Riemann integration

Riemann integration approximates area by upper and lower sums over finite partitions.

Changing the value of a bounded Riemann-integrable function at one point preserves its Riemann integral. The difference is supported at that point: partitions isolating it in intervals of arbitrarily small total length make both Darboux sums tend to zero. Thus a nonnegative single-point spike can have zero integral without vanishing everywhere. Its indefinite integral is identically zero, so the fundamental theorem of calculus derivative identity can fail at the changed point despite differentiability of the primitive.

For continuous $f$ and an integrator $g$ of bounded variation, the Riemann-Stieltjes integral is the limit of tagged sums $\sum f({\xi }_j)(g(t_{j+1})-g(t_j))$ as the mesh tends to zero. It exists and equals integration against the signed measure induced by $g$. Continuous vector-valued integrators are handled componentwise, which defines path signatures of continuous bounded variation paths.

The difference between a weighted integral and a discrete approximation is a quadrature error. For a Hölder continuous function on a partition of an interval with mesh $\Delta$, a left endpoint approximation has error of order ${\Delta }^{\alpha }$. For a normalized window indicator function, only the two endpoint cells contribute and the error is bounded by a constant times $\Delta /h$.

A partition of $[a,b]$ is a finite increasing sequence $a=t_0<t_1<\cdots <t_n=b$. Its mesh is ${\max }_i(t_i-t_{i-1})$, and a sequence of partitions is refining when each partition contains the preceding one.

The mesh of a partition of an interval is its largest subinterval length. A growing number of subintervals does not alone force the mesh to tend to zero: one subinterval could retain a fixed positive length.

A refinement of a finite interval partition inserts additional points without removing original points. The union of two partitions is a common refinement. Refinement makes each subinterval smaller, increasing its infimum and decreasing its supremum; this proves Darboux sum refinement monotonicity.

A dyadic partition of $[0,1]$ at level $J$ consists of $2^J$ equal cells of length $2^{-J}$. Each cell splits into two cells at the next level. Use left-closed, right-open cells, adding the endpoint $1$ to the final cell when a pointwise representative is needed. These nested partitions underlie the Haar scaling functions and Haar wavelets.

A dyadic interval is one cell in a dyadic partition. Its children are the left and right half intervals. Disjoint dyadic intervals at a fixed level produce orthogonal indicator functions and independent Brownian motion increments.

Define a finite colouring of the positive integers by $c(n)=\lfloor {\log }_{2}n\rfloor (\mathrm{mod}2)$. No increasing infinite sequence has all the numbers $x_i+2x_j$, $i\ne j$, in one colour class.

Write $\delta (y)=2^{\lfloor {\log }_{2}y\rfloor +1}-y$. If $\delta$ is unbounded on the sequence, fix a term $x$ and choose a later term $y>2x$ with $\delta (y)>2x$. With $L=2^{\lfloor {\log }_{2}y\rfloor }$, the numbers $y+2x$ and $2y+x$ lie respectively in the adjacent dyadic intervals $[L,2L)$ and $[2L,4L)$, so their colours differ. If $\delta$ is bounded by $D$ on the sequence, choose $x>2D$ and then $y>2x$. Now $y+2x=2L-\delta (y)+2x$ lies in $[2L,4L)$, whereas $2y+x=4L-2\delta (y)+x$ lies in $[4L,8L)$. Their colours again differ. The finite colouring therefore excludes the symmetric weighted-pair configuration.

For a tagged partition $a=x_0<\cdots <x_n=b$ with $t_i\in [x_{i-1},x_i]$, the corresponding Riemann sum isIts limits over increasingly fine partitions define the Riemann integral.

A right Riemann sum on a partition $a=x_0<\cdots <x_N=b$ samples each subinterval at its right endpoint. For a continuous function its limit as the mesh tends to zero is the integral, by the Riemann sum convergence theorem. On a uniform partition of width $h$, comparison of endpoint weights in the composite trapezoidal rule gives $R_N-T_N=h(f(b)-f(a))/2$. Thus $N^{-5}{\sum }_{j=1}^N{j}^4$ is a right Riemann sum for ${\int }_0^1{x}^{4}dx$, and its limit is $1/5$.

For a bounded function on a fixed closed interval, Riemann integrability with Riemann integral $I$ is equivalent to convergence of all tagged Riemann sums to $I$ as the mesh of a partition tends to zero. Precisely, for every $\epsilon >0$ there is $\delta >0$ such that every partition of an interval of mesh less than $\delta$, with any tags in the corresponding closed subintervals, has its Riemann sum within $\epsilon$ of $I$.

On the geometric progression $x_i=(1+ϵ{)}^i$, with $ϵ>0$, the interval widths are $ϵ(1+ϵ{)}^i$. The left-endpoint Riemann sum for the decreasing integrand $x^{-1-\alpha }$ is the geometric seriesThe right-endpoint sum is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$. Integral comparison on each interval gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds tend to $1/\alpha$ as $ϵ\to 0^{+}$, proving the improper integral by the squeeze theorem. Although the largest interval width is unbounded on the infinite domain, these explicit upper and lower bounds justify the limit without incorrectly applying a finite-interval mesh argument to the entire half-line.

Every Riemann-integrable function on a compact interval can be approximated in $L^1$ by continuous functions. Consequently their integrals converge uniformly over all subintervals.

The total variation of a function $f:[a,b]\to \mathbb{R}$ measures its accumulated rises and falls, rather than just its change between the endpoints. For each partition of an interval $\Pi =(a=t_0<\cdots <t_n=b)$, add the absolute values of its sampled increments. Then take the supremum over all such partitions of an interval:The function has bounded variation when $V_a^b(f)<\infty$. This definition uses its value at every point of the interval, including both endpoints.^[ref]

For a nondecreasing function, all increments are nonnegative and the sum telescopes to $f(b)-f(a)$. More generally, the triangle inequality gives $|f(b)-f(a)|\le V_a^b(f)$. Equality can fail dramatically when the function changes direction: an object can travel a long distance and finish near its starting position. Here we measure movement of the function's value; arc length of its function graph also includes horizontal movement.

On the vector space of continuous functions $f:[0,1]\to \mathbb{R}$ of bounded variation with $f(0)=0$, the total variation of a function is a norm. Its triangle inequality and absolute homogeneity of a norm follow by applying the corresponding absolute value identities to every partition of an interval, then taking the supremum. The partition of an interval $(0,t,1)$ gives $|f(t)|\le V_0^1(f)$, so zero total variation of a function forces $f=0$. The same inequality gives $\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_V$, proving that convergence for this norm implies uniform convergence. Without $f(0)=0$, every constant function would have zero total variation of a function.

Define a piecewise linear function by joining the five points $(0,0),(1,2),(2,-1),(3,3),(4,1)$. Its net change is $1$, but the increments at these knots are $2,-3,4,-2$. Their absolute values giveThis is its exact total variation, not merely one lower bound. Given any partition of an interval, insert all five knots. The triangle inequality ensures that this partition refinement cannot decrease the sum. On each resulting monotone function segment, the absolute increments telescope, so every refined sum is $11$. The knot partition of an interval itself attains $11$.

Sampling only at $0,4$ gives $1$. Sampling at $0,1,3,4$ gives $2+1+2=5$. Inserting the remaining knot at $2$ reveals the descent and rise that were hidden inside $[1,3]$, giving $11$. These are nested partitions of an interval: their sums increase because each is a partition refinement of the preceding one. Arbitrarily chosen unrelated partitions of an interval need not produce an increasing sequence of sums.

For each positive integer $n$, put $f_n(t)=\sin (2\pi nt)/n$ on $[0,1]$. Its supremum norm is exactly $1/n$, so this sequence of functions converges to zero by uniform convergence. Nevertheless,Each of its $n$ complete sine oscillations contributes $4/n$: rise from zero to $1/n$, fall through zero to $-1/n$, then return to zero. Taking a partition of an interval through all extrema attains $4$. Every further partition refinement merely subdivides monotone function segments, so the same reasoning as in partition example of total variation proves the upper bound too.

For a continuously differentiable function on a closed bounded interval, total variation also satisfiesApplied here, the derivative is $2\pi \cos (2\pi nt)$, and integrating its absolute value over $n$ periods gives $4$. The differentiability hypothesis matters: the partition definition applies much more widely than this derivative formula.^[ref] Small pointwise size and small accumulated movement are different approximation goals.

Let $f$ have bounded variation, and let $g,h$ be continuous functions on $[a,b]$. In any tagged sum for the Riemann-Stieltjes integral,Taking the limit proves the displayed error bound. Thus approximating $g$ uniformly to accuracy $\epsilon$ controls this integration error by $\epsilon V_a^b(f)$, even when the integrator is not differentiable. For the piecewise linear function in partition example of total variation, accuracy $0.001$ gives an error at most $0.011$. Total variation supplies exactly the movement budget that an endpoint difference would miss.

Camille Jordan developed bounded variation as part of his work in real analysis; his treatment of functions and arc length was part of the influence of his analysis textbooks.^[ref] The Jordan decomposition of a function of bounded variation expresses the underlying idea especially well: complicated changes of direction can be separated into two functions that only increase.

For $p\ge 1$, the displayed quantity measures the accumulated oscillation of a path in a metric space. For $p=1$ it is total variation. A Hölder continuous path of exponent $\alpha$ has finite $p$-variation on a compact interval if $p\alpha \ge 1$. Applying this construction to the Carnot-Carathéodory distance gives the defining regularity of a weak geometric p-rough path.

For a càdlàg function, replace the last sampled value at $2^{-n}\lceil 2^{n}t\rceil$ by $X_t$. Right continuity makes this change tend to zero. The corrected sum is bounded by the total variation of a function. Conversely every finite partition of an interval can be approximated by dyadic points from the right, retaining the endpoint $t$. The triangle inequality bounds that partition's limiting sum by the dyadic sums. Taking the supremum proves convergence, including the value $+\infty$.

If continuously differentiable $F_j,F$ converge uniformly on a compact interval and have uniformly bounded total variation of a function, then $\int b{F}_j^{\prime }\to \int b{F}^{\prime }$ for every continuous $b$. For continuously differentiable test functions this is integration by parts. Uniform approximation by polynomials extends the conclusion to continuous tests using the variation bound.

Every real function on $[a,b]$ has bounded variation if and only if it is a difference of two nondecreasing functions. For a normalized version, let $v(t)=V_a^t(f)$ and $v(a)=0$. Its accumulated positive and negative variations areBoth are nondecreasing functions, $P(a)=N(a)=0$, andThis is the Jordan decomposition of a function of bounded variation.^[ref] Equivalently one may write $f=f_{+}-f_{-}$ with $f_{+}=(v+f)/2$ and $f_{-}=(v-f)/2$; those functions need not vanish at $a$.

For the piecewise linear function in partition example of total variation, the successive values of $P$ are $0,2,2,6,6$, while those of $N$ are $0,0,3,3,5$. At $t=4$, the total rise is $6$ and the total fall is $5$: their difference gives the net change $1$, and their sum gives the total variation $11$. A declining segment increases $N$ while $P$ stays constant; an ascending segment does the opposite.

For a short proof outline, total variation is additive over adjacent intervals, because any two partitions of an interval have a common partition refinement. Thus $v(t)-v(s)=V_s^t(f)\ge |f(t)-f(s)|$ for $s<t$. Substituting this inequality into the formulas for $P$ and $N$ shows their increments are nonnegative. Conversely, if $f=g-h$ for two nondecreasing functions, the triangle inequality bounds each partition sum by $g(b)-g(a)+h(b)-h(a)$. This proves finite total variation without restricting the number of changes of direction.

A proper Riemann integral on a compact interval requires a bounded integrand, but a derivative can be unbounded. For $f(x)=x^2\sin (1/x^2)$ away from zero and $f(0)=0$, the difference quotient gives $f^{\prime }(0)=0$, whereas $f^{\prime }(x)=2x\sin (1/x^2)-2\cos (1/x^2)/x$ for $x\ne 0$. At $x_n=(2\pi n{)}^{-1/2}$, this derivative is $-2/x_n$ and is unbounded. The fundamental theorem of calculus version that integrates a derivative therefore needs an explicit integrability hypothesis.

For real continuous functions $f,g$ on a compact interval with $g\ge 0$, some $\alpha$ in the interval satisfies $\int fg=f(\alpha )\int g$. If $\int g>0$, the weighted average lies between the minimum and maximum of $f$, and the intermediate value theorem supplies $\alpha$. If $\int g=0$, the bound $|\int fg|\le \sup |f|\int g$ makes both sides zero. Nonnegativity of the weight is essential to this argument.

If $f,g:[a,b]\to \mathbb{R}$ are continuous functions, $g\ge 0$, and $f$ is a nondecreasing monotonic function, there is $d\in [a,b]$ satisfying the identity in the title. It expresses the weighted integral as a mixture of the two endpoint values of $f$, with the split occurring in the accumulated weight.

Put $G={\int }_a^{b}g$ and $I={\int }_a^{b}fg$. The monotonicity of the Riemann integral gives $f(a)G\le I\le f(b)G$. The function $H(x)={\int }_a^{x}g$ is continuous; indeed, $|H(y)-H(x)|\le {\max }_{[a,b]}|g||y-x|$ by the extreme value theorem and the triangle inequality for the Riemann integral. Thereforeis continuous, with $K(a)=f(b)G$ and $K(b)=f(a)G$. The intermediate value theorem gives $K(d)=I$. If the endpoint values coincide, the bounds force $I$ to have that common value and an endpoint works. The proof covers zero total weight and a degenerate interval without division.

A function $g$ on $[0,\infty )$ is directly Riemann integrable when its upper and lower sums on an equal-width mesh are absolutely finite and converge to the same finite integral as the mesh width tends to zero. A locally absolutely continuous integrable function with integrable derivative has this property: the upper-minus-lower sum is bounded by mesh width times its total variation. The derivative bound therefore gives bounded variation as well as tail control. This stronger form of integrability is a hypothesis of the key renewal theorem.

An improper integral is defined by taking a limit when an integration endpoint is infinite or the integrand is unbounded. Its convergence must be established before ordinary integral manipulations are applied.

Let $h$ be measurable, integrable over a positive period $L$, and have mean zero there. Its positive and negative parts have equal integrals $H$. On the $k$-th period the signed integral of $h(x)/x$ has absolute value at mostThis follows by bounding the positive and negative weighted integrals between $H/((k+1)L)$ and $H/(kL)$. The bounds are summable, so the integrals at period endpoints converge. A leftover partial period has absolute integral at most $\Vert h{\Vert }_{L^1(0,L)}/(kL)$, tending to zero, which proves convergence for arbitrary upper endpoints too.

In contrast, each full-period integral of $|h(x)|/x$ is at least $2H/((k+1)L)$. It diverges in total unless $H=0$, equivalently $h=0$ almost everywhere. Thus a nonzero mean-zero periodic function gives a convergent improper integral here but no Lebesgue integrability.

For $p>0$, the function $x\mapsto \sin x/x^p$ is Lebesgue integrable on $(0,\infty )$ exactly when $1<p<2$. Near zero, $\sin x/x\to 1$, so absolute integrability is equivalent to that of $x^{1-p}$, which requires $p<2$. At infinity, $p>1$ gives the upper bound $x^{-p}$. For $0<p\le 1$, on each interval $[k\pi ,(k+1)\pi ]$,and the corresponding P-series diverges.

The signed improper integral nevertheless converges for every $0<p<2$: near zero the same local comparison applies, and at infinity the Dirichlet test for improper integrals uses the bounded antiderivative of sine and the decreasing factor $x^{-p}\to 0$. Thus $0<p\le 1$ gives conditional convergence, while $1<p<2$ gives absolute integrability. In particular, substituting $t=1/x$ turns ${\int }_0^1|x^{-1}\sin (1/x)|dx$ into the divergent tail ${\int }_1^{\infty }|\sin t|/tdt$.

For a locally integrable real or complex function, the displayed tail condition is equivalent to ordinary convergence of its improper integral at infinity. Apply the Cauchy criterion for a function limit to its truncated integrals, using completeness of the real or complex numbers. The necessity follows by subtracting two nearby-to-limit truncations; sufficiency follows by first taking a convergent sequence of truncations and then bounding all other late truncations against it. The analogous criterion applies at a finite singular endpoint. Bounding the modulus of every tail by its absolute-value integral proves that an absolutely convergent improper integral converges ordinarily.

An improper integral whose absolute-value integral is finite. The triangle inequality for an integral bounds each truncated tail by the corresponding tail of $|f|$, so the Cauchy criterion for an improper integral gives ordinary convergence. Absolute convergence is stronger than a Cauchy principal value, which may exist only because two divergent sides cancel. The continuous sine-product integrand $\sin (ax)\sin (bx)/x^2$ has a removable value $ab$ at zero and an integrable $1/x^2$ tail; its auxiliary complex contour integrand can nevertheless require a principal value.

If $F(t)={\int }_a^{t}f(x)dx$ is bounded and $g$ is a nonnegative differentiable decreasing function tending to zero, then ${\int }_a^{\infty }f(x)g(x)dx$ converges as an improper integral. Integration by parts bounds the tail ${\int }_A^{B}fg$ by a constant multiple of $g(A)$, because $F$ is bounded and ${\int }_A^B|g^{\prime }|=g(A)-g(B)$. In particular ${\int }_1^{\infty }{x}^{-\alpha }\sin xdx$ converges for every $\alpha >0$, even though absolute integrability of that tail requires $\alpha >1$.

A logarithmic divergence grows like a logarithm as a cutoff is removed. For example ${\int }_{ϵ}^{1}dx/x=\log (1/ϵ)\to \infty$ as $ϵ\to 0^{+}$. An integrand $f(x)=C/x+O(1)$ near zero gives $C\log (1/ϵ)+O(1)$. This behavior distinguishes logarithmic sensitivity to an infrared or ultraviolet cutoff from power-law divergence.

The improper integral ${\int }_1^{\infty }{u}^{-p}du$ converges exactly when $p>1$, in which case its value is $1/(p-1)$.

A function has an integrable singularity at a point if it becomes unbounded there but its absolute value has a finite improper integral over a sufficiently small surrounding interval, one-sided at an endpoint. The endpoint behavior of the arcsine distribution's probability density function is an example.

An endpoint integrable singularity with $f(x)\sim C/\sqrt{x-x_0}$ as $x\downarrow x_0$, for finite nonzero $C$. It is an integrable singularity because ${\int }_{x_0}^{x_0+ϵ}(x-x_0{)}^{-1/2}dx=2\sqrt{ϵ}$. One common origin is a stationary change of variables: if $y-y_0=b{z}^2+O(z^4)$ with $b>0$ and a smooth nonzero probability density for $z$ at zero, the two inverse branches give a probability density for $y$ proportional to $(y-y_0{)}^{-1/2}$.

If Riemann-integrable functions satisfy $f(x)\le g(x)$ throughout $[a,b]$, thenThis follows directly by comparing their lower and upper Darboux sums, or their Riemann sums on a common sequence of refining partitions.

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.

A lower Darboux sum uses the infimum of a function on each partition interval, while an upper sum uses the supremum.

For a bounded real function, its oscillation on an interval is its full range width there. An upper Darboux sum minus a lower Darboux sum is the sum of these widths multiplied by their interval lengths. This gives the Riemann integrability criterion in terms of oscillation.

For a bounded real function, inserting partition points increases its lower Darboux sum and decreases its upper Darboux sum. This follows interval by interval because infima increase on smaller sets while suprema decrease, and the new widths sum to the original width. A common partition refinement consequently proves $s(f,D_1)\le S(f,D_2)$ for arbitrary two partitions.

For a bounded function $f$ and a partition of an interval $P$, the upper Darboux sum multiplies the supremum of $f$ on each subinterval by its length and adds the results.

For a bounded function $f$ and a partition of an interval $P$, the lower Darboux sum multiplies the infimum of $f$ on each subinterval by its length and adds the results.

For a nonnegative Riemann-integrable function and an interval partition,Every factor is nonnegative, so multiplication of $1+t\le e^t$ is legitimate. The final step uses the lower-sum bound on the Riemann integral.

For a bounded function,The function is Riemann integrable exactly when these values agree.

The lower Darboux integral of a bounded real function on a closed interval is the supremum of its lower Darboux sums over all partitions of an interval. Refinement increases a lower Darboux sum. A common refinement of two arbitrary partitions of an interval proves that every lower Darboux sum is at most every upper Darboux sum, hence the lower Darboux integral never exceeds the upper Darboux integral.

The upper Darboux integral of a bounded real function on a closed interval is the infimum of its upper Darboux sums over all partitions of an interval. Refinement decreases an upper Darboux sum, and the function is Riemann integrable precisely when this infimum agrees with its lower Darboux integral.

For a nonnegative unbounded function, one truncation convention sets $f_r=\min (f,r)$ and asks that every $f_r$ be Riemann integrable and that $\int f_r$ have a finite limit as $r\to \infty$.

Under uniform length on $[0,1]$, the position $K$ of the first occurrence of a fixed decimal digit hasThe set of expansions in which the digit never occurs has length zero, and