Riemann integral of a function with Banach space values My question is: how to prove that Riemann integral of a continuous function $f\colon [a,b]\to Y$, where $Y$ is a Banach space, is independent of the choice of intermediate points?
Let $\{t_0,...,t_n\}\subset [a,b]$ be such that
$a=t_0\le t_1\le...\le t_n=b$ and $\Delta Z_n=\max_{i\in\{i,...n\}}(t_i-t_{i-1})\to0$ as $n\to\infty$. Let $S_{Z_{n}}=\sum_{i=1}^{n}f(x_i)(t_i-t_{i-1})$, where $x_i\in [t_i-t_{i-1}]$. I define the Riemann integral by
$$\int_{a}^{b}f(t)\,dt :=\lim_{n\to\infty}S_{Z_{n}}.$$
First of all we notice that the function $[a,b]\ni t\mapsto\|f(t)\|$ is continuous on a compact set, so it is bounded. Hence
$$\lim_{n\to\infty}S_{Z_{n}}\le \sup_{t\in [a,b]}\|f(t)\|\cdot\lambda([a,b])=M<\infty,$$
where $\lambda$ is meant to be Lebesgue measure. In order to justify the definition of Riemann integral I have to do two things. First, I need to prove that the limit $\lim_{n\to\infty}S_{Z_{n}}$ exists and secondly that the integral  is independent of the choice of intermediate points. We can prove it at one go by showing that the sequance $(S_{Z_{n}})$ is Cauchy I suppose. We shall prove that $\|S_{Z_{n}}-S_{Z_{m}}\|\to 0$. Consider
$$\|S_{Z_{n}}-S_{Z_{m}}\|=\|\sum_{i=1}^{n}f(x_{n,i})(t_{n,i}-t_{n, i-1})-\sum_{j=1}^{m}f(x_{m,j})(t_{m,j}-t_{m, i-j})\|\le\\
\le\sum_{i=1}^{n}\sum_{j=1}^{m}\|f(x_{n,i})-f(x_{m,j})\|\,\lambda([t_{n,i}-t_{n,i-1}]\cap [t_{m,j}-t_{m,j-1}])\le\\
\le \sum_{i=1}^{n}\sum_{j=1}^{m}\|f(x_{n,i})-f(x_{m,j})\|\max\{\Delta Z_{n}, \Delta Z'_{m}\}.
$$
Unfortunately, I can't deduce the desired convergance, since I have no info about the ratio of convergance of $\max\{\Delta Z_{n}, \Delta Z'_{m}\}$. As concerns $\|f(x_{n,i})-f(x_{m,j})\|$, we can easily estimate it from above or maybe it would be more sufficient to use the fact that $t\mapsto\|f(t)\|$ is uniformly continuous. What shall I do?
 A: Convergence of Riemann sums of Banach space valued continuous function can be shown in a similar way showing convergence of Riemann sums of real valued continuous functions (using uniform continuity). To prove uniqueness, assume $\phi\in Y^*$ is given. One can notice that $\phi(S_{Z_n})$ is equal to the Riemann sum of $\phi\circ f:[a,b]\to\Bbb C$ which is a complex valued continuous function. Since a continuous function is Riemann integrable, we have that
$$
\phi(S_{Z_n})\xrightarrow{n\to\infty}\int_a^b\phi(f(t))\ dt.\tag{*}
$$ If $S_{Z_n}$ and $S_{Z'_n}$ have different limits in $Y$, then there is $\phi\in Y^*$ such that $\lim_{n\to\infty}\phi(S_{Z_n})\ne\lim_{n\to\infty}\phi(S_{Z'_n}).$ But this contradicts $(*)$, so it must be $\lim_{n\to\infty}S_{Z_n}=\lim_{n\to\infty}S_{Z'_n}$ proving uniqueness.
A: It can be done in a very easy way. Let $F\colon [0,1]\to \mathbb{R}$ be defined by
$$F(t)=\|f(t)\|.$$
Then,
$$M=\int_{[0,1]}F(t)\,dt=\lim_{n\to\infty}\sum_{i=1}^{n}\|f(x_i)\|(t_{i}-t_{i-1}).$$
This implies that the sequance $a_n=\sum_{i=1}^{n}\|f(x_i)\|(t_{i}-t_{i-1})$ is bounded. Moreover, $a_{n}\le a_{n+1}$, hence $(a_n)$is convergent. This implies however that there exists the following limit
$$\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)(t_{i}-t_{i-1})=\int_{[0,1]}f(t)\,dt =y.$$
So, if a function is continuous on a compact subset, then it is Riemman integrable (with values in Banach space). To prove uniqueness, observe that
$$\|S_{Z_n}-S_{Z'_{m}}\|=\|S_{Z_n}-y+y-S_{Z'_{m}}\|\le \|S_{Z_n}-y\|+\|y-S_{Z'_{m}}\|\to 0.$$
