Darboux Integral

• categories: Real Analysis, Definition

Definition

Darboux Integrability is a method for defining the integral of a function based on the concept of upper and lower sums. It is particularly useful for determining the integrability of functions that may not be Riemann integrable.

Darboux Upper and Lower Sums

For a function $f:[a,b]\to \mathbb{R}$ and a partition $P=\{x_0,x_1,\dots ,x_n\}$ of the interval $[a,b]$, where $a=x_0<x_1<\cdots <x_n=b$:

• Lower Sum ( L(f, P) ):

$$L(f,P)=\sum _{i=0}^{n-1}{m}_i(x_{i+1}-x_i)$$

where

$$m_i=\underset{x\in [x_i,x_{i+1}]}{\inf }f(x).$$

• Upper Sum ( U(f, P) ):

$$U(f,P)=\sum _{i=0}^{n-1}{M}_i(x_{i+1}-x_i)$$

where

$$M_i=\underset{x\in [x_i,x_{i+1}]}{\sup }f(x).$$

Integrability Condition

A function $f$ is said to be Darboux integrable on the interval $[a,b]$ if the lower and upper sums can be made arbitrarily close as the partition gets finer. Formally, $f$ is Darboux integrable if:

$$\underset{||P||\to 0}{\lim }(U(f,P)-L(f,P))=0$$

where ( ||P|| ) is the norm of the partition, defined as:

$$||P||=\underset{i}{\max }(x_{i+1}-x_i).$$

Relationship to Riemann Integrability

• A function is Riemann integrable if and only if it is Darboux integrable.
• The concepts of upper and lower sums provide an alternative perspective on Riemann integration, allowing one to analyze functions with potentially discontinuous behavior.

Properties

1. Monotonic Functions: If $f$ is monotonic on $[a,b]$, then $f$ is Darboux integrable.

2. Continuous Functions: If $f$ is continuous on $[a,b]$, then it is Darboux integrable.

3. Discontinuities: A bounded function that has a finite number of discontinuities on $[a,b]$ is Darboux integrable.

Example

Consider the function $f(x)=x$ on the interval $[0,1]$.

1. Choose a partition $P=\{0,\frac{1}{2},1\}$

   • Lower sum
   $$L(f,P)=m_0(x_1-x_0)+m_1(x_2-x_1)=0\cdot \frac{1}{2}+\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}.$$
   • Upper sum
   $$U(f,P)=M_0(x_1-x_0)+M_1(x_2-x_1)=1\cdot \frac{1}{2}+1\cdot \frac{1}{2}=1.$$

2. Refine the partition to make the upper and lower sums converge.

Ultimately, as the partition is refined, the lower and upper sums will converge to the same value, which is the definite integral:

$${\int }_0^{1}f(x)dx=\frac{1}{2}.$$