> [!NOTE] Theorem > Let $[a,b]$ be a [[Closed Real Interval|closed real interval]]. > > Let $f:[a,b]\to \mathbb{R}$ be a [[Monotone Real Function|monotone]] & [[Bounded Real Function|bounded]] [[Real Function|real function]]. Then $f$ is [[Riemann integration|Riemann integrable]]. **Proof**: WLOG suppose $f$ is [[Increasing Real Function|increasing]] since if $f$ is decreasing then $-f$ is increasing. By [[Riemann's criterion for integrability]], it suffices to check that for all $\varepsilon>0,$ there exists a [[Finite Partition of Closed Real Interval|finite partition]] of $[a,b],$ denote $P,$ such that that $U(f,P)-L(f,P)<\varepsilon$where $U(f,P)$ and $L(f,P)$ denote the [[Riemann integration|upper]] and [[Riemann integration|lower Darboux sums]] of $f$ with respect to $P.$ Let $n\in \mathbb{N}^{+}$ and $P=\{ x_{0},x_{1},\dots,x_{n} \}$ be a finite partition of $[a,b]$ into $n$ intervals of equal length $\frac{b-a}{n}.$ Then for each $i=1,\dots,n,$ $\sup\{ f(x) : x_{i-1}\leq x\leq x_{i} \}=f(x_{i})$ and $\inf\{ f(x) : x_{i-1}\leq x\leq x_{i} \}=f(x_{i-1})$ since $f$ is strictly increasing. Therefore $\begin{align} U(f,P)-L(f,P)&= \sum_{i=1}^{n} (f(x_{i})-f(x_{i-1}))(x_{i}-x_{i-1}) \\ &= \frac{b-a}{n} \sum_{i=1}^{n} (f(x_{i})-f(x_{i-1})) \\ &= \frac{b-a}{n} (f(x_{n})-f(x_{0})) \\ &= \frac{b-a}{n} (f(b)-f(a)) \end{align}$ By [[Archimedean Property of Real Numbers]], choose $n> \frac{(b-a)(f(b)-f(a))}{\varepsilon}.$ Then $U(f,P)-L(f,P)<\varepsilon.$ $\square$ **Proof**: Note that we can also choose $P$ so that its [[Mesh Size of Finite Partition of Closed Real Interval|mesh size]] is strictly less than $\frac{\varepsilon}{(f(b)-f(a))}.$