Darboux integral

A partition of an interval [a, b] is a finite sequence of values x_i such that

${\displaystyle a=x_{0}<x_{1}<\cdots <x_{n}=b.\,\!}$

Each interval [x_i−1, x_i] is called a subinterval of the partition. Let ƒ: [a, b] → ℝ be a bounded function, and let

${\displaystyle P=(x_{0},\ldots ,x_{n})\,\!}$

be a partition of [a, b]. Let

${\displaystyle {\begin{aligned}M_{i}=\sup _{x\in [x_{i-1},x_{i}]}f(x),\\m_{i}=\inf _{x\in [x_{i-1},x_{i}]}f(x).\end{aligned}}}$

Lower (green) and upper (green plus lavender) Darboux sums for four subintervals

The upper Darboux sum of ƒ with respect to P is

${\displaystyle U_{f,P}=\sum _{i=1}^{n}(x_{i}-x_{i-1})M_{i}.\,\!}$

The lower Darboux sum of ƒ with respect to P is

${\displaystyle L_{f,P}=\sum _{i=1}^{n}(x_{i}-x_{i-1})m_{i}.\,\!}$

The lower and upper Darboux sums are often called the lower and upper sums.

The upper Darboux integral of ƒ is

${\displaystyle U_{f}=\inf\{U_{f,P}\colon P{\text{ is a partition of }}[a,b]\}.\,\!}$

The lower Darboux integral of ƒ is

${\displaystyle L_{f}=\sup\{L_{f,P}\colon P{\text{ is a partition of }}[a,b]\}.\,\!}$

In some literature an integral symbol with an underline and overline represent the lower and upper Darboux integrals respectively.

${\displaystyle {\begin{aligned}L_{f}\equiv {\underline {\int _{a}^{b}}}f(x)\,\mathrm {d} x&\quad U_{f}\equiv {\overline {\int _{a}^{b}}}f(x)\,\mathrm {d} x\end{aligned}},}$

and like Darboux sums they are sometimes simply called the lower and upper integrals.

If U_ƒ = L_ƒ, then we call the common value the Darboux Integral.[4] We also say that ƒ is Darboux-integrable or simply integrable and set

${\displaystyle \int _{a}^{b}{f(t)\,dt}=U_{f}=L_{f},\,\!}$

An equivalent and sometimes useful criterion for the integrability of f is to show that for every ε > 0 there exists a partition P_ε of [a, b] such that[5]

${\displaystyle U_{f,P_{\epsilon }}-L_{f,P_{\epsilon }}<\varepsilon .}$

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 kth 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

${\displaystyle {\begin{aligned}L_{f,P_{n}}&=\sum _{k=1}^{n}f(x_{k-1})(x_{k}-x_{k-1})\\&=\sum _{k=1}^{n}{\frac {k-1}{n}}\cdot {\frac {1}{n}}\\&={\frac {1}{n^{2}}}\sum _{k=1}^{n}[k-1]\\&={\frac {1}{n^{2}}}\left[{\frac {(n-1)n}{2}}\right]\end{aligned}}}$

similarly, the upper Darboux sum is given by

${\displaystyle {\begin{aligned}U_{f,P_{n}}&=\sum _{k=1}^{n}f(x_{k})(x_{k}-x_{k-1})\\&=\sum _{k=1}^{n}{\frac {k}{n}}\cdot {\frac {1}{n}}\\&={\frac {1}{n^{2}}}\sum _{k=1}^{n}k\\&={\frac {1}{n^{2}}}\left[{\frac {(n+1)n}{2}}\right]\end{aligned}}}$

Since

${\displaystyle {\begin{aligned}U_{f,P_{n}}-L_{f,P_{n}}&={\frac {1}{n}}\end{aligned}}}$

Thus for given any ε > 0, we have that any partition P_n with n > 1/ε satisfies

${\displaystyle {\begin{aligned}U_{f,P_{n}}-L_{f,P_{n}}&<\epsilon \end{aligned}}}$

which shows that f is Darboux integrable. To find the value of the integral note that

${\displaystyle {\begin{aligned}\int _{0}^{1}f(x)\,dx&=\lim _{n\to \infty }U_{f,P_{n}}=\lim _{n\to \infty }L_{f,P_{n}}={\frac {1}{2}}\end{aligned}}}$

Darboux sums

Darboux upper sums of the function y = x²

Darboux lower sums of the function y = x²

Suppose we have the function f: [0, 1] → ℝ defined as

${\displaystyle {\begin{aligned}f(x)&={\begin{cases}0,&{\text{if }}x{\text{ is rational}}\\1,&{\text{if }}x{\text{ is irrational}}\end{cases}}\end{aligned}}}$

Since the rational and irrational numbers are both dense subsets of ℝ, it follows that f takes on the value of 0 and 1 on every subinterval of any partition. Thus for any partition P we have

${\displaystyle {\begin{aligned}L_{f,P}&=\sum _{k=1}^{n}(x_{k}-x_{k-1})\inf _{x\in [x_{k-1},x_{k}]}f=0\\U_{f,P}&=\sum _{k=1}^{n}(x_{k}-x_{k-1})\sup _{x\in [x_{k-1},x_{k}]}f=1\end{aligned}}}$

from which we can see that the lower and upper Darboux integrals are unequal.