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\noindent
\parbox{2.25in}{\bf Math 3220}
\hfill {\Large \bf Darboux Integration} \hfill 
\parbox{2.25in}{\hfill \bf March 2019}

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Jean Gaston Darboux provides us with a useful characterization of Riemann integrability. Instead of stating an integrability condition based on a generalized kind of limit, Darboux's condition is stated in terms of supremums and infimums.

{\bf Notation:} Let $a,b \in \mathbb{R}$ with $a<b$. Suppose that $f:[a,b] \to \mathbb{R}$ is a {\bf bounded} function. Given a partition $\mathcal{P}=\{x_0,\dots,x_n\}$ of $[a,b]$ (i.e., $a=x_0<x_1<\cdots<x_n=b$), let $\Delta x_i = x_i-x_{i-1}$ (i.e., the width of the $i^\text{th}$-subinterval). $\|P\| = \max \{\Delta x_1,\dots,\Delta x_n \}$ is the norm  (or mesh size) of the partition. A sampling for $\mathcal{P}$ is a set $\mathcal{S} = \{s_1,\dots,s_n\}$ where for each $i$, $s_i \in [x_{i-1},x_i]$. Finally, $\RS(f,\mathcal{P},\mathcal{S}) = \sum\limits_{i=1}^n f(s_i)\Delta x_i$ is the Riemann sum of $f$ relative to the partition $\mathcal{P}$ and sampling $\mathcal{S}$.

\vspace{0.05in}

\noindent {\bf Definition:} Let $\mathcal{P} = \{x_0,\dots,x_n \}$ be a partition with $m_i = \inf\limits_{x \in [x_{i-1},x_i]} f(x)$ and $M_i = \sup\limits_{x \in [x_{i-1},x_i]} f(x)$ (these numbers exist because $f$ is bounded). We define\dots

\vspace{-0.15in}

$$\text{Upper Darboux Sum:} \quad U(f,\mathcal{P}) = \sum\limits_{i=1}^n M_i\Delta x_i \qquad \qquad
    \text{Lower Darboux Sum:} \quad L(f,\mathcal{P}) = \sum\limits_{i=1}^n m_i\Delta x_i$$

Let $\mathcal{S}$ be some sampling relative to $\mathcal{P}$. Then $m_i \leq f(s_i) \leq M_i$ (by definition of infimum and supremum), so $L(f,\mathcal{P}) \leq \RS(f,\mathcal{P},\mathcal{S}) \leq U(f,\mathcal{P})$.

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\noindent {\bf Lemma:} Given a partition $\mathcal{P}$, for every $\epsilon>0$ there exists a sampling $\mathcal{S}_1$ such that $U(f,\mathcal{P})- \RS(f,\mathcal{P},\mathcal{S}_1) < \epsilon$ and a sampling $\mathcal{S}_2$ such that $\RS(f,\mathcal{P},\mathcal{S}_2) - L(f,\mathcal{P}) < \epsilon$.

\bproof Consider $\epsilon/(n\Delta x_i) >0$. There exists some $s_i \in [x_{i-1},x_i]$ such that $M_i - f(s_i) < \epsilon/(n\Delta x_i)$ because $M_i = \sup\limits_{x \in [x_{i-1},x_i]} f(x)$. This gives us our sampling $\mathcal{S}_1  = \{s_1,\dots,s_n\}$.

\vspace{-0.15in}

$$U(f,\mathcal{P}) - \RS(f,\mathcal{P},\mathcal{S}_1) = \sum\limits_{i=1}^n M_i\Delta x_i - \sum\limits_{i=1}^n f(s_i)\Delta x_i = \sum\limits_{i=1}^n (M_i-f(s_i))\Delta x_i < \sum\limits_{i=1}^n \dfrac{\epsilon}{n\Delta x_i}\Delta x_i = \sum\limits_{i=1}^n \dfrac{\epsilon}{n} = \epsilon$$
The proof for the lower sum is similar. \eproof

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\noindent {\bf Definition:} 

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$$\text{Upper Darboux Integral:} \quad \overline{\int_a^b} f = U(f) = \inf \{ U(f,\mathcal{P} \;|\; \mathcal{P} \text{ is a partition}\}$$
$$\text{Lower Darboux Integral:} \quad \underline{\int_a^b} f = L(f) = \sup \{ L(f,\mathcal{P} \;|\; \mathcal{P} \text{ is a partition}\}$$

Notice that the set of upper sums is bounded below by lower sums and vice-versa. Therefore, these upper and lower integrals {\it always exist}. We can think of $U(f,\mathcal{P})$ as an overestimate of the integral of $f$ and $L(f,\mathcal{P})$ as an underestimate. So more-or-less, this makes $U(f)$ the lowest of all of the overestimates and $L(f)$ the highest of all the underestimates.

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\noindent {\bf Definition:} Adding more points to a partition yields a {\bf refinement}. In particular, if $\mathcal{P}$ and $\mathcal{P'}$ are partitions such that $\mathcal{P} \subseteq \mathcal{P}'$, we say that $\mathcal{P}'$ is a {\it refinement} of $\mathcal{P}$.  

Notice that if $\mathcal{P}_1$ and $\mathcal{P}_2$ are partitions, then $\mathcal{P}_1 \cup \mathcal{P}_2$ is a {\it common refinement} of both $\mathcal{P}_1$ and $\mathcal{P}_2$.
Also, it should be obvious that if $M=\sup\limits_{x\in [z_0,z_m]} f(x)$ and $M_i = \sup\limits_{x \in [z_{i-1},z_i]} f(x)$, then $M = \max \{ M_1,\dots,M_n\}$. This means that $M(z_n-z_0) = \sum\limits_{i=1}^n M\Delta z_i \geq \sum\limits_{i=1}^n M_i\Delta z_i$. A similar statement follows for infimums. Thus we have that\dots

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\noindent {\bf Lemma:} Given a refinement of partitions $\mathcal{P} \subseteq \mathcal{P}'$, we have\dots
$$L(f,\mathcal{P}) \leq L(f,\mathcal{P}') \leq U(f,\mathcal{P}') \leq U(f,\mathcal{P})$$
In particular, given any two partitions, $\mathcal{P}_1$ and $\mathcal{P}_2$, we have\dots
$$L(f,\mathcal{P}_1) \leq L(f,\mathcal{P}_1 \cup \mathcal{P}_2) \leq U(f,\mathcal{P}_1 \cup \mathcal{P}_2) \leq U(f,\mathcal{P}_2)$$
Therefore, every lower sum is a lower bound for all upper sums and every upper sum is an upper bound for all lower sums. This then implies that $L(f,\mathcal{P}_1) \leq U(f)$ and
$L(f) \leq U(f,\mathcal{P}_2)$ (by the definitions of infimum and supremum). Thus $U(f)$ is an upper bound for all lower sums and $L(f)$ is a lower bound for all upper sums. Thus because $L(f)$ and $U(f)$ are greatest lower and least upper bounds, so for all partitions $\mathcal{P}_1$ and $\mathcal{P}_2$\dots
$$L(f,\mathcal{P}_1) \leq L(f) \leq U(f) \leq U(f,\mathcal{P}_2)$$

Next, we a technical lemma. Its proof doesn't use anything fancy, it's just takes some effort to write down.

\newpage

\noindent {\bf Lemma:} Given a partition $\mathcal{P}'$ and $\epsilon>0$, there is some $\delta>0$ such that for every partition $\mathcal{P}$ with $\|\mathcal{P}\|<\delta$ we have $U(f,\mathcal{P})-U(f,\mathcal{P}\cup\mathcal{P}')<\epsilon$. Similarly there is some $\delta>0$ such that for every partition $\mathcal{P}$ with $\|\mathcal{P}\|<\delta$ we have $L(f,\mathcal{P} \cup \mathcal{P}')-L(f,\mathcal{P})<\epsilon$. Putting these facts together (pick the minimum of two $\delta$'s), there is some $\delta>0$  such that for every partition $\mathcal{P}$ with $\|\mathcal{P}\|<\delta$ we have both $U(f,\mathcal{P})-U(f,\mathcal{P}\cup\mathcal{P}')<\epsilon$ and $L(f,\mathcal{P} \cup \mathcal{P}')-L(f,\mathcal{P})<\epsilon$.

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\bproof Let $M \in \mathbb{R}$ such that $|f(x)|<M$ for all $x \in [a,b]$ ($M$ exists because we assumed that $f$ is bounded on $[a,b]$). Let $\mathcal{P}' = \{x_0,\dots,x_n\}$. Suppose $\epsilon>0$ and let $\delta = \min \{ \epsilon/(2nM), \Delta x_1,\dots,\Delta x_n \}$. 

Consider any partition $\mathcal{P} = \{y_0,\dots,y_m\}$ such that $\|\mathcal{P}\| < \delta$. We want to analyze $\mathcal{P} \cup \mathcal{P}' = \{ z_0,\dots,z_\ell \}$. Now $\|\mathcal{P}\| < \delta \leq \min \{ \Delta x_1,\dots, \Delta x_n \}$ says that the largest gap between two $y_j$'s is smaller than the smallest gap between two $x_k$'s. This means that in $\mathcal{P} \cup \mathcal{P}'$ every $x_k \in \mathcal{P}'$ (other than $x_0$ and $x_n$) is surrounded by $y_j$'s from $\mathcal{P}$. In particular, our partition looks like:

\vspace{-0.15in}

$$a=x_0=y_0 < y_1 < \cdots < y_{i_1} \leq x_1 < y_{i_1+1} <\dots y_{i_2} \leq x_2 < y_{i_2+1} < \cdots < y_m = x_n = b$$

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Let's consider each subinterval $[z_{j-1},z_j]$.

\noindent {\it Case 1:} Suppose we move between $y_k$'s. That is, suppose $[z_{j-1},z_j] = [y_{k-1},y_k]$ for some $k$. Then\dots

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$$ \sup\limits_{x \in [z_{j-1},z_j]} f(x)\Delta z_j = \sup\limits_{x \in [y_{k-1},y_k]} f(x)\Delta y_k $$

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So the corresponding terms in the summations $U(f,\mathcal{P})$ and $U(f,\mathcal{P}\cup\mathcal{P}')$ cancel out.

 \noindent {\it Case 2:} Suppose we jump over an $x_k$. That is, $z_{j-1}=y_{i_k} < x_k=z_j < y_{i_k+1}=z_{j+1}$.
 
 Let $M'''=\sup\limits_{x\in [y_{i_k},y_{i_k+1}]} f(x) =\sup\limits_{x\in [z_{j-1},z_{j+1}]} f(x)$, $M'=\sup\limits_{x\in [y_{i_k},x_{k}]} f(x) =\sup\limits_{x\in [z_{j-1},z_j]} f(x)$, and  $M''=\sup\limits_{x\in [x_{k},y_{i_k+1}]} f(x) =\sup\limits_{x\in [z_{j},z_{j+1}]} f(x)$. Then the $(i_k+1)$-st term in $U(f,\mathcal{P})$ is $M'''\Delta y_{i_k+1} = M'''(\Delta z_j + \Delta z_{j+1}) = M'''\Delta z_j + M'''\Delta z_{j+1}$ and the $j$-th and $(j+1)$-st terms in $U(f,\mathcal{P}\cup\mathcal{P}')$ are $M' \Delta z_j$ and $M'' \Delta z_{j+1}$. Now since $M'$ and $M''$ are bounded above by $M'''$ and the function itself is bounded by $M$, we have $M'''-M' < M$ and $M'''-M'' < M$. Thus the difference between the $(i_k+1)$-st term in $U(f,\mathcal{P})$ and the $j$-th and $(j+1)$-st terms in $U(f,\mathcal{P}\cup\mathcal{P}')$ is $(M'''-M')\Delta z_j + (M'''-M'')\Delta z_{j+1} < M\Delta z_j + M\Delta z_{j+1} < 2M\|\mathcal{P} \cup \mathcal{P}'\| \leq 2M \|\mathcal{P} \| < 2M\delta \leq 2M \dfrac{\epsilon}{2nM} = \dfrac{\epsilon}{n}$. 
 
 So in the difference of upper sums, each case 1 occurrence contributes a $0$ and each case 2 occurrence contributes at most $\epsilon/n$. But case 2 can only occur (at most) $n$-times (each occurrence must involve some new $x_k$). Thus $U(f,\mathcal{P})-U(f,\mathcal{P}\cup\mathcal{P}') < \dfrac{\epsilon}{n}\cdot n = \epsilon$. 
 
 The proof for lower sums is analogous. \eproof

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\noindent {\bf Theorem:} Let $f:[a,b] \to \mathbb{R}$ be a bounded function. Then $f$ is Riemann integrable on $[a,b]$ if and only if $\underline{\int_a^b f} = \overline{\int_a^b f}$. Moreover, in such a case, $\int_a^b f = \underline{\int_a^b f} = \overline{\int_a^b f}$.

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\bproof Let $\underline{I} = \underline{\int_a^b f} = L(f)$ and $\overline{I} = \overline{\int_a^b f} = U(f)$.

Suppose $f$ be Riemann integrable on $[a,b]$ and let $I=\int_a^b f$. Suppose $\epsilon>0$. There exists some partition $\mathcal{P}'$ such that $U(f,\mathcal{P}')-\overline{I} < \epsilon/3$ since $\overline{I}$ is a supremum. $f$ is Riemann integrable so there exists some $\delta>0$ such that $\|\mathcal{P}\|<\delta$ implies $|\RS(f,\mathcal{P},\mathcal{S})-I|<\epsilon/3$. Let $\mathcal{P}'' \supseteq \mathcal{P}'$ be a refinement such that $\|\mathcal{P}''\|<\delta$ (such refinements always exist -- just union $\mathcal{P}'$ with a suitable standard partition). Thus $|\RS(f,\mathcal{P}'',\mathcal{S}'')-I|<\epsilon/3$. We have $\overline{I} \leq U(f,\mathcal{P}'') \leq U(f,\mathcal{P}')$ so that $U(f,\mathcal{P}'')-\overline{I} < \epsilon/3$. A previous lemma guarantees that there is some sampling $\mathcal{S}''$ such that $U(f,\mathcal{P}'')- \RS(f,\mathcal{P}'',\mathcal{S}'') < \epsilon/3$. Therefore,

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$$\begin{array}{rcl} |I-\overline{I}| & = & |I-\RS(f,\mathcal{P}'',\mathcal{S}'')+\RS(f,\mathcal{P}'',\mathcal{S}'')-U(f,\mathcal{P}'')+U(f,\mathcal{P}'')-\overline{I}| \\
 & \leq & |I-\RS(f,\mathcal{P}'',\mathcal{S}'')|+|\RS(f,\mathcal{P}'',\mathcal{S}'')-U(f,\mathcal{P}'')|+|U(f,\mathcal{P}'')-\overline{I}| < \dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3} = \epsilon \end{array}
$$

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This holds for all $\epsilon>0$. Therefore, $I=\overline{I}$. Similarly, $I=\underline{I}$ so that $I=\overline{I}=\underline{I}$.

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Now suppose that $\underline{I}=\overline{I}$. Call this number $I=\underline{I}=\overline{I}$. Let $\epsilon>0$. Because the lower and upper sums are defined as a supremum and infimum, there are partitions $\mathcal{P}_1$ and $\mathcal{P}_2$ such that $U(f,\mathcal{P}_1)<I+\epsilon/2$ and $L(f,\mathcal{P}_2)>I-\epsilon/2$. Let $\mathcal{P}' = \mathcal{P}_1 \cup \mathcal{P}_2$. Then $I-\epsilon/2 < L(f,\mathcal{P}_2) \leq L(f,\mathcal{P}') \leq U(f,\mathcal{P}') \leq U(f,\mathcal{P}_1) < I+\epsilon/2$. 

Let $\delta>0$ be the quantity given in our previous technical lemma such that for all partitions $\mathcal{P}$ with $\|\mathcal{P}\| < \delta$ we have that both $U(f,\mathcal{P})-U(f,\mathcal{P}\cup\mathcal{P}')<\epsilon/2$ and $L(f,\mathcal{P} \cup \mathcal{P}')-L(f,\mathcal{P})<\epsilon/2$. 

So let $\mathcal{P}$ be a partition with $\|\mathcal{P}\|<\delta$ and let $\mathcal{S}$ be a sampling. Then 
$$\RS(f,\mathcal{P},\mathcal{S}) \leq U(f,\mathcal{P}) < U(f,\mathcal{P}\cup\mathcal{P}')+\dfrac{\epsilon}{2} \leq U(f,\mathcal{P}')+\dfrac{\epsilon}{2} < I+\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=I+\epsilon$$

Likewise, $I-\epsilon < \RS(f,\mathcal{P},\mathcal{S})$ so that $I-\epsilon < \RS(f,\mathcal{P},\mathcal{S}) < I+\epsilon$. We have shown that for every $\epsilon>0$ there is some $\delta>0$ such that for all partitions $\mathcal{P}$ with $\|\mathcal{P}\|<\delta$ and samplings $\mathcal{S}$, we have $|\RS(f,\mathcal{P},\mathcal{S})-I|<\epsilon$. Therefore, $f$ is Riemann integrable with $\int_a^b f = I = \overline{I}=\underline{I}$. \eproof

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