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Let $|f|\le M$ on $[0,1]$ and let $P$ have $r$ interior division points. At most $r$ cells of the equal-length partition $D_n$ have one of those points in their interior. Their total length is at most $r/n$, and their oscillation is at most $2M$. Every other cell lies in a single cell of $P$. ThereforeThis comparison of Darboux sums avoids the incorrect assumption that $D_n$ refines $P$ for all large $n$.

If $f$ is Riemann integrable and $g$ is Lipschitz on an interval containing the range of $f$, then $g\circ f$ is Riemann integrable. With Lipschitz constant $K$, the oscillation on each partition cell satisfies $\mathrm{osc}(g\circ f)\le K\mathrm{osc}(f)$, and hence the gap between Darboux sums is multiplied by at most $K$. A continuously differentiable $g$ is Lipschitz on each closed bounded interval by the mean value theorem and the extreme value theorem applied to $g^{\prime }$.

For a nondecreasing function, the endpoint bounds show it is bounded. On the uniform partition into $N$ intervals, its Darboux sums satisfyFor a nonincreasing function apply the same calculation to $-f$, or use $|f(1)-f(0)|/N$. Thus every monotone function on $[0,1]$ is Riemann integrable.

For the unheaded rational-weight example, each summand is nonnegative, and increasing $x$ only adds indices. Hence its sum is nondecreasing and lies between zero and one, including the separately specified value at zero. It is a left-continuous cumulative function of an atomic measure.

If $q_j\in [0,1)$ and $q_j<x\le 1$, the index $j$ appears in the sum at $x$ but not at $q_j$. All other changes are nonnegative, so $f(x)-f(q_j)\ge 2^{-j}$. This remains true for arbitrarily small positive $x-q_j$, proving discontinuity at $q_j$, including a right discontinuity at zero when that rational is encountered. Every interval of positive length contains such a rational in its interior. The discontinuities are therefore dense, although the function is Riemann integrable by the monotone-function result just proved.

More concretely, truncate after the first $N$ weights. The resulting finite sum of step functions is integrable, and the uniform tail is at most $2^{-N}$. This also proves integrability and gives, if its value is desired,The integral formula follows by integrating the finite sums and using the uniform tail bound. Here “every interval” has the usual nondegenerate-interval meaning; a singleton is not being asserted to contain a discontinuity.

Assume $a_n\to 0$. The function is bounded, and every interval has lower Darboux sum zero because it contains an irrational. Given $\epsilon >0$, choose $N$ so that $a_n<\epsilon /2$ for $n>N$. Put small intervals around the finitely many points $q_1,\dots ,q_N$ whose total length makes their contribution to the upper sum less than $\epsilon /2$, and refine their endpoints to a partition. On every remaining subinterval the function is at most $\epsilon /2$, so the upper sum is less than $\epsilon$. The Riemann integrability criterion proves that $f$ is Riemann integrable with integral zero.

The condition $a_n\to 0$ is sufficient but not necessary. Assign $a_n=1$ when $q_n=1/k$ for some positive integer $k$, and $a_n=1/n$ otherwise. Then $(a_n)$ does not tend to zero. Outside a short interval near zero only finitely many unit spikes occur and can be enclosed in intervals of arbitrarily small total length; all remaining values can be made uniformly small except for finitely many further points. The same Darboux-sum argument proves that this $f$ is Riemann integrable with integral zero.

It is enough to treat an increasing function; replacing $f$ by $-f$ handles a decreasing one. Let $\mathcal{R}$ be the common refinement of dissections $\mathcal{D}$ and ${\mathcal{D}}^{\prime }$. Refinement raises lower Darboux sums and lowers upper ones, so

For the uniform dissection ${\mathcal{D}}_n=\{k/n:0\le k\le n\}$, monotonicity gives the telescoping differenceThis can be made smaller than any $\epsilon >0$, so the Riemann integrability criterion proves that $f$ is integrable.

The integral lies between the lower and upper sums, and the displayed sum in the question is the right-endpoint upper sum. Hence the generally valid sharp estimate isThe strict inequality printed in the question is false for arbitrary monotone functions: if $f(x)=0$ for $x<1$ and $f(1)=1$, the two sides of (1) are both $1/n$.

For the final claim, writeSince $F^{\prime }$ is continuous on a compact interval, it is uniformly continuous. Uniformly for $0\le t\le 1$,After the substitution $x=(k-1+t)/n$, summing the uniform errors givesThe right-hand sum is a Riemann integral, so the fundamental theorem of calculus yields