Clearly contradicting the construction of $P_n^1,P_n^2$.
