Riemann integration

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

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.

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.

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.

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 $f:[a,b]\to \mathbb{R}$ isThe function has bounded variation when this quantity is finite.

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 $f$ of bounded variation is the difference of two nondecreasing functions. Writing $V_f$ for its cumulative total variation, one may takeup to constants chosen for the desired normalization.

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.

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.

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 $1/(p-1)$.

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 $U(f,P)-L(f,P)<\epsilon$.

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, 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.

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