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

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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: \begin{equation} \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 \end{equation} 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?

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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).