# 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})

(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}. 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}), but as $x_i\in\mathcal{P}$ is arbitrary, we have that $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}), 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.