Darboux integral

Properties

• 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).

  ${\displaystyle (b-a)\underset{x\in [a,b]}{inf}f(x)\le L_{f,P}\le U_{f,P}\le (b-a)\underset{x\in [a,b]}{sup}f(x)}$

• The lower and upper Darboux integrals satisfy

  ${\displaystyle \underset{\_}{{\int }_a^b}f(x)dx\le \overline{{\int }_a^b}f(x)dx}$

• Given any c in (a, b)

  ${\displaystyle \begin{array}{cc}\underset{\_}{{\int }_a^b}f(x)dx& =\underset{\_}{{\int }_a^c}f(x)dx+\underset{\_}{{\int }_c^b}f(x)dx\\ \overline{{\int }_a^b}f(x)dx& =\overline{{\int }_a^c}f(x)dx+\overline{{\int }_c^b}f(x)dx\end{array}}$

• 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:

  ${\displaystyle \begin{array}{cc}\underset{\_}{{\int }_a^b}f(x)dx+\underset{\_}{{\int }_a^b}g(x)dx& \le \underset{\_}{{\int }_a^b}(f(x)+g(x))dx\\ \overline{{\int }_a^b}f(x)dx+\overline{{\int }_a^b}g(x)dx& \ge \overline{{\int }_a^b}(f(x)+g(x))dx\end{array}}$

• For a constant c ≥ 0 we have

  ${\displaystyle \begin{array}{cc}\underset{\_}{{\int }_a^b}cf(x)dx& =c\underset{\_}{{\int }_a^b}f(x)dx\\ \overline{{\int }_a^b}cf(x)dx& =c\overline{{\int }_a^b}f(x)dx\end{array}}$

• For a constant c ≤ 0 we have

  ${\displaystyle \begin{array}{cc}\underset{\_}{{\int }_a^b}cf(x)dx& =c\overline{{\int }_a^b}f(x)dx\\ \overline{{\int }_a^b}cf(x)dx& =c\underset{\_}{{\int }_a^b}f(x)dx\end{array}}$

• Consider the function

  ${\displaystyle \begin{array}{cc}& F:[a,b]\to \mathbb{R}\\ & F(x)=\underset{\_}{{\int }_a^x}f(t)dt,\end{array}}$

then F is Lipschitz continuous. An identical result holds if F is defined using an upper Darboux integral.

Examples

A Darboux-integrable function

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

Now since ${\displaystyle f(x)=x}$ is strictly increasing on ${\displaystyle [0,1]}$, 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 ${\displaystyle k}$-th subinterval in ${\displaystyle P_n}$ is ${\displaystyle (k-1)/n}$ and the end point is ${\displaystyle k/n}$. Thus the lower Darboux sum on a partition ${\displaystyle P_n}$ is given by

${\displaystyle \begin{array}{cc}{L}_{f,P_n}& =\sum _{k=1}^{n}f(x_{k-1})(x_k-x_{k-1})\\ & =\sum _{k=1}^n\frac{k-1}{n}\cdot \frac{1}{n}\\ & =\frac{1}{n^2}\sum _{k=1}^n[k-1]\\ & =\frac{1}{n^2}\left[\frac{(n-1)n}{2}\right]\\ & =\frac{1}{2}-\frac{1}{2n}\end{array}}$

similarly, the upper Darboux sum is given by

${\displaystyle \begin{array}{cc}{U}_{f,P_n}& =\sum _{k=1}^{n}f(x_k)(x_k-x_{k-1})\\ & =\sum _{k=1}^n\frac{k}{n}\cdot \frac{1}{n}\\ & =\frac{1}{n^2}\sum _{k=1}^{n}k\\ & =\frac{1}{n^2}\left[\frac{(n+1)n}{2}\right]\\ & =\frac{1}{2}+\frac{1}{2n}\end{array}}$

Since

${\displaystyle U_{f,P_n}-L_{f,P_n}=\frac{1}{n}}$

Thus for given any ${\displaystyle \epsilon >0}$, we have that any partition ${\displaystyle P_n}$ with ${\displaystyle n>\frac{1}{\epsilon }}$ satisfies

${\displaystyle U_{f,P_n}-L_{f,P_n}<\epsilon }$

which shows that ${\displaystyle f}$ is Darboux integrable. To find the value of the integral note that

${\displaystyle {\int }_0^{1}f(x)dx=\underset{n\to \mathrm{\infty }}{lim}{U}_{f,P_n}=\underset{n\to \mathrm{\infty }}{lim}{L}_{f,P_n}=\frac{1}{2}}$

A nonintegrable function

Suppose we have the Dirichlet function ${\displaystyle f:\mathbb{R}\to [0,1]}$ defined as

${\displaystyle \begin{array}{cc}f(x)& =\{\begin{array}{ll}1& \text{if }x\text{ is rational}\\ 0& \text{if }x\text{ is irrational}\end{array}\end{array}}$

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

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

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

Refinement of a partition and relation to Riemann integration

A refinement of the partition ${\displaystyle x_0,\dots ,x_n}$ is a partition ${\displaystyle y_0,\dots ,y_m}$ such that for all i = 0, …, n there is an integer r(i) such that

${\displaystyle x_i=y_{r(i)}.}$

In other words, to make a refinement, cut the subintervals into smaller pieces and do not remove any existing cuts.

If ${\displaystyle P^{\prime }=(y_0,\dots ,y_m)}$ is a refinement of ${\displaystyle P=(x_0,\dots ,x_n),}$ then

${\displaystyle U_{f,P}\ge U_{f,P^{\prime }}}$

and

${\displaystyle L_{f,P}\le L_{f,P^{\prime }}.}$

If P₁, P₂ are two partitions of the same interval (one need not be a refinement of the other), then

${\displaystyle L_{f,P_1}\le U_{f,P_2},}$

and it follows that

${\displaystyle L_f\le U_f.}$

Riemann sums always lie between the corresponding lower and upper Darboux sums. Formally, if ${\displaystyle P=(x_0,\dots ,x_n)}$ and ${\displaystyle T=(t_1,\dots ,t_n)}$ together make a tagged partition

${\displaystyle x_0\le t_1\le x_1\le \cdots \le x_{n-1}\le t_n\le x_n}$

(as in the definition of the Riemann integral), and if the Riemann sum of ${\displaystyle f}$ is equal to ${\displaystyle R_{f,P}}$ corresponding to P and T, then

${\displaystyle L_{f,P}\le R_{f,P}\le U_{f,P}.}$

From the previous fact, Riemann integrals are at least as strong as Darboux integrals: if the Darboux integral exists, then the upper and lower Darboux sums corresponding to a sufficiently fine partition will be close to the value of the integral, so any Riemann sum over the same partition will also be close to the value of the integral. There is (see below) a tagged partition that comes arbitrarily close to the value of the upper Darboux integral or lower Darboux integral, and consequently, if the Riemann integral exists, then the Darboux integral must exist as well.