For the Dirichlet function every uniform partition with rational tags gives Riemann sum $1$, so the sums converge along that sequence of tagged partitions although the function is not integrable: the mesh condition of the Riemann definition quantifies over all tagged partitions and cannot be weakened to one sequence

Refuted: that a bounded $f$ on $[a,b]$ is Riemann integrable with integral $I$ as soon as there is one sequence $(P_N,\xi^N)$ of tagged partitions with $\|P_N\| \to 0$ and $S(f,P_N,\xi^N) \to I$ (Tagged partitions of $[a,b]$, with a tag $\xi_i$ in each subinterval, and the Riemann sum $S(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i$, The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$, Darboux integrability as their equality, and the notation $\int_a^b f$).

The witness is again the Dirichlet function on $[0,1]$ (The Dirichlet function $1_{\mathbb{Q}}$, and Thomae's function $t$ with $t(x) = 1/q$ at a rational $x = p/q$ in lowest terms with $q \ge 1$ and $t(x) = 0$ at every irrational $x$). Take $P_N := U_N$, the uniform partition into $N \ge 1$ parts, and tag each subinterval by its left endpoint $\xi^N_i := \iota(i)/\iota(N)$, a rational. Then

$\|U_N\| = \frac{1}{\iota(N)} \longrightarrow 0, \qquad S(\mathbf{1}_{\mathbb{Q}}, U_N, \xi^N) = 1 \ \text{ for every } N \ge 1,$

so the Riemann sums converge, to $1$; and yet $\mathbf{1}_{\mathbb{Q}}$ is not Riemann integrable on $[0,1]$ (The Dirichlet function on $[0,1]$ has lower Darboux integral $0$ and upper Darboux integral $1$, so it is bounded and not Riemann integrable).

What this shows about The Darboux and Riemann definitions agree: a bounded $f$ on $[a,b]$ is Darboux integrable with integral $I$ if and only if for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that $|S(f,P,\xi) - I| < \varepsilon$ for every tagged partition of mesh below $\delta$. Condition 2 there quantifies over every tagged partition of mesh below $\delta$, tags included, and the quantifier cannot be replaced by the existence of one good sequence. Tagging the very same partition by irrationals instead gives Riemann sum $0$, so for each $N$ the two taggings of $U_N$ give the two values $1$ and $0$; no real number is within $1/2$ of both.

• What survives of the naive formulation. If $f$ is integrable then every sequence of tagged partitions with meshes tending to $0$ has Riemann sums tending to $\int_a^b f$, by The Darboux and Riemann definitions agree: a bounded $f$ on $[a,b]$ is Darboux integrable with integral $I$ if and only if for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that $|S(f,P,\xi) - I| < \varepsilon$ for every tagged partition of mesh below $\delta$; the failure is only in the converse. So sequences of Riemann sums are a legitimate way to compute an integral once integrability is known, and no way at all to establish it.

• The Darboux sums see the difference immediately. For every partition $P$ of $[0,1]$ one has $L(g,P) = 0$ and $U(g,P) = 1$ (The Dirichlet function on $[0,1]$ has lower Darboux integral $0$ and upper Darboux integral $1$, so it is bounded and not Riemann integrable), and by Tagged partitions of $[a,b]$, with a tag $\xi_i$ in each subinterval, and the Riemann sum $S(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i$ every Riemann sum over $P$ lies between them. The rational tags realise the upper end and the irrational tags the lower end; the choice of tags moves the sum across the whole gap.

• Choosing the irrational tags costs nothing. Step 4.1 selects one irrational in each of finitely many subintervals, which is a finite family; the same observation as in What this page costs in choice: Riemann's criterion, the Darboux-Riemann equivalence and integrability of a monotone function are theorems of ZF; integrability of a continuous function inherits the single use of countable choice inside Heine-Cantor; and only the forward half of the Lebesgue criterion spends countable choice, once, at the countable union of null sets applies, and no choice principle is used.