# If $f$ is integrable, then $\| f\|$ is also integrable.

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 $$f.$$ $$\square$$

• 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/…
– avs
Mar 13 '19 at 22:23
• 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. Mar 13 '19 at 22:39