Let $f:[a,b] \rightarrow \mathbb{R}$ be a monotone function and $P_k = {a = x_0 < x_1 < \dots < x_k = b }$ be a uniform partition of $[a,b]$, i.e., such that $x_j - x_{j - 1} = \frac{b - a}{k}$ for $1 \le j \le k$. Show that
\[\int^b_a f(x)dx = \lim_{k \rightarrow + \infty} s(f:P_k) = \lim_{k \rightarrow + \infty} S(f:P_k)\]$Proof$:
lower sum:
\[s(f:P_k) = \sum^k_{j = 1} m_j \Delta x, \quad m_j = \inf_{x \in [x_{j-1},x_j]} f(x)\]upper sum:
\[S(f:P_k) = \sum^k_{j = 1} M_j \Delta x, \quad M_j = \sup_{x \in [x_{j-1},x_j]} f(x)\]As $k \rightarrow +\infty$, $\Delta x \rightarrow 0$, and the partition $P_k$ becomes infinitely fine.
Since $f$ is monotone, it is Riemann integrable on $[a,b]$ (monotone functions are integrable).
Therefore, both the lower and upper sums conerge to the Riemann integral:
\[\lim_{k \to +\infty} s(f:P_k) = \int_a^b f(x)dx, \quad \lim_{k \to +\infty} S(f:P_k) = \int_a^b f(x)dx.\]Conclusion:
\[\int^b_a f(x)dx = \lim_{k \rightarrow + \infty} s(f:P_k) = \lim_{k \rightarrow + \infty} S(f:P_k)\]$\square$
(I’m not sure whether the explanation above is entirely correct. If you happen to notice any mistakes, I would really appreciate it if you could leave your suggestions in the comments when you have a moment. Thank you very much for your help and corrections.)