# Showing that the norm of a Riemann-Integrable Banach-valued function is a real-valued, Riemann-Integrable function

Given a Banach space $(X,||\cdot||_{X})$ over $\mathbb{C}$ and a Riemann Integrable $X$-valued function $f$ on $[a,b]$, I am trying to show that $||f||_{X}$ is a real-valued, Riemann Integrable function on $[a,b]$.

Since $||f||_{X}$ is real-valued, we may show that it's R.I. by showing that for every $\epsilon>0$, there exists a partition of $[a,b]$, $P=(p_{k})_{k=0}^{n}$, such that: $$\sum_{k=1}^{n}\Big{(}\underset{p_{k-1}\leq x\leq p_{k}}\sup||f(x)||_{X}-\underset{p_{k-1}\leq x\leq p_{k}}\inf||f(x)||_{X}\Big{)}\triangle P_{k}<\epsilon$$ In particular, fixing $\epsilon>0$, we may find $\delta>0$ such that for any two tagged partitions of $[a,b]$, $(P,Q),(P',Q')$ with $||P||,||P'||<\delta$, the norm of the difference of the Riemann Sums on these tagged partitions is less than $\epsilon$. (This is the "Cauchy Criterion" as $f$ is R.I.)

Now, choosing $P=(p_{k})_{k=0}^{n}$ with $||P||<\delta$, we may approximate the above sum as: \begin{multline*} \sum_{k=1}^{n}\Big{(}\underset{p_{k-1}\leq x\leq p_{k}}\sup||f(x)||_{X}-\underset{p_{k-1}\leq x\leq p_{k}}\inf||f(x)||_{X}\Big{)}\triangle P_{k} \\ \leq \sum_{k=1}^{n}\Big{(}||f(t^{\sup}_{k})||_{X}-||f(t^{\inf}_{k})||_{X}\Big{)}\triangle P_{k} + 2\epsilon(b-a)\end{multline*} via the approximation property of the sup and inf on each interval $[p_{k-1},p_{k}]$. This estimate may then be bounded by: $$\sum_{k=1}^{n}||f(t^{\sup}_{k})-f(t^{\inf}_{k})||_{X}\triangle P_{k} +2\epsilon(b-a)$$ This is as far as I've gotten. There is a similar question on this site where they try to bound the last sum above, but I am not sure their answer is correct. Could someone give me a hint on where to go from here?

I would say that since $f$ is Riemann-integrable, it has at most a countable number of discontinuities (according to Froda's theorem). Since the norm is continuous, that would make $||f||_{X}$ continuous except maybe for a countable number of points, which implies that $f$ is Riemann-integrable (since it is in particular continuous almost everywhere).
• Froda's Theorem does not work here. It describes the set of discontinuities of monotone real-valued functions. Above, $f$ is a Banach-valued function where the Banach space is an arbitrary Banach space over $\mathbb{C}$, so we may not even have the notion of monotonicity of $f$. More generally, one could investigate whether or not Lebesgue's criterion for Riemann-Integrability of real-valued functions will hold in arbitrary Banach spaces, and the answer is no. – JWP_HTX Oct 9 '17 at 22:56
• I believe that Froda's Theorem extends to functions $[a,b] \rightarrow B$ where $B$ is a Banach space and $f$ is a uniform limit of step functions ($f$ thus admits a left- and right-limit at each point, and the same proof as for monotone functions applies). This is the case here since $f$ is Riemann-integrable with values in a Banach space (and thus can be approached uniformly by step functions). Please do correct me if I am wrong, thanks! – Julien Oct 10 '17 at 16:11
• Hi Julien, looking at the proof of Froda's Theorem on Wikipedia, it relies on the notion that we can compare values of $f$, which we don't have in an arbitrary Banach space. I am skeptical that such a proof could extend to Banach spaces, especially since Riemann-Integrability in an arbitrary Banach spaces does not imply that the set of discontinuities has measure zero. Could you maybe provide a link to a proof of Froda's Thm for Banach spaces? – JWP_HTX Oct 10 '17 at 17:44
• If $f: [a;b] \rightarrow M$ ((M,d) any metric space) admits a jump discontinuity of size $\epsilon>0$ at point $x_{0}$, then $\exists \eta>0, \forall x, d(x_{0},x) \leq \eta \Rightarrow Variation(f(]x_{0}-\eta;x_{0}[) < \epsilon$ and $Variation(f, ]x_{0};x_{0}+\eta[) < \epsilon$. This shows that $x_{0}$ is the only jump discontinuity with size $>\epsilon$ on $]x_{0}-\eta;x_{0}+\eta[$, thus the set of jump discontinuities of $f$ is at most countable. Now every Riemann-integrable function in a Banach is a uniform limit of step functions (and thus has a left/right limit at every point)... – Julien Oct 10 '17 at 20:46
• "Riemann-Integrability in an arbitrary Banach spaces does not imply that the set of discontinuities has measure zero": in this case maybe we do not use the same construct of the Riemann integral in a Banach space? I see it as the sum of step functions, extended to the closure of the set of all step functions with values in the Banach space (for the norm $||.||_{\infty}$). – Julien Oct 10 '17 at 20:56