I'm (recreationally) trying to see to what extent Riemann integration can be extended to functions of the form $f : [a, b] \to B$ where $[a, b] \subseteq \mathbb{R}$ is some compact interval and $(B, \lVert . \rVert)$ is a Banach space. I've been able to show that the first fundamental theorem of calculus still holds but, interestingly, have not been able to show all of the basic properties.
The definition I'm using is basically just the Wikipedia definition of the Riemann integral. This definition requires only a norm (absolute value) to be defined on the image space, instead of a total order. However, most proofs of basic properties rely on the Darboux definition of the integral, which does require the total order on $\mathbb{R}$. I haven't found any proofs of the needed basic properties that do not rely on suprema and infima in $\mathbb{R}$.
Using the definition with Riemann sums, I have been able to show the $\mathbb{R}$-linearity of the integral and also that changing a finite number of values of $f$ does not change the value of the Riemann integral.
I am struggling to show that, if $f : [a, b] \to B$ is Riemann integrable, then so are $f \vert_{[a, s]}$ and $f \vert_{[s, b]}$ for any $s \in ]a, b[$ and we have $$ \int_a^bf(t)dt = \int_a^sf(t)dt + \int_s^bf(t)dt. $$ I would also want to show that all continuous functions are integrable.
Any hints on how to do this? I think my main problem is that I really wouldn't know how to use the definition to show that a function is integrable if I do not know what the integral should be.
By definition, $f: [a,b] \to B$ is Riemann integrable on $[a,b]$ with integral $L \in B$ if for each $\epsilon > 0$ there is $\delta > 0$ such that for every tagged partition $x_0,\ldots,x_n$, $t_0, \ldots, t_{n-1}$ of $[a,b]$ with mesh $<\delta$, $$\left\| \sum_{i=0}^{n-1} (x_{i+1}-x_i) f(t_i) - L \right\| < \epsilon \tag{1}$$
I claim that $f$ is Riemann integrable iff for every $\epsilon > 0$ there is $\delta > 0$ such that for any tagged partitions $x_0, \ldots, x_m$, $t_0, \ldots, t_{m-1}$ and $x'_0,\ldots, x'_n$, $t'_0, \ldots, t'_{n-1}$, both with mesh $< \delta$, $$ \left\| \sum_{i=0}^{m-1} (x_{i+1}-x_i) f(t_i) - \sum_{i=0}^{n-1} (x'_{i+1}-x'_i) f(t'_i)\right\| < \epsilon \tag{2} $$
If $f$ is Riemann integrable, the Triangle Inequality gives you (2) (with $\epsilon$ replaced by $2\epsilon$, but $\epsilon$ was arbitrary).
Conversely if my condition holds, then the Riemann sums for any sequence of tagged partitions with mesh $\to 0$ form a Cauchy sequence, which converges since $B$ is complete, and the limit of (2) with $(x_0',\ldots,x'_n, t'_0,\ldots,t'_{n-1})$ in that sequence of tagged partitions gives you (1) (with $<$ replaced by $\le$, but again $\epsilon$ was arbitrary).
Now suppose $f$ is Riemann integrable on $[a,b]$, and for some $\epsilon > 0$ take $\delta > 0$ such that my criterion holds. Let $a < s < b$, and let $x_0, \ldots, x_m, t_0, \ldots, t_{m-1}$ and $x'_0, \ldots, x'_n, t'_0, \ldots, t'_{n-1}$ be tagged partitions of $[a,s]$ with mesh $< \delta$. By adding points of $[s,b]$ to these we can form tagged partitions $x_0, \ldots, x_M, t_0, \ldots, t_{M-1}$ and $x'_0, \ldots, x'_N, t'_0, \ldots, t'_{N-1}$ of $[a,b]$, still with mesh $< \delta$. Suppose we add the same points to both tagged partitions. Then $$ \left\| \sum_{i=0}^{m-1} (x_{i+1}-x_i) f(t_i) - \sum_{i=0}^{n-1} (x'_{i+1}-x'_i) f(t'_i)\right\| = \left\| \sum_{i=0}^{M-1} (x_{i+1}-x_i) f(t_i) - \sum_{i=0}^{N-1} (x'_{i+1}-x'_i) f(t'_i)\right\| < \epsilon $$ We conclude that $f$ is Riemann integrable on $[a,s]$. Similarly it is Riemann integrable on $[s,b]$.
Since the sum of a Riemann sum for $[a,s]$ and a Riemann sum for $[s,b]$, both with mesh $< \delta$, is a Riemann sum for $a,b]$ with mesh $< \delta$, we can then take limits and get $$ \int_a^s f(x)\; dx + \int_s^b f(x)\; dx = \int_a^b f(x)\; dx$$
EDIT: Here's how to prove a continuous function $f$ is Riemann integrable. By uniform continuity, take $\delta > 0$ such that $\|f(x) - f(y)\| < \epsilon/(b-a)$ when $|x-y| <\delta$. Consider two tagged partitions $ x_0,\ldots, x_m, s_0, \ldots, s_{m-1}$ and $y_0, \ldots, y_n, t_0,\ldots,t_{n-1}$ with mesh $< \delta/2$, and the corresponding Riemann sums $S$ and $S'$. Let $a = z_0 < \ldots < z_k = b$ be the common refinement of the two partitions, i.e. take $\{x_0, \ldots, x_m\} \cup \{y_0,\ldots,y_n\}$ and sort it in increasing order. Then we can write $S = \sum_{i=0}^k (z_{i+1}-z_i) f(s'_i)$ where each $s'_i$ is one of the $s_i$: it's not necessarily in the interval $[z_i, z_{i+1}]$, but its distances to $z_i$ and $z_{i+1}$ are less than $\delta/2$. Similarly, $S' = \sum_{i=0}^k (z_{i+1} - z_i) f(t'_i)$. And then $|s'_i - t'_i| < \delta$ so $$ \|S - S'\| \le \sum_{i=0}^k (z_{i+1}-z_i) \|f(s'_i) - f(t'_i)\| < \sum_{i=0}^k (z_{i+1}-z_i) \dfrac{\epsilon}{b-a} = \epsilon$$