Introduction to Darboux Integral

Published: 2024-05-17

Partition

Def 1: Let $a < b$. A set $P \subseteq [a,b]$ is a partition of $[a,b]$ if $|P| \in \mathbb{N}$.

Refinement

Def 2: Let $P$ and $P'$ be partitions of $[a,b]$, then $P'$ is refinement of $P$ if $P \subseteq P'$.

Lower Darboux Sum

Def 3: For a partition $P$ of $[a,b]$ and $|P| = n + 1$, the lower Darboux Sum of $f: [a,b] \to \mathbb{R}$ over $P$, $\mathcal{L}(f,P)$, is

$$\sum^{n}_{i=1} (p_{i+1} - p_{i}) \inf_{x \in (p_{i},p_{i+1})} f(x)$$

where $\exists!p_{n} \in\ P: p_{1} \leq p_{2} \leq \dots \leq p_{n+1}$.

Upper Darboux Sum

Def 4: For a partition $P$ of $[a,b]$ and $|P| = n + 1$, the upper Darboux Sum of $f: [a,b] \to \mathbb{R}$ over $P$, $\mathcal{U}(f,P)$, is

$$\sum^{n}_{i=1} (p_{i+1} - p_{i}) \sup_{x \in (p_{i},p_{i+1})} f(x)$$

where $\exists!p_{n} \in P: p_{1} \leq p_{2} \leq \dots \leq p_{n+1}$.

Refinement Inequality Theorem

Lemma 5: For an ordered set $S$ and its subset $B \subseteq S$, $\inf B \geq \inf S$ and $\sup B \leq \sup S$.

Lemma 6: For $f:[a,b] \to R$, given a partition $P$ of $[a,b]$ and its refinement $P' = P \cup \{ c \}$, $\mathcal{L}(f,P) \leq \mathcal{L}(f,P’)$ and $\mathcal{U}(f,P) \geq \mathcal{U}(f,P')$.

Let $p_{n-1},p_{n},p_{n+1}$ where $p_{n} = c$, then

For simplicity, we assume $i_{0} = \inf_{x \in (p_{n-1}, p_{n+1})} f(x)$, $i_{1} = \inf_{x \in (p_{n}, p_{n+1})} f(x)$ and $i_{2} = \inf_{x \in (p_{n-1}, p_{n})} f(x)$, then

Without the loss of generality, assume that $i_{1} = i_{0}$ (since $(p_{n-1},p_{n}) \cup (p_{n},p_{n+1}) = (p_{n-1},p_{n+1})$ implies that $i_{0}$ is contained in one of those two intervals), then

By Lemma 5, $i_{2} -i_{0} \geq 0$ and by definition $p_{n}-p_{n-1}> 0$; therefore,

Equivalently, it can be shown that $\mathcal{U}(f,P') \leq \mathcal{U}(f,P)$.

Theorem 7: Let $f: [a,b] \to \mathbb{R}$ and $P$ be a partition of $[a,b]$. If $P'$ is a refinement of $P$, then $\mathcal{L}(f,P) \leq \mathcal{L}(f,P')$ and $\mathcal{U}(f,P) \geq \mathcal{U}(f,P')$

Theorem 7 can be proven by induction of lemma 6.

Upper-Lower Sum Comparison

Thm 8: Let $f: [a,b] \to \mathbb{R}$, $P$ and $P'$ be arbitrary partitions of $[a,b]$, then $\mathcal{L}(f,P) \leq \mathcal{U}(f,P')$

Let $P'' = P \cup P'$, then $P,P'$ are refinements of $P''$, so by Refinement Inequality Theorem,

[!Question]- Why is $\mathcal{L}(f,P'') \leq \mathcal{U}(f,P'')$. Intuitively, since upper sum always overshoots and lower sum over undershoots the actual integral, upper sum will always be bigger or equal to the lower sum; however, if you want a rigorous proof, here it is:

For $f: [a,b] \to \mathbb{R}$ and Partition $P$, we define

$$\Delta = \mathcal{U}(f,P) - \mathcal{L}(f,P),$$

which, by definition of upper and lower Darboux sums is equal to

$$\Delta = \sum^{n}_{i=1} \left(\sup_{x \in (p_{i},p_{i+1})} f(x) - \sup_{x \in (p_{i},p_{i+1})} f(x) \right).$$

By definition of supremum and infimum,

$$\forall i \in [1\dots n],\left(\sup_{x \in (p_{i},p_{i+1})} f(x) - \inf_{x \in (p_{i},p_{i+1})} f(x)\right) \geq 0;$$

therefore,

$$\Delta \geq 0 \implies \mathcal{U}(f,P) - \mathcal{L}(f,P) \geq 0 \implies \mathcal{U}(f,P) \geq \mathcal{L}(f,P) \implies \mathcal{L}(f,P) \leq \mathcal{U}(f,P). \square$$

[!Note] “Forcing Equality”> If we prove there exists $\mathcal{L} \geq \mathcal{U}$, then the equality $\mathcal{L} = \mathcal{U}$ is forced.

Darboux Integral

Def 9: Let $\mathcal{P}$ be the collection of partitions on $[a,b]$ and $f: [a,b] \to \mathbb{R}$. The lower (upper) Darboux integral, $\mathcal{L}(f)$ ($\mathcal{U}(f)$) is

$$\begin{gather} \mathcal{L}(f) = \sup_{P \in \mathcal{P}} \mathcal{L}(f,P) \\ \left( \mathcal{U}(f) = \inf_{P \in \mathcal{P}} \mathcal{U}(f,P) \right) \end{gather}$$

[!Note] The idea comes form the Refinement Inequality Theorem. We’re basically “taking the limit” here to get to the most optimal sum.

Def 10: If $\mathcal{L}(f) = \mathcal{U}(f)$, then the Darboux integral exists and is equal to both $\mathcal{L}(f)$ and $\mathcal{U}(f)$.

Integral Comparison Theorem

Thm 11: $\mathcal{L}(f) \leq \mathcal{U}(f)$.

Let $P \in \mathcal{P}$. Since $\mathcal{U}(f,P)$ is an upper bound of $\{ \mathcal{L}(f,Q) : Q \in \mathcal{P} \}$, $\mathcal{U}(f,P) \geq \mathcal{L}(f)$ (by Upper-Lower Sum Comparison. Since $P$ is arbitrary, $\mathcal{L}(f)$ is a lower bound of $\{ \mathcal{U}(f,Q) : Q \in \mathcal{P} \}$ (by Upper-Lower Sum Comparison), so $\mathcal{L}(f) \leq \mathcal{U}(f)$.

[!Note] Theorem 11, combined with the theorem 8, is basically the same the following lemma: $A \leq B \implies \sup A \leq \inf B$

Forcing Equality (Important Corollary)

$$\mathcal{U}(f) \leq \mathcal{L}(f,P) \implies \mathcal{U}(f) = \mathcal{L}(f)$$

Proof of the corollary:

By definition of Lower Darboux Integral, $\mathcal{L}(f,P) \leq \mathcal{L}(f)$, then

By Integral Comparison Theorem, $\mathcal{L}(f) \leq \mathcal{U}(f)$, hence

Darboux Integrable Theorem

Thm 12: Let $f: [a,b] \to \mathbb{R}$ be continuous, then $f$ is Darboux integrable.

[!Note] From continuity on a closed interval, necessary (for the proof) uniform continuity and boundness conditions are implied.

Let $\varepsilon > 0$, then by definition of continuity $\exists \delta > 0$ such that

Let $n$ be such that $\frac{b-a}{n} < \delta$, then let partition

[!Note] $P$ gives us the values of $x$ of columns equal in distance across $[a,b]$, i.e. the columns the good-old Reinman integral would split the sum onto.

then

where $S_{k} = \sup_{x \in (p_{k},p_{k+1})} f(x)$, $I_{k} = \inf_{x \in (p_{k},p_{k+1})} f(x)$. Since $\varepsilon$ is multiplied by a constant expression $(b-a)$, it follows

[!Question]- Why is $\mathcal{U}(f) - \mathcal{L}(f) \leq \varepsilon$ a sufficient condition for Darboux Integrability? It is a pretty important theorem.

Let $\mathcal{U}(f) - \mathcal{L}(f) \leq \varepsilon$. By Integral Comparison Theorem, it is sufficient to prove $\mathcal{U}(f) \leq \mathcal{L}(f)$ in order to show that Darboux integral exists.

For the sake of contradiction, assume that $\mathcal{U}(f) \not\leq \mathcal{L}(f)$, then $\exists m >0$ such that

$$\mathcal{U}(f) - \mathcal{L}(f) = m$$

However since $\varepsilon$ is arbitrary, let $\varepsilon = \frac{m}{2}$, therefore

$$\mathcal{U}(f) - \mathcal{L}(f) \leq \frac{m}{2}.$$

This is a contradiction since $m$ cannot be less than or equal than $\frac{m}{2}$; as such, Darboux integral exists.

You can also read about it here: Condition for Darboux Integrability - ProofWiki.

%% Since $\varepsilon$ is arbitrary, by Refinement Inequality Theorem $\exists P$ such that

which by the Forcing Equality Corollary to Integral Comparison Theorem implies that $\mathcal{U}(f) = \mathcal{L}(f)$ and Darboux integral exists. $\square$ %%

[!Note]

$$\begin{align} \mathcal{U}(f,P) - \mathcal{L}(f,P) &= \sum_{k=1}^{n} (p_{k+1} - p_{k}) \sup_{x \in (p_{k},p_{k+1})} f(x) - \sum_{k=1}^{n} (p_{k+1} -p_{k}) \inf_{x \in (p_{k},p_{k+1})} f(x) \\ &= \sum^{n}_{k=1} (p_{k+1}-p_{k})(S_{k}-I_{k}) \end{align}$$

From definition of $P$, it follows that $p_{k} = (k-1) \frac{b-a}{n} + a$, so $p_{k+1} - p_{k} = \frac{b-a}{n}$, so

$$= \sum^{n}_{k=1} \left(\frac{b-a}{n}\right) (S_{k} - I_{k})$$

[!Question]- Why is $(S_{k} - I_{k}) \leq \varepsilon$?

Let $k \in [1\dots n]$, recall $\frac{b-a}{n} < \delta$, then

$$\begin{align} S_{k} - I_{k} &= \sup_{x \in ((k-1) \frac{b-a}{n} + a,k \frac{b-a}{n} + a)} f(x) - \inf_{x \in ((k-1) \frac{b-a}{n} + a,k \frac{b-a}{n} + a)} f(x) \end{align}$$

Let $s,t \in ((k-1) \frac{b-a}{n} + a,k \frac{b-a}{n} + a)$, then

$$\begin{align} |s-t| &< \left|k \frac{b-a}{n} + a - \left((k-1) \frac{b-a}{n} + a\right)\right| \\ &= \left|\frac{b-a}{n}\right| \\ &< \delta. \end{align}$$

(the worst-case scenario for $|s-t|$ is still less than $\delta$). By definition,

$$|s-t| < \delta \implies |f(s)-f(t)| < \varepsilon \stackrel{\text{(*)}}{\implies} |S_{k} - I_{k}| < \varepsilon$$

$(*)$: Since $\forall s,t$, $|f(s) - f(t)| < \varepsilon$ and $S_{k},I_{k} \in \{ f(s) : s\}, \{ f(t) : t\}$.

Finally, from the definition of supremum and infimum $S_{k} \geq I_{k}$, hence

$$|S_{k} - I_{k}| < \varepsilon \implies S_{k} - I_{k} < \varepsilon. \square$$

[!Question]- Why is $\mathcal{U}(f) - \mathcal{L}(f) \leq \mathcal{U}(f,P) - \mathcal{L}(f,P)$?

Intuitively, as we refine the sums, $\mathcal{L}$ gets larger and $\mathcal{U}$ gets smaller (since $\mathcal{L}$ undershoots and $\mathcal{U}$ overshoots the sum), so their difference gets smaller, “the window shrinks.”

Formal Proof:

Lemma 12.1.1: For any $A,A_{1},B,B_{1} \in \mathbb{R}$, $(A \leq A_{1}) \land (B \leq B_{1}) \implies (A + B) \leq (A_{1} + B_{1})$.

Intuitively, if you have of length $A$ and $B$, where segment $A$ is smaller than some segment $A_{1}$ and segment $B$ is smaller than segment $B_{1}$. If you put $A$ and $B$ side by side and compare them to $A_{1}$ and $B_{1}$ side by side, then the length of $AB$ will be smaller than the length of $A_{1}B_{1}$.

Since $\forall C,C_{1} \in \mathbb{R}$

$$C \leq C_{1} \iff \exists m \in \mathbb{R}_{\geq 0} : C + m = C_{1},$$

$$\begin{cases} A \leq A_{1} \\ B \leq B_{1} \end{cases} \iff \begin{cases} \exists m\geq0 : A + m = A_{1} \\ \exists m_{1}\geq0 : B + m_{1} = B_{1} \end{cases} \implies (A+B) + (m + m_{1}) = A_{1} + B_{1}.$$

Since $(m + m_{1}) \in \mathbb{R}_{\geq 0}$,

$$A + B \leq A_{1} + B_{1}. \square$$

Lemma 12.1.2: For any $A,B \in \mathbb{R}$, $A \leq B \iff -A \geq -B$.

Since $\forall C,C_{1} \in \mathbb{R}$

$$\begin{gather} C \leq C_{1} \iff \exists m \in \mathbb{R}_{\geq 0} : C + m = C_{1}, \\ C \geq C_{1} \iff \exists m \in \mathbb{R}_{\geq 0} : C = C_{1} + m, \end{gather}$$

$$A \leq B \iff \exists m \in \mathbb{R}_{\geq 0} : A + m = B \iff \exists m \in \mathbb{R}_{\geq 0} : -A = -B + m \iff -A \geq -B. \square$$

Lemma 12.1: $\mathcal{U}(f) - \mathcal{L}(f) \leq \mathcal{U}(f,P) - \mathcal{L}(f,P)$.

By definition of Darboux Integral, supremum and infimum, for $f:[a,b] \to \mathbb{R}$ and all partitions $P$,

$$\begin{cases} \mathcal{L}(f) \geq \mathcal{L}(f,P) \\ \mathcal{U}(f) \leq \mathcal{U}(f,P) \end{cases} \implies$$

By Lemma 12.1.2,

$$\implies \begin{cases} -\mathcal{L}(f) \leq -\mathcal{L}(f,P) \\ \mathcal{U}(f) \leq \mathcal{U}(f,P) \end{cases} \implies$$

By Lemma 12.1.1,

$$\implies \mathcal{U}(f) - \mathcal{L}(f) \leq \mathcal{U}(f,P) - \mathcal{L}(f,P). \square$$

Non-Integrability of Indicator Functions

Thm 13: $f(x) = \begin{cases} x = 1 & x \in \mathbb{Q} \\ x = 0 & x \not\in \mathbb{Q}\end{cases}$ is not Darboux-integrable in $\mathbb{R}$.

Without formality, due to the fact that rationals and irrationals are dense in reals, in every partition $P$, $\mathcal{L}(f,P) = 0$ and $\mathcal{U}(f,P) = 1$; as such, the Darboux integral does not exist, since $\mathcal{L}(f) \neq \mathcal{U}(f)$.

[!abstract] References

• Continuous Real Function is Darboux Integrable - ProofWiki
  • Condition for Darboux Integrability - ProofWiki