Riemann integrability criterion

A bounded function is Riemann integrable exactly when for every $\epsilon >0$ some partition $P$ satisfies $U(f,P)-L(f,P)<\epsilon$.

Enumerate the rational points of a compact nondegenerate interval as $q_1,q_2,\dots$. Each singleton indicator function $f_n=1_{\{q_n\}}$ is Riemann integrable with integral zero. Their pointwise supremum is the Dirichlet function, which is not Riemann integrable. Thus even a uniformly bounded countable supremum need not preserve integrability.

The indicator of the rational numbers is not Riemann integrable on a nondegenerate interval: every subinterval contains both rational and irrational points, so its lower sum is zero and its upper sum is the interval length.

For a common interval partition, the oscillation of a function on an interval satisfies ${\mathrm{osc}}_I(f+g)\le {\mathrm{osc}}_{I}f+{\mathrm{osc}}_{I}g$. Refining two good Darboux sums therefore proves integrability of the sum. Lipschitz composition preserves Riemann integrability, so absolute values preserve it too. The identity $\max (f,g)=(f+g+|f-g|)/2$ then proves the assertion for the maximum.

Let $f$ be bounded on $[a,b]$ and continuous on $(a,b]$. For any positive error tolerance, isolate $[a,a+\delta ]$. Its contribution to the difference of upper Darboux sums and lower Darboux sums is at most $2\Vert f{\Vert }_{\infty }\delta$. On $[a+\delta ,b]$, uniform continuity supplies a partition of an interval with arbitrarily small remaining oscillation sum. The Riemann integrability criterion then proves integrability, even when the values oscillate without a limit at the endpoint.

Let $K$ consist of numbers whose decimal digits all lie in a fixed set $D\subseteq \{1,\dots ,8\}$ with $m<10$ digits. The $m^N$ allowed prefixes give closed decimal intervals of length $10^{-N}$ covering $K$, with no endpoints in $K$. A partition of an interval at those endpoints gives an upper Darboux sum at most $(m/10{)}^N$. Every cell contains a terminating decimal outside $K$, so the lower Darboux sum is zero. Thus the indicator function is Riemann integrable with Riemann integral zero. Such sets illustrate how an uncountable set may have an integrable indicator function of zero integral.

Let $|f|\le M$ on $[0,1]$. Suppose finitely many nonoverlapping closed intervals, of total length at least $1-\delta$, can be chosen for each $\delta >0$, with $f$ Riemann integrable on each. Refine the intervals to make their total Darboux sum gap less than $\epsilon /2$. The remaining length contributes at most $2M\delta$. Taking $\delta <\epsilon /(4M)$ proves global Riemann integrability by the Riemann integrability criterion.

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 $g^{\prime }$.

A bounded function on $[0,1]$ is Riemann integrable if and only if its upper Darboux sum minus its lower Darboux sum on the equal-length partition $D_n$ tends to zero. Sufficiency follows directly from the Riemann integrability criterion. Necessity follows by comparing with a fixed partition whose gap is small and using the finite bad-cell estimate for Darboux sums. The equal-length partitions need not be nested.

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 $n$.

A bounded function on a closed interval is Riemann-integrable when its upper and lower Darboux integrals agree.

A left-continuous cumulative function of an atomic measure with positive masses at a dense countable set has dense discontinuities. Nevertheless it is bounded and monotone on a compact interval, hence a Riemann-integrable function. Finite step-function sums approximate it uniformly because the positive masses are summable. Thus dense discontinuities do not imply failure of the Riemann integral.

A continuous function on a compact interval is uniformly continuous. A partition with sufficiently small mesh then makes the oscillation on every subinterval small, so its upper and lower Darboux sums can be made arbitrarily close.