[proofplan] We use the Darboux characterization of [Riemann integrability](/page/Riemann%20Integrability): choose one fixed partition $Q$ for which the gap between the upper and lower Darboux sums is small. An arbitrary sufficiently fine tagged partition $P$ is then compared to $Q$ by separating the subintervals of $P$ that lie inside intervals of $Q$ from the few subintervals that cross points of $Q$. The inside intervals are controlled by the oscillation appearing in $U(f,Q)-L(f,Q)$, while the crossing intervals have small total length because the mesh of $P$ is small. [/proofplan]

[step:Choose a Darboux partition with small oscillation gap]Let \begin{align*} I := \int_a^b f\, d\mathcal{L}^1 \end{align*} denote the [Riemann integral](/page/Riemann%20Integral) of $f$ over $[a,b]$. Since $f$ is Riemann integrable on the compact interval $[a,b]$, the function $f$ is bounded. Choose $M \geq 0$ such that $|f(x)| \leq M$ for every $x \in [a,b]$. If $M = 0$, then $f(x)=0$ for every $x \in [a,b]$, so every Riemann sum and the integral are equal to $0$. The conclusion follows for any $\delta > 0$. Hence assume $M>0$. By the Darboux characterization of Riemann integrability, there exists a partition \begin{align*}Q = \{q_0,q_1,\dots,q_m\}\end{align*} with \begin{align*}a=q_0<q_1<\cdots<q_m=b\end{align*} such that the corresponding upper and lower Darboux sums, defined below, have gap smaller than $\varepsilon/2$. For each $j \in \{1,\dots,m\}$, define \begin{align*} M_j := \sup\{f(x):x \in [q_{j-1},q_j]\} \end{align*} and \begin{align*} m_j := \inf\{f(x):x \in [q_{j-1},q_j]\}. \end{align*} Define the Darboux sums for $Q$ by \begin{align*} U(f,Q) := \sum_{j=1}^m M_j(q_j-q_{j-1}) \end{align*} and \begin{align*} L(f,Q) := \sum_{j=1}^m m_j(q_j-q_{j-1}). \end{align*} Then \begin{align*} U(f,Q)-L(f,Q)=\sum_{j=1}^m (M_j-m_j)(q_j-q_{j-1}) \end{align*} and, by the choice of $Q$, this quantity is smaller than $\varepsilon/2$.[/step]

[guided]The goal is to compare every sufficiently fine tagged sum with the integral. Since $f$ is Riemann integrable, the Darboux characterization says that upper and lower Darboux sums can be made arbitrarily close. We therefore choose one fixed partition \begin{align*}Q = \{q_0,q_1,\dots,q_m\}\end{align*} with \begin{align*}a=q_0<q_1<\cdots<q_m=b\end{align*} whose Darboux gap will be smaller than $\varepsilon/2$ after the upper and lower sums are defined below. We also use boundedness. Since every Riemann integrable function on a compact interval is bounded, there is a number $M \geq 0$ such that $|f(x)| \leq M$ for all $x \in [a,b]$. If $M=0$, then $f$ is identically zero, so every tagged Riemann sum is $0$ and the Riemann integral is $0$. Thus the theorem is immediate in that case. For the rest of the proof we assume $M>0$. For each interval of the fixed partition $Q$, we record the largest and smallest possible values of $f$ on that interval. For $j \in \{1,\dots,m\}$, define \begin{align*} M_j := \sup\{f(x):x \in [q_{j-1},q_j]\} \end{align*} and \begin{align*} m_j := \inf\{f(x):x \in [q_{j-1},q_j]\}. \end{align*} These numbers are finite because $f$ is bounded. The Darboux gap for $Q$ is exactly \begin{align*} U(f,Q)-L(f,Q)=\sum_{j=1}^m (M_j-m_j)(q_j-q_{j-1}). \end{align*} This sum measures the total oscillation of $f$ at the scale of the partition $Q$. By the choice of $Q$, it satisfies \begin{align*} U(f,Q)-L(f,Q) < \frac{\varepsilon}{2}. \end{align*}[/guided]

[step:Choose the mesh so crossing intervals have small total length]Define \begin{align*} \eta := \min\{q_j-q_{j-1}:1 \leq j \leq m\}. \end{align*} Since $Q$ is a finite partition and each interval length is positive, $\eta>0$. Choose \begin{align*} \delta := \min\left\{\eta,\frac{\varepsilon}{8M(m-1)}\right\} \end{align*} if $m \geq 2$, and choose $\delta := \eta$ if $m=1$. Let $(P,t)$ be a tagged partition of $[a,b]$ with \begin{align*}P=\{x_0,x_1,\dots,x_n\}, \qquad a=x_0<x_1<\cdots<x_n=b,\end{align*} and tags $t_i \in [x_{i-1},x_i]$ for $i \in \{1,\dots,n\}$. Write \begin{align*} \Delta x_i := x_i-x_{i-1}. \end{align*} Let $B$ be the set of indices $i \in \{1,\dots,n\}$ such that the interval $[x_{i-1},x_i]$ contains at least one point of $\{q_1,\dots,q_{m-1}\}$ in its interior. Let $G:=\{1,\dots,n\}\setminus B$. Because $|P|<\eta$, each interval $[x_{i-1},x_i]$ contains at most one interior point of $Q$. Hence the map sending $i \in B$ to the unique point of $\{q_1,\dots,q_{m-1}\}$ contained in $(x_{i-1},x_i)$ is injective, so $|B| \leq m-1$. Therefore, if $m \geq 2$, \begin{align*} \sum_{i \in B}\Delta x_i \leq (m-1)|P| < (m-1)\delta \leq \frac{\varepsilon}{8M}. \end{align*}[/step]

[guided]We choose the mesh bound so that intervals of the arbitrary partition $P$ cannot cross too much of the fixed Darboux partition $Q$. Define \begin{align*} \eta := \min\{q_j-q_{j-1}:1 \leq j \leq m\}. \end{align*} Because $Q$ has finitely many subintervals and each has positive length, $\eta>0$. If $m\geq 2$, set \begin{align*} \delta := \min\left\{\eta,\frac{\varepsilon}{8M(m-1)}\right\}, \end{align*} and if $m=1$, set $\delta:=\eta$. Let $(P,t)$ be a tagged partition with $|P|<\delta$. Write \begin{align*} P=\{x_0,x_1,\dots,x_n\}, \qquad a=x_0<x_1<\cdots<x_n=b, \end{align*} with tags $t_i\in[x_{i-1},x_i]$, and define $\Delta x_i:=x_i-x_{i-1}$. Let $B$ be the set of indices $i$ for which $[x_{i-1},x_i]$ contains at least one point of $\{q_1,\dots,q_{m-1}\}$ in its interior, and let $G:=\{1,\dots,n\}\setminus B$. Since $|P|<\eta$, no interval of $P$ can contain two distinct interior points of $Q$: two such points would be separated by at least $\eta$, while the interval length is at most $|P|<\eta$. Thus each bad interval contains a unique interior point of $Q$, and the map from $B$ to $\{q_1,\dots,q_{m-1}\}$ sending a bad interval to that point is injective. Hence $|B|\leq m-1$. Therefore, when $m\geq 2$, \begin{align*} \sum_{i\in B}\Delta x_i \leq (m-1)|P| < (m-1)\delta \leq \frac{\varepsilon}{8M}. \end{align*} If $m=1$, there are no interior points of $Q$, so $B=\varnothing$.[/guided]

[step:Control the contribution from intervals contained inside the Darboux partition]For each $i \in G$, the interval $[x_{i-1},x_i]$ is contained in some interval $[q_{j-1},q_j]$ of the partition $Q$. Define $j(i) \in \{1,\dots,m\}$ to be an index such that \begin{align*} [x_{i-1},x_i]\subset [q_{j(i)-1},q_{j(i)}]. \end{align*} Since the tag satisfies $t_i \in [x_{i-1},x_i]$, we have \begin{align*} m_{j(i)} \leq f(t_i) \leq M_{j(i)}. \end{align*} Multiplying by the non-negative length $\Delta x_i$ gives \begin{align*} m_{j(i)}\Delta x_i \leq f(t_i)\Delta x_i \leq M_{j(i)}\Delta x_i. \end{align*} Consequently, \begin{align*} \left|\sum_{i\in G} f(t_i)\Delta x_i-\sum_{i\in G} m_{j(i)}\Delta x_i\right| \leq \sum_{i\in G}(M_{j(i)}-m_{j(i)})\Delta x_i. \end{align*} Grouping the terms according to $j(i)$ and using that the total length of good intervals contained in $[q_{j-1},q_j]$ is at most $q_j-q_{j-1}$, we obtain \begin{align*} \sum_{i\in G}(M_{j(i)}-m_{j(i)})\Delta x_i \leq \sum_{j=1}^m (M_j-m_j)(q_j-q_{j-1}) < \frac{\varepsilon}{2}. \end{align*}[/step]

[guided]For a good interval $i\in G$, the interval $[x_{i-1},x_i]$ does not contain an interior point of $Q$. Since the endpoints of $Q$ partition $[a,b]$, this means there is an index $j(i)\in\{1,\dots,m\}$ such that \begin{align*} [x_{i-1},x_i]\subset [q_{j(i)-1},q_{j(i)}]. \end{align*} The tag $t_i$ lies in this same interval, so the definitions of $m_{j(i)}$ and $M_{j(i)}$ give \begin{align*} m_{j(i)} \leq f(t_i) \leq M_{j(i)}. \end{align*} Multiplying by the non-negative number $\Delta x_i$ preserves the inequalities: \begin{align*} m_{j(i)}\Delta x_i \leq f(t_i)\Delta x_i \leq M_{j(i)}\Delta x_i. \end{align*} Subtracting the lower comparison term and using the upper bound gives \begin{align*} \left|\sum_{i\in G} f(t_i)\Delta x_i-\sum_{i\in G} m_{j(i)}\Delta x_i\right| \leq \sum_{i\in G}(M_{j(i)}-m_{j(i)})\Delta x_i. \end{align*} Now group the indices $i\in G$ by the value of $j(i)$. For a fixed $j$, the good intervals contained in $[q_{j-1},q_j]$ are disjoint subintervals of $[q_{j-1},q_j]$, so their total length is at most $q_j-q_{j-1}$. Therefore \begin{align*} \sum_{i\in G}(M_{j(i)}-m_{j(i)})\Delta x_i \leq \sum_{j=1}^m (M_j-m_j)(q_j-q_{j-1}) = U(f,Q)-L(f,Q) < \frac{\varepsilon}{2}. \end{align*}[/guided]

[step:Sandwich the tagged sum between Darboux bounds up to a small crossing error]The Riemann sum is \begin{align*} S(P,t;f):=\sum_{i=1}^n f(t_i)\Delta x_i. \end{align*} For each bad interval $i \in B$, boundedness gives \begin{align*} |f(t_i)\Delta x_i| \leq M\Delta x_i. \end{align*} Hence \begin{align*} \left|\sum_{i\in B} f(t_i)\Delta x_i\right| \leq M\sum_{i\in B}\Delta x_i. \end{align*} If $m=1$, then $B=\varnothing$. If $m\geq 2$, the choice of $\delta$ gives \begin{align*} M\sum_{i\in B}\Delta x_i < \frac{\varepsilon}{8}. \end{align*} The lower Darboux sum satisfies \begin{align*} L(f,Q)=\sum_{j=1}^m m_j(q_j-q_{j-1}), \end{align*} and the upper Darboux sum satisfies \begin{align*} U(f,Q)=\sum_{j=1}^m M_j(q_j-q_{j-1}). \end{align*} For each $i\in B$, split the interval $[x_{i-1},x_i]$ at the points of $\{q_1,\dots,q_{m-1}\}$ that it contains. Each resulting piece lies in a single interval of $Q$, and the total length of all such bad pieces is $\sum_{i\in B}\Delta x_i$. For each $j\in\{1,\dots,m\}$, the part of $[q_{j-1},q_j]$ not covered by good intervals is contained in the union of bad pieces. Since $|m_j|\leq M$ and $|M_j|\leq M$, deleting those pieces changes the lower and upper Darboux contributions by at most $M$ times their total length. Hence \begin{align*} L(f,Q)-M\sum_{i\in B}\Delta x_i \leq \sum_{i\in G}m_{j(i)}\Delta x_i \end{align*} and \begin{align*} \sum_{i\in G}M_{j(i)}\Delta x_i \leq U(f,Q)+M\sum_{i\in B}\Delta x_i. \end{align*} The good-interval inequalities give \begin{align*} \sum_{i\in G}m_{j(i)}\Delta x_i \leq \sum_{i\in G}f(t_i)\Delta x_i \leq \sum_{i\in G}M_{j(i)}\Delta x_i. \end{align*} Adding the bad part of the Riemann sum, whose absolute value is bounded by $M\sum_{i\in B}\Delta x_i$, yields \begin{align*} L(f,Q)-2M\sum_{i\in B}\Delta x_i \leq S(P,t;f) \leq U(f,Q)+2M\sum_{i\in B}\Delta x_i. \end{align*}[/step]

[guided]We now compare the arbitrary tagged sum with the fixed Darboux sums. The Riemann sum is \begin{align*} S(P,t;f):=\sum_{i=1}^n f(t_i)\Delta x_i, \end{align*} where $\Delta x_i=x_i-x_{i-1}$. The good intervals are controlled by the oscillation of $f$ on the corresponding interval of $Q$. The only obstruction is that some intervals of $P$ may cross a point of $Q$. These intervals are the bad intervals $i \in B$. On a bad interval we do not try to use oscillation estimates, because the interval may touch two adjacent intervals of $Q$. Instead we use the crude bound $|f|\leq M$: \begin{align*} |f(t_i)\Delta x_i| \leq M\Delta x_i. \end{align*} Summing over $i \in B$ gives \begin{align*} \left|\sum_{i\in B} f(t_i)\Delta x_i\right| \leq M\sum_{i\in B}\Delta x_i. \end{align*} The previous mesh choice gives \begin{align*} M\sum_{i\in B}\Delta x_i < \frac{\varepsilon}{8} \end{align*} when $m\geq 2$, and if $m=1$ then there are no interior points of $Q$, so $B=\varnothing$. Now consider the good intervals. If $i\in G$, then $[x_{i-1},x_i]$ lies inside one of the intervals $[q_{j-1},q_j]$. Calling that index $j(i)$, the tag condition $t_i\in[x_{i-1},x_i]$ implies \begin{align*} m_{j(i)} \leq f(t_i) \leq M_{j(i)}. \end{align*} Multiplying by $\Delta x_i\geq 0$ gives \begin{align*} m_{j(i)}\Delta x_i \leq f(t_i)\Delta x_i \leq M_{j(i)}\Delta x_i. \end{align*} If we sum these inequalities over all good intervals, we get a lower and upper estimate for the good part of the Riemann sum. The missing pieces are exactly the bad intervals. Because $|m_j|\leq M$ and $|M_j|\leq M$ for every $j$, deleting bad subintervals from the Darboux lower and upper sums changes either comparison by at most \begin{align*} M\sum_{i\in B}\Delta x_i. \end{align*} The bad part of the Riemann sum itself contributes at most the same quantity in absolute value. Therefore the whole tagged sum satisfies \begin{align*} L(f,Q)-2M\sum_{i\in B}\Delta x_i \leq S(P,t;f) \leq U(f,Q)+2M\sum_{i\in B}\Delta x_i. \end{align*} This is the central estimate: all dependence on the arbitrary partition $P$ has been compressed into the total length of the bad intervals, which is small because the mesh is small.[/guided]

[step:Use the integral between the Darboux sums to finish the estimate]Since $I$ is the Riemann integral of $f$, every lower Darboux sum is at most $I$ and every upper Darboux sum is at least $I$: \begin{align*} L(f,Q)\leq I \leq U(f,Q). \end{align*} From the estimate in the previous step, \begin{align*} S(P,t;f)-I \leq U(f,Q)-L(f,Q)+2M\sum_{i\in B}\Delta x_i \end{align*} and \begin{align*} I-S(P,t;f) \leq U(f,Q)-L(f,Q)+2M\sum_{i\in B}\Delta x_i. \end{align*} Therefore \begin{align*} |S(P,t;f)-I| \leq U(f,Q)-L(f,Q)+2M\sum_{i\in B}\Delta x_i. \end{align*} By the choice of $Q$ and $\delta$, \begin{align*} |S(P,t;f)-I| < \frac{\varepsilon}{2}+2M\cdot \frac{\varepsilon}{8M} = \frac{3\varepsilon}{4} < \varepsilon. \end{align*} Thus every tagged partition with mesh smaller than $\delta$ has its Riemann sum within $\varepsilon$ of the Riemann integral, which proves the theorem.[/step]

[guided]The Riemann integral $I$ lies between every lower and upper Darboux sum, so for the fixed partition $Q$ we have \begin{align*} L(f,Q)\leq I \leq U(f,Q). \end{align*} The previous sandwich estimate gives \begin{align*} S(P,t;f)\leq U(f,Q)+2M\sum_{i\in B}\Delta x_i \end{align*} and \begin{align*} S(P,t;f)\geq L(f,Q)-2M\sum_{i\in B}\Delta x_i. \end{align*} Using $I\geq L(f,Q)$ in the [first inequality](/theorems/2897) gives \begin{align*} S(P,t;f)-I \leq U(f,Q)-L(f,Q)+2M\sum_{i\in B}\Delta x_i. \end{align*} Using $I\leq U(f,Q)$ in the [second inequality](/theorems/2136) gives \begin{align*} I-S(P,t;f) \leq U(f,Q)-L(f,Q)+2M\sum_{i\in B}\Delta x_i. \end{align*} Therefore \begin{align*} |S(P,t;f)-I| \leq U(f,Q)-L(f,Q)+2M\sum_{i\in B}\Delta x_i. \end{align*} The Darboux gap is smaller than $\varepsilon/2$. If $m=1$, then $B=\varnothing$; if $m\geq 2$, the mesh choice gives $\sum_{i\in B}\Delta x_i<\varepsilon/(8M)$. In either case, \begin{align*} |S(P,t;f)-I| < \frac{\varepsilon}{2}+2M\cdot \frac{\varepsilon}{8M} = \frac{3\varepsilon}{4} < \varepsilon. \end{align*} This proves that every tagged partition with mesh smaller than $\delta$ has Riemann sum within $\varepsilon$ of the Riemann integral.[/guided]