Proof of Lebesgue criterion in Analysis 2 by Zorich In the book "Analysis 2" by Vladimir Zorich, the author gives a proof of Lebesgue criterion for integrability. Later in Remark 2 on page 144 the following is stated:
"Since the Cauchy criterion for existence of a limit is valid in any complete metric space, the sufficiency (but not the necessity part), as the proof shows, holds for functions in any complete normed vector space"
I have worked through the proof that is given thoroughly, and I understand it well. One thing I can't seem to pinpoint is "but not the necessity part". In the proof I can not find anything that would fail in an arbitrary Banach space. Since $\mathbb{R}^n$ is isomorphic to any n-dimensional normed vector space I assume that the proof doesn't necessarily hold for infinite dimensional spaces. Note that I am aware that some points in the proof require further explaining (such as the $1/2n_0$ inequality).
Here is the proof given in the book:
Let $f:I\to \mathbb R$ where $I$ is an n-dimensional interval.
Necessity: It is already known that $f\in \mathcal{R}(I)$ implies that $f$ is bounded. Suppose $|f|\leq M$ for some $M\geq 0$.
We shall now verify that $f$ is continous at almost all points of $I$. To do this, we shall show that if the set $E$ of its points of discontinuity does not have measure zero, then $f \notin \mathcal{R}(I)$.
Indeed, representing $E$ in the form $E = \bigcup_{n=1}^\infty E_n$ where $E_n=\{x\in I | \omega(f;x)\geq \frac{1}{n}\}$, we conclude that if $E$ does not have measure zero, then there exists an index $n_0$ such that $E_{n_0}$ is also not a set of measure zero (It is known that a countable union of sets of measure zero is of measure zero). Let $P$ be an arbitrary partition of the interval $I$ into intervals $\{I_i\}$. We break partition $P$ into two groups of intervals $A$ and $B$, where
$$
A = \bigg{\{} I_i \in P \bigg| I_i \cap E_{n_0} \neq \emptyset \land \omega(f, I_i) \geq \frac{1}{2n_0} \bigg{\}}; B=P\backslash A
$$
The system of intervals $A$ froms a covering of set $E_{n_0}$. In fact, each point of $E_{n_0}$ lies either in the interior of some interval $I_i \in P$ (Because the oscillation at an interior point of a set $S$ is always less that or equal to the oscillation over the entire set $S$), in which case obviously $I_i \in A$, or on the boundary of several intervals of the partition $P$ (This required a little explanation, but doesn't seem to fail for infinite-dimensional spaces). In the latter case, the oscillation of the function must be at least $\frac{1}{2n_0}$ on at least one of these intervals (because of the triangle inequality), and that interval belongs to the system $A$.
We shall now show that by choosing the set $\xi$ of distinguished points in the intervals of the partition $P$ in different ways we can change the value of the Riemann sum significantly.
To be specific, we chose the sets of points $\xi ^\prime$ and $\xi ^{\prime\prime}$ such that in the intervals of the system $B$ the distinguished points are the same, while in the intervals $I_i$ of the system $A$, we choose the points $\xi_i^\prime$ and $\xi_i ^{\prime\prime}$ so that $f(\xi_i^\prime) - f(\xi_i^{\prime \prime}) > \frac{1}{3n_0}$. We then have
$$
|\sigma(f,P,\xi^\prime) - \sigma(f,P,\xi^{\prime \prime})| = \bigg|\sum_{I_i\in A}(f(\xi_i^\prime) - f(\xi_i^{\prime \prime}))|I_i|\bigg| > \frac{1}{3n_0} \sum_{I_i \in A}|I_i| > c > 0
$$
The existence of such a constant c follows from the fact that the intervals of the system $A$ from a covering of the set $E_{n_0}$, which by hypothesis is not a set of measure zero.
Since $P$ was an arbitrary partition of the interval I, we conclude from the Cauchy criterion that the Riemann sums $\sigma(f,P,\xi)$ cannot have a limit as $\lambda (P) \to 0$, that is $f\notin \mathcal {R}. \Box$
 A: The problem is with summing large number of discontinuties to one large enough. In $\mathbb R$, we have that if $|x_{i, 0} - x_{i, 1}| \geq 1$ for all $i$, we can find $\alpha_i \in \{0, 1\}$ s.t. $|\sum_{i = 1}^n t_i \cdot (x_{i, \alpha_i} - x_{i, 1 - \alpha_i})| \geq \sum_{i=1}^n t_i$.
Similarly, for finite-dimension spaces, if $\|x_{i, 0} - x_{i, 1}\| \geq 1$, for some $\alpha_i$ we have $\|\sum_{i = 1}^n t_i \cdot (x_{i, \alpha_i} - x_{i, 1 - \alpha_i})\| \geq c \cdot \sum_{i=1}^n t_i$ (where $c$ depends on space, but is positive).
In infinite-dimension case, it's no longer the case. For example, in $c_0$, $x_{i, 0} = 0$, $x_{i, 1} = e_i$, for any choice of $\alpha$, we will get $\sum_{i=1}^n t_i \cdot (x_{i, \alpha_i} - x_{i, 1 - \alpha_i}) = (\pm t_1, \pm t_2, \ldots, \pm t_n, \ldots)$ and thus norm of this sum is at most $\max t_i$.
This sequence can be used to make a Riemann-integrable function discontinuous everywhere. Let $q_i$ be any enumeration of rational points in $[0, 1]$ and define
$$f(x) = \begin{cases} e_i, x = q_i\\ 0, x \notin \mathbb{Q} \end{cases}$$
For any partition of $[0, 1]$ with mesh less than $\varepsilon$ and any choice of points, we will have all coordinates of Riemann sum have absoulute value of at most $\varepsilon$, so $f$ is Riemann integrable with integral $0$.
