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Riemann Sums

Rectangles become thinner, making the rectangular approximation approach the area under the graph.
A Riemann sum approximates the area under a curve by splitting an interval into small subintervals and replacing the curved region above each subinterval by a rectangle. If the interval [a,b][a,b] is divided into nn pieces of width Δx\Delta x, and xk∗x_k^* is a sample point in the kk-th subinterval, then the approximation has the form
∑k=1nf(xk∗) Δx.\begin{equation*}\sum_{k=1}^{n} f(x_k^*)\,\Delta x.\end{equation*}
As the partition becomes finer, the rectangle width Δx\Delta x tends to zero. When the limit exists, the Riemann sums converge to the definite integral
lim⁡n→∞∑k=1nf(xk∗) Δx=∫abf(x) dx.\begin{equation*}\lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*)\,\Delta x=\int_a^b f(x)\,\mathrm{d}x.\end{equation*}
To make Riemann sums precise, we first need to specify how the interval is divided.
PARTITION
A partition PP of [a,b][a,b] is a finite set of points from [a,b][a,b] that includes both endpoints. We list the points in increasing order,
P={x0,x1,x2,…,xn}.\begin{equation*}P=\{x_0,x_1,x_2,\dots,x_n\}.\end{equation*}
This means that
a=x0<x1<x2<⋯<xn=b.\begin{equation*}a=x_0<x_1<x_2<\cdots <x_n=b.\end{equation*}
Each pair of neighboring partition points determines a subinterval [xk−1,xk][x_{k-1},x_k]. On this subinterval, define the lower height
mk=inf⁡{f(x):x∈[xk−1,xk]}.\begin{equation*}m_k=\inf\{f(x):x\in[x_{k-1},x_k]\}.\end{equation*}
Upper and lower rectangles for a partition $P$ of $[a,b]$.
Upper and lower rectangles for a partition PP of [a,b][a,b].
LOWER SUM
The lower sum of ff with respect to PP is
L(f,P)=∑k=1nmk(xk−xk−1).\begin{equation*}L(f,P)=\sum_{k=1}^{n} m_k(x_k-x_{k-1}).\end{equation*}
Similarly, define the upper height
Mk=sup⁡{f(x):x∈[xk−1,xk]}.\begin{equation*}M_k=\sup\{f(x):x\in[x_{k-1},x_k]\}.\end{equation*}
UPPER SUM
The upper sum of ff with respect to PP is
U(f,P)=∑k=1nMk(xk−xk−1).\begin{equation*}U(f,P)=\sum_{k=1}^{n} M_k(x_k-x_{k-1}).\end{equation*}
For a fixed partition PP, we always have L(f,P)≤U(f,P)L(f,P)\leq U(f,P) because mk≤Mkm_k\leq M_k on every subinterval. The lower sum approximates the area from below, while the upper sum approximates it from above.

The integrability criterion follows Abbott2015, Theorem 7.2.8.

Upper and lower sums give two complementary approximations. For a fixed partition, the lower sum estimates the area from below and the upper sum estimates it from above. As we add more partition points, the lower sums can only improve upward and the upper sums can only improve downward.
Let P\mathcal{P} denote the collection of all partitions of [a,b][a,b].
UPPER AND LOWER INTEGRALS
Let ff be a bounded function on [a,b][a,b]. The upper integral of ff is
U(f)=inf⁡{U(f,P):P∈P}.\begin{equation*}U(f)=\inf\{U(f,P):P\in\mathcal{P}\}.\end{equation*}
The lower integral of ff is
L(f)=sup⁡{L(f,P):P∈P}.\begin{equation*}L(f)=\sup\{L(f,P):P\in\mathcal{P}\}.\end{equation*}
Thus U(f)U(f) is the best possible upper estimate obtained from partitions, while L(f)L(f) is the best possible lower estimate. One important fact is that for every bounded function ff on [a,b][a,b],
L(f)≤U(f).\begin{equation*}L(f) \leq U(f).\end{equation*}
A function should be integrable exactly when these two ways of trapping the area lead to the same number.
RIEMANN INTEGRABILITY
A bounded function ff on [a,b][a,b] is Riemann integrable if
U(f)=L(f).\begin{equation*}U(f)=L(f).\end{equation*}
In this case, the definite integral is this common value:
∫abf(x) dx=U(f)=L(f).\begin{equation*}\int_a^b f(x)\,\mathrm{d}x=U(f)=L(f).\end{equation*}
The following criterion gives a practical way to prove integrability: instead of computing U(f)U(f) and L(f)L(f) directly, it is enough to find one partition whose upper and lower sums are arbitrarily close.
INTEGRABILITY CRITERION
A bounded function ff on [a,b][a,b] is Riemann integrable if and only if for every ε>0\varepsilon>0 there exists a partition PεP_\varepsilon of [a,b][a,b] such that
U(f,Pε)−L(f,Pε)<ε.\begin{equation*}U(f,P_\varepsilon)-L(f,P_\varepsilon)<\varepsilon.\end{equation*}
Proof. First suppose that such a partition exists for every ε>0\varepsilon>0. Since L(f)≤U(f)L(f)\leq U(f) and
U(f)−L(f)≤U(f,Pε)−L(f,Pε)<ε,\begin{equation*}U(f)-L(f)\leq U(f,P_\varepsilon)-L(f,P_\varepsilon)<\varepsilon,\end{equation*}
the nonnegative number U(f)−L(f)U(f)-L(f) is smaller than every positive ε\varepsilon. Therefore U(f)=L(f)U(f)=L(f).
Conversely, suppose ff is Riemann integrable, so U(f)=L(f)U(f)=L(f). Choose a partition P1P_1 such that
U(f,P1)<U(f)+ε2.\begin{equation*}U(f,P_1)<U(f)+\frac{\varepsilon}{2}.\end{equation*}
Also choose P2P_2 so that
L(f,P2)>L(f)−ε2.\begin{equation*}L(f,P_2)>L(f)-\frac{\varepsilon}{2}.\end{equation*}
Let Pε=P1∪P2P_\varepsilon=P_1\cup P_2 be the common refinement. Refinement decreases upper sums and increases lower sums, so
U(f,Pε)−L(f,Pε)≤U(f,P1)−L(f,P2).\begin{equation*}U(f,P_\varepsilon)-L(f,P_\varepsilon) \leq U(f,P_1)-L(f,P_2).\end{equation*}
Therefore
U(f,Pε)−L(f,Pε)<(U(f)+ε2)−(L(f)−ε2)=U(f)−L(f)+ε=ε.\begin{equation*}U(f,P_\varepsilon)-L(f,P_\varepsilon)<\left(U(f)+\frac{\varepsilon}{2}\right)-\left(L(f)-\frac{\varepsilon}{2}\right)=U(f)-L(f)+\varepsilon = \varepsilon.\end{equation*}
This proves the criterion. □\Box

The theorem that continuous functions on closed intervals are integrable follows Abbott2015, Theorem 7.2.9.

Since we have at our disposal Integrability criterion we can prove that the continuous function on [a,b][a,b] are integrable. The main idea of the proof is to use fact that continuum function on interval [a,b][a,b] are uniform continuos, it means for fixed ε\varepsilon
∃δ>0,∣x−y∣≤δ→∣f(x)−f(y)∣≤ε.\begin{equation*}\exists \delta > 0, | x - y |\leq \delta \rightarrow |f(x) - f(y)| \leq \varepsilon.\end{equation*}
We can use it to controll difference Mk−mkM_k - m_k on each subinterval.
On a subinterval $[x_{k-1},x_k]$, uniform continuity gives $M_k-m_k \leq \frac{\varepsilon}{b-a}$
On a subinterval [xk−1,xk][x_{k-1},x_k], uniform continuity gives Mk−mk≤εb−aM_k-m_k \leq \frac{\varepsilon}{b-a}
CONTINUOUS FUNCTIONS ARE INTEGRABLE
If ff is continuous on [a,b][a,b], then ff is Riemann integrable on [a,b][a,b].
Proof. Since ff is continuous on the compact interval [a,b][a,b], it is bounded and uniformly continuous. Let ε>0\varepsilon>0. By uniform continuity, there exists δ>0\delta>0 such that whenever x,y∈[a,b]x,y\in[a,b] and ∣x−y∣<δ|x-y|<\delta, we have
∣f(x)−f(y)∣<εb−a.\begin{equation*}|f(x)-f(y)|<\frac{\varepsilon}{b-a}.\end{equation*}
Choose a partition P={x0,x1,…,xn}P=\{x_0,x_1,\dots,x_n\} of [a,b][a,b] whose subinterval widths satisfy
xk−xk−1<δ\begin{equation*}x_k-x_{k-1}<\delta\end{equation*}
for every k=1,…,nk=1,\dots,n.
On each subinterval [xk−1,xk][x_{k-1},x_k], the Extreme Value Theorem gives points zkz_k and yky_k such that
Mk=f(zk) and mk=f(yk).\begin{equation*}M_k=f(z_k) \text{ and } m_k=f(y_k).\end{equation*}
Because zkz_k and yky_k lie in the same subinterval, we have
∣zk−yk∣<δ⟹Mk−mk=f(zk)−f(yk)<εb−a.\begin{equation*}|z_k-y_k|<\delta \quad \Longrightarrow \quad M_k-m_k=f(z_k)-f(y_k)<\frac{\varepsilon}{b-a}.\end{equation*}
It follows that
U(f,P)−L(f,P)=∑k=1n(Mk−mk)(xk−xk−1)<εb−a∑k=1n(xk−xk−1)=ε.\begin{equation*}U(f,P)-L(f,P) = \sum_{k=1}^{n}(M_k-m_k)(x_k-x_{k-1})< \frac{\varepsilon}{b-a}\sum_{k=1}^{n}(x_k-x_{k-1}) = \varepsilon.\end{equation*}
By the integrability criterion, ff is Riemann integrable. □\Box

References

  • [Abbott2015]Abbott, Stephen. Understanding Analysis. Second edition. Springer, 2015.