Darboux integral

In the branch of mathematics known as real analysis, the Darboux integral is constructed using Darboux sums and is one possible definition of the integral of a function. Darboux integrals are equivalent to Riemann integrals, meaning that a function is Darboux-integrable if and only if it is Riemann-integrable, and the values of the two integrals, if they exist, are equal.^[1] The definition of the Darboux integral has the advantage of being easier to apply in computations or proofs than that of the Riemann integral. Consequently, introductory textbooks on calculus and real analysis often develop Riemann integration using the Darboux integral, rather than the true Riemann integral.^[2] Moreover, the definition is readily extended to defining Riemann–Stieltjes integration.^[3] Darboux integrals are named after their inventor, Gaston Darboux (1842–1917).

For any given partition, the upper Darboux sum is always greater than or equal to the lower Darboux sum. Furthermore, the lower Darboux sum is bounded below by the rectangle of width (b−a) and height inf(f) taken over [a, b]. Likewise, the upper sum is bounded above by the rectangle of width (b−a) and height sup(f).
$(b-a)\inf _{x\in [a,b]}f(x)\leq L_{f,P}\leq U_{f,P}\leq (b-a)\sup _{x\in [a,b]}f(x)$

The lower and upper Darboux integrals satisfy
${\underline {\int _{a}^{b}}}f(x)\,dx\leq {\overline {\int _{a}^{b}}}f(x)\,dx$

Given any c in (a, b)
${\begin{aligned}{\underline {\int _{a}^{b}}}f(x)\,dx&={\underline {\int _{a}^{c}}}f(x)\,dx+{\underline {\int _{c}^{b}}}f(x)\,dx\\[6pt]{\overline {\int _{a}^{b}}}f(x)\,dx&={\overline {\int _{a}^{c}}}f(x)\,dx+{\overline {\int _{c}^{b}}}f(x)\,dx\end{aligned}}$

The lower and upper Darboux integrals are not necessarily linear. Suppose that g:[a, b] → R is also a bounded function, then the upper and lower integrals satisfy the following inequalities:
${\begin{aligned}{\underline {\int _{a}^{b}}}f(x)\,dx+{\underline {\int _{a}^{b}}}g(x)\,dx&\leq {\underline {\int _{a}^{b}}}(f(x)+g(x))\,dx\\[6pt]{\overline {\int _{a}^{b}}}f(x)\,dx+{\overline {\int _{a}^{b}}}g(x)\,dx&\geq {\overline {\int _{a}^{b}}}(f(x)+g(x))\,dx\end{aligned}}$

For a constant c ≥ 0 we have
${\begin{aligned}{\underline {\int _{a}^{b}}}cf(x)\,dx&=c{\underline {\int _{a}^{b}}}f(x)\,dx\\[6pt]{\overline {\int _{a}^{b}}}cf(x)\,dx&=c{\overline {\int _{a}^{b}}}f(x)\,dx\end{aligned}}$

For a constant c ≤ 0 we have
${\begin{aligned}{\underline {\int _{a}^{b}}}cf(x)\,dx&=c{\overline {\int _{a}^{b}}}f(x)\,dx\\[6pt]{\overline {\int _{a}^{b}}}cf(x)\,dx&=c{\underline {\int _{a}^{b}}}f(x)\,dx\end{aligned}}$

Consider the function
${\begin{aligned}&{}F:[a,b]\to \mathbb {R} \\&{}F(x)={\underline {\int _{a}^{x}}}f(t)\,dt,\end{aligned}}$

Examples[edit]

A Darboux-integrable function[edit]

Suppose we want to show that the function $f(x)=x$ is Darboux-integrable on the interval $[0,1]$ and determine its value. To do this we partition $[0,1]$ into $n$ equally sized subintervals each of length $1/n$ . We denote a partition of $n$ equally sized subintervals as $P_{n}$ .

Now since $f(x)=x$ is strictly increasing on $[0,1]$ , the infimum on any particular subinterval is given by its starting point. Likewise the supremum on any particular subinterval is given by its end point. The starting point of the $k$ -th subinterval in $P_{n}$ is $(k-1)/n$ and the end point is $k/n$ . Thus the lower Darboux sum on a partition $P_{n}$ is given by