Last modified on 7 September 2009, at 19:05

Real Analysis/Darboux Integral

Real Analysis
Darboux Integral

Another popular definition of "integration" was provided by Jean Gaston Darboux and is often used in more advanced texts. In this chapter, we will define the Darboux integral, and demonstrate the equivalence of Riemann and Darboux integrals.

Upper and Lower SumsEdit

DefinitionEdit

Let f:[a,b]\to\mathbb{R}

Let \mathcal{P} be a (finite) partition of [a,b]


For every x_i\in\mathcal{P} define:

m_i\dot{=}\inf \{f(x)|x\in [x_{i-1},x_i]\} and

M_i\dot{=}\sup \{f(x)|x\in [x_{i-1},x_i]\}


The Upper Sum of f with respect to \mathcal{P} is defined as U(f,\mathcal{P})=\sum_{i=1}^n M_i(x_i-x_{i-1})

The Lower Sum of f with respect to \mathcal{P} is defined as L(f,\mathcal{P})=\sum_{i=1}^n m_i(x_i-x_{i-1})

DefinitionEdit

A partition \mathcal{P}^* is said to be a Refinement for a given partition \mathcal{P} iff \mathcal{P}\subset\mathcal{P}^*

TheoremEdit

Let f:[a,b]\to\mathbb{R}

Let \mathcal{P} be a partition and let \mathcal{P}^* be a refinement of \mathcal{P}. Then,

(i) L(f,\mathcal{P})<L(f,\mathcal{P}^*)

(ii) U(f,\mathcal{P})>U(f,\mathcal{P}^*)

ProofEdit

Let x_i,x_{i-1}\in\mathcal{P} and let x^*\in\mathcal{P}^*\setminus\mathcal{P} be such that x_{i-1}<x^*<x_i. Also, let m_i=\inf \{f(x)|x\in [x_{i-1},x_i]\}, m'_i=\inf \{f(x)|x\in [x_{i-1},x^*]\} and m''_i=\inf \{f(x)|x\in [x^*,x_i]\}

Obviously, m_i(x_i-x_{i-1})<m'_i(x^*-x_{i-1})+m''_i(x_i-x^*), but as x_i\in\mathcal{P} is arbitrary, we have that L(f,\mathcal{P})<L(f,\mathcal{P}^*)

Similarly, we can prove U(f,\mathcal{P})>U(f,\mathcal{P}^*)

Darboux IntegrationEdit

Let f:[a,b]\to\mathbb{R}

We say that f is Darboux Integrable on [a,b] if and only if

\sup_{\mathcal{P}}L(f,\mathcal{P})=\inf_{\mathcal{P}}U(f,\mathcal{P}), where the supremum is taken over the Set of all partitions on that interval

L=\sup_{\mathcal{P}}L(f,\mathcal{P}) is also written as \int_a^b f=L

TheoremEdit

Let f:[a,b]\to\mathbb{R}

f is Darboux integrable over [a,b] if and only if for every \varepsilon>0, there exists a partition \mathcal{P} on [a,b] such that U(f,\mathcal{P})-L(f,\mathcal{P})<\varepsilon

ProofEdit

(\Rightarrow)Let A=\int_a^b f and let \varepsilon>0 be given. Thus, by Gap Lemma, there exists a partition \mathcal{P} such that both U(f,\mathcal{P}),L(f,\mathcal{P})\in V_{\frac{\varepsilon}{2}}(A), and hence U(f,\mathcal{P})-L(f,\mathcal{P})<\varepsilon

(\Leftarrow)Let \mathcal{P}_0 be any partition on [a,b]. Observe that L(f,\mathcal{P}_0) is a lower bound of the set \mathcal{U}=\{U(f,\mathcal{P})|\mathcal{P} is any partition\} and that U(f,\mathcal{P}_0) is an upper bound of the set \mathcal{L}=\{L(f,\mathcal{P})|\mathcal{P} is any partition\}

Thus, let \alpha=\sup\mathcal{L} and \beta=\inf\mathcal{U}. As L(f,\mathcal{P})<U(f,\mathcal{P}), we have that \alpha>\beta cannot be true. Also, as \alpha,\beta are a supremum and infimum respectively, \alpha<\beta is also not possible. Hence, \alpha=\beta=L (say).

As L=\sup_{\mathcal{P}} L(f,\mathcal{P})=\inf_{\mathcal{P}} U(f,\mathcal{P}), we have that \int_a^b f=L

Equivalence of Riemann and Darboux IntegralsEdit

At first sight, it may appear that the Darboux integral is a special case of the Riemann integral. However, this is illusionary, and indeed the two are equivalent.

LemmaEdit

(1) Let f:[a,b]\rightarrow\mathbb{R} be Darboux Integrable, with integral L

Define function \varepsilon (\delta)=\sup\{|L-S(f,\dot{P})|:\|\dot{P}\|=\delta\}

(2) Then \delta_1<\delta_2 \Rightarrow \varepsilon(\delta_1)<\varepsilon(\delta_2)

ProofEdit

Let \delta_1<\delta_2. Consider set T of tagged partitions \dot{P} such that \varepsilon(\delta_1)\leq|L-S(f,\dot{P})|

Let T' be the set of \dot{P'} where \dot{P'}\subset\dot{P}\in T and \|\dot{P'}\|=\delta_2>\delta_1

note that T'\neq T and that the set T' indeed contains all partitions \dot{P'} with \|\dot{P'}\|=\delta_2

Now, for \dot{P}\in T, we can construct \dot{P'}\in T' such that |L-S(f,\dot{P'})|>|L-S(f,\dot{P})|

Hence, \displaystyle\sup_{\dot{P}\in T}\{|L-S(f,\dot{P})|\}<\sup_{\dot{P'}\in T'}\{|L-S(f,\dot{P'})|\}

i.e. \varepsilon(\delta_2)>\varepsilon(\delta_1)

TheoremEdit

Let f:[a,b]\rightarrow\mathbb{R}

(1)f is Riemann integrable on [a,b] iff

(2)f is Darboux Integrable on [a,b]

ProofEdit

(\Rightarrow) Let \epsilon>0 be given.

(1)\Rightarrow \exists tagged partition \dot{P} such that |S(f,\dot{P})-L|<\frac{\epsilon}{2}.

Let partitions P_1 and P_2 be the same refinement of \dot{P} but with different tags.

Therefore, |S(f,P_1)-S(f,P_2)|<\epsilon \forall P_1 and P_2

i.e., by the triangle inequality, |S(f,P_1)|-|S(f,P_2)|<\epsilon

Gap Lemma \Rightarrow U(f,P)-L(f,P)<\epsilon,

\epsilon>0 being arbitrary, using Theorem 2.1, we have that f is Darboux Integrable.

(\Leftarrow)Let \epsilon>0 be given.

(2), Theorem 2.1 \Rightarrow \exists partition P such that U(f,P)-L(f,P)<\epsilon

Hence, |L-S(f,P)|<\epsilon as L(f,P)\leq S(f,P)\leq U(f,P)

By Lemma 3.1, |L-S(f,P')|<\epsilon if \|P'\|<\|P\|

Thus, if we put \delta=\|P\|, we have (1)

We note here that the crucial element in this proof is Lemma 3.1, as it essentially is giving an order relation between \varepsilon and \delta, which is not directly present in either the Riemann or Darboux definition.