Real Analysis/Darboux Integral

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Darboux Integral		 Generalized Integration→

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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.

[edit] Upper and Lower Sums

[edit] Definition

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})

[edit] Definition

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

[edit] Theorem

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}^*)

[edit] Proof

Let x_i,x_{i-1}\in\mathcal{P} and let x^*\in\mathcal{P}^*\setminus\mathcal{P} be such that xi − 1 < x * < xi. 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, mi(xi − xi − 1) < m'i(x * − xi − 1) + m''i(xi − 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}^*)

[edit] Darboux Integration

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

[edit] Theorem

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

[edit] Proof

(\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 α > β cannot be true. Also, as α,β are a supremum and infimum respectively, α < β is also not possible. Hence, α = β = 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

[edit] Equivalence of Riemann and Darboux Integrals

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.

[edit] Lemma

(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 δ1 < δ2 \Rightarrow \varepsilon(\delta_1)<\varepsilon(\delta_2)

[edit] Proof

Let δ1 < δ2. 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)

[edit] Theorem

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

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

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

[edit] Proof

(\Rightarrow) Let ε > 0 be given.

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

Let partitions P1 and P2 be the same refinement of \dot{P} but with different tags.

Therefore, | S(f,P1) − S(f,P2) | < ε \forall P1 and P2

i.e., by the triangle inequality, | S(f,P1) | − | S(f,P2) | < ε

Gap Lemma \Rightarrow U(f,P) − L(f,P) < ε,

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

(\Leftarrow)Let ε > 0 be given.

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

Hence, | L − S(f,P) | < ε as L(f,P)\leq S(f,P)\leq U(f,P)

By Lemma 3.1, | L − S(f,P') | < ε 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 δ, which is not directly present in either the Riemann or Darboux definition.

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