As usual, a partition of a compact interval $[a, b]$ is, by definition, an strictly increasing family $\Pi = (t_k)_{k = 0}^m$ ($m \geq 0$) of points in the interval such that $t_0 = a$ and $t_m = b;$ $\tilde\Pi$ is refining $\Pi$ if $\Pi$ is subfamily of $\tilde\Pi$ and $\tilde\Pi$ is also a partition. A valid selection of tags $\tau$ for the partition $\Pi = (t_k)_{k = 0}^m$ is, by definition, a family $(\tau_k)_{k = 1}^m$ for which $\tau_k \in [t_{k - 1}, t_k],$ as $k$ runs from $1$ until $m.$

Definition of Banach space valued Riemann integral. A function $f:[a, b] \to \mathrm{X},$ $\mathrm{X}$ being a Banach space, is Riemann-integrable if there exists a vector $I \in \mathrm{X}$ obeying the following law: for all $\varepsilon > 0,$ there exists a partition $\Pi_\varepsilon$ such that for whatever the partition $\Pi$ of $[a, b]$ refining $\Pi_\varepsilon$ and whatever valid selection of tags $\tau$ for the partition $\Pi$ may be, the Riemman sum of $f$ associated with the partition $\Pi$ under the valid selection of tags $\tau,$ $S(f, \Pi, \tau) = \sum\limits_{k = 1}^m f(\tau_k)(t_k - t_{k - 1}),$ satisfies $\|I - S(f, \Pi, \tau)\| < \varepsilon.$

Problem. How to show that $\|f\|$ will be integrable whenever $f$ is?

There is the following "fundamental criterion" for Riemann-integration that may be useful but I just couldn't find a way to apply it.

Fundamental criterion for existence. For the function $f:[a, b] \to \mathrm{X}$ to be Riemann-integrable it is necessary and sufficient that the following conditions should hold, for every $\varepsilon > 0,$ there exists a partition $\Pi_\varepsilon,$ such that $\left\| S \left(f, \Pi^{(1)}, \tau^{(1)} \right) - S \left(f, \Pi^{(2)}, \tau^{(2)} \right) \right\| < \varepsilon$ for all refinements $\Pi^{(1)}$ and $\Pi^{(2)}$ of $\Pi_\varepsilon$ and all corresponding valid selections of tags $\tau^{(1)}$ and $\tau^{(2)}.$

Sketch of proof of the fundamental criterion. Necessity is obvious. For sufficiency, consider, for each $n,$ a partition $\Pi^{(n)},$ refining all previous $\Pi^{(k)},$ and consider valid tags $\tau^{(n)}$ such that $$\left\| S \left(f, \Pi^{(n)}, \tau^{(n)} \right) - S \left(f, \Pi^{(k)}, \tau^{(k)} \right) \right\| < \dfrac{1}{k}.$$ Set then $I_n = S \left(f, \Pi^{(n)}, \tau^{(n)} \right)$ and notice this defines a fundamental sequence is $\mathrm{X}$ and therefore, converging to some vector $I \in \mathrm{X},$ which can be shown, using triangle inequality, to be the Riemann integral of $f.$ $\square$

  • $\begingroup$ Consider this: given a Riemann-intergrable $f : [a, b] \rightarrow X$ and a continuous function $g : X \rightarrow {\mathbb R}$, is the function $$ g \circ f : [a, b] \rightarrow {\mathbb R} $$ Riemann-integrable? In this context, see mathoverflow.net/questions/20045/… $\endgroup$
    – avs
    Mar 13 '19 at 22:23
  • $\begingroup$ I believe when $\mathbf{X} = \mathbf{R},$ as $f$ is bounded (you can show that if $f$ is unbounded, in my general context, then it cannot be Riemann-integrable by contrapositive), $g$ can be assumed uniformly continuous and then the magic will happen. But for general Banach space, I do not think the same idea holds. $\endgroup$
    – Will M.
    Mar 13 '19 at 22:39

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