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 $\mathrm{\parallel }{P}_N\mathrm{\parallel }\to 0$ 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 $S(f,P,\xi )={\sum }_{i}f({\xi }_i){\mathrm{\Delta }}_i$, 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 $t(x)=1\mathrm{/}q$ at a rational $x=p\mathrm{/}q$ 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 ${\xi }_i^N:=\iota (i)\mathrm{/}\iota (N)$, a rational. Then

$\mathrm{\parallel }{U}_N\mathrm{\parallel }=\frac{1}{\iota (N)}⟶0,S({\mathbf{1}}_{\mathbb{Q}},U_N,{\xi }^N)=1\text{ for every }N\ge 1,$

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 $\epsilon >0$ there is a real $\delta >0$ such that $\mathrm{\mid }S(f,P,\xi )-I\mathrm{\mid }<\epsilon$ 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 $1\mathrm{/}2$ 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 $\epsilon >0$ there is a real $\delta >0$ such that $\mathrm{\mid }S(f,P,\xi )-I\mathrm{\mid }<\epsilon$ 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 $S(f,P,\xi )={\sum }_{i}f({\xi }_i){\mathrm{\Delta }}_i$ 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.