Proof that Darboux and Riemann integrals agree

Proof

Let $a < b$ and let $f \colon [a, b] \to \mathbb{R}$ be bounded. Fix $K > 0$ with $|f(x)| \le K$ for all $x \in [a, b]$. Write $U(P)$ and $L(P)$ for the upper and lower Darboux sums of $f$ for a partition $P$. Write $S(P, t)$ for the Riemann sum with tags $t$. Let $U^*$ and $L^*$ be the upper and lower Darboux integrals. We first prove four facts, then the two directions.

1. Let $P = (x_0, \dots, x_n)$ with tags $t_i \in [x_{i-1}, x_i]$. Then $m_i \le f(t_i) \le M_i$ for each $i$. Multiply by $x_i - x_{i-1} > 0$ and add: $L(P) \le S(P, t) \le U(P)$.
2. Let $R$ be a refinement of a partition $Q = (y_0, \dots, y_m)$. Each subinterval of $R$ lies in exactly one subinterval $[y_{j-1}, y_j]$ of $Q$, because no point of $Q$ lies strictly between two consecutive points of $R$. The subintervals of $R$ inside $[y_{j-1}, y_j]$ have total length $y_j - y_{j-1}$.
3. A supremum over a subset is at most the supremum over the whole set, and an infimum over a subset is at least the infimum. Group the terms of $U(R)$ by step 2. This gives $U(R) \le U(Q)$, and in the same way $L(R) \ge L(Q)$.
4. For any partitions $P$ and $Q$, let $R$ consist of the points of both. By steps 3 and 1, $L(P) \le L(R) \le U(R) \le U(Q)$. Taking the supremum over $P$ and then the infimum over $Q$ gives $L^* \le U^*$.
5. For each $\delta > 0$ there is a partition with mesh less than $\delta$. Indeed, take $n$ with $n \delta > b - a$, by the Archimedean property, and $x_i = a + i (b - a)/n$.

First assume that $f$ is Riemann integrable with integral $I$.

6. Let $\varepsilon > 0$. Take $\delta > 0$ such that $|S(P, t) - I| < \varepsilon / 2$ for each tagged partition of mesh less than $\delta$. By step 5, fix such a partition $P = (x_0, \dots, x_n)$.
7. For each $i$, the number $M_i - \frac{\varepsilon}{2(b - a)}$ is not an upper bound of $f$ on $[x_{i-1}, x_i]$. So there is $t_i$ there with $f(t_i) > M_i - \frac{\varepsilon}{2(b - a)}$.
8. Multiply by $x_i - x_{i-1}$ and add. This gives $S(P, t) > U(P) - \varepsilon/2$. So $U^* \le U(P) < S(P, t) + \varepsilon / 2 < I + \varepsilon$.
9. In the same way, choose tags $t'_i$ with $f(t'_i) < m_i + \frac{\varepsilon}{2(b - a)}$. This gives $L^* \ge L(P) > S(P, t') - \varepsilon/2 > I - \varepsilon$.
10. So $I - \varepsilon < L^* \le U^* < I + \varepsilon$ for each $\varepsilon > 0$, using step 4. Hence $L^* = U^* = I$.

Now assume that $L^* = U^*$, and call this number $D$.

11. Let $\varepsilon > 0$. Since $D + \varepsilon / 8$ is not a lower bound of the upper sums, there is $Q_1$ with $U(Q_1) < D + \varepsilon/8$. In the same way, there is $Q_2$ with $L(Q_2) > D - \varepsilon / 8$.
12. Let $Q$ consist of the points of $Q_1$ and $Q_2$. By step 3, $U(Q) - L(Q) \le U(Q_1) - L(Q_2) < \varepsilon / 4$. Let $m$ be the number of subintervals of $Q$, and put $\delta = \frac{\varepsilon}{8 m K}$.
13. Let $P = (x_0, \dots, x_n)$ have mesh less than $\delta$, and let $P'$ consist of the points of $P$ and $Q$. Then $P'$ refines both $P$ and $Q$.
14. Call a subinterval $[x_{k-1}, x_k]$ of $P$ split if a point of $Q$ lies strictly inside it. Only the $m - 1$ points of $Q$ other than $a$ and $b$ can do this. Each lies inside at most one subinterval of $P$. So at most $m - 1$ subintervals are split.
15. Group the terms of $U(P')$ by the subintervals of $P$, as in step 2. A subinterval of $P$ that is not split is also a subinterval of $P'$, and its terms cancel in $U(P) - U(P')$.
16. For a split subinterval $J$ of length $|J| < \delta$, its term in $U(P)$ is at most $K |J|$. Its terms in $U(P')$ add up to at least $-K |J|$, since every supremum is at least $-K$. So it adds at most $2 K |J| < 2 K \delta$ to $U(P) - U(P')$.
17. By steps 14 to 16, $U(P) - U(P') \le 2 (m - 1) K \delta < \varepsilon / 4$. With step 3 for $P'$ and $Q$, $U(P) < U(P') + \varepsilon/4 \le U(Q) + \varepsilon / 4$.
18. The same argument with infima gives $L(P) > L(P') - \varepsilon/4 \ge L(Q) - \varepsilon / 4$.
19. By steps 17, 18 and 12, $U(P) - L(P) < U(Q) - L(Q) + \varepsilon / 2 < \varepsilon$.
20. By the definitions of $L^*$ and $U^*$, we have $L(P) \le D \le U(P)$. By step 1, also $L(P) \le S(P, t) \le U(P)$ for all tags $t$. So $|S(P, t) - D| \le U(P) - L(P) < \varepsilon$.
21. This holds for every tagged partition of mesh less than $\delta$. So $f$ is Riemann integrable, and its integral is $D$.

Steps 10 and 21 prove both directions. In both cases the Riemann integral equals the common value of the Darboux integrals.