> [!NOTE] Theorem (Integrability of continuous functions) >Let $[a,b]$ be a [[Closed Real Interval|closed real interval]]. > > Let $f:[a,b]\to \mathbb{R}$ be a [[Continuous Real Function|continuous real function]]. Then $f$ is [[Darboux Integrable Function|Darboux integrable]]. **Proof**: It follows from [[Image of Closed Real Interval under Continuous Real Function is Closed Real Interval]], that $f$ [[Bounded Real Function|bounded]]. By [[Necessary Condition for Darboux 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 $U(f,P)-L(f,P)<\varepsilon$where $U(f,P)$ and $L(f,P)$ denote the [[Upper Darboux Sum|upper]] and [[Lower Darboux Sum|lower Darboux sums]] of $f$ with respect to $P.$ By [[Continuous Real Function on Closed Real Interval is Uniformly Continuous]], $f$ is [[Uniformly Continuous Real Function|uniformly continuous]] on $[a,b].$ Let $\varepsilon>0.$ In definition of [[Uniformly Continuous Real Function|uniform continuity]], choose $\delta$ so that for all $x,y\in[a,b]$ $|x-y|<\delta \implies |f(x)-f(y)|< \frac{\varepsilon}{b-a}.$ ?Now let $P=\{ x_{0},x_{1},\dots, x_{n} \}$ be a finite partition of $[a,b]$ whose [[Mesh Size of Finite Partition of Closed Real Interval|mesh size]] ($\max_{1\leq i\leq n}(x_{i}-x_{i-1})$) is strictly less than $\delta.$ Then for each $i=1,\dots,n,$ $M_{i}-m_{i}:= \sup\{ f(x): x\in [x_{i-1},x_{i}] \}-\inf\{ f(x): x\in [x_{i-1},x_{i}]\} \leq \frac{\varepsilon}{b-a}$ Therefore $\begin{align} U(f,P)-L(f,P) &= \sum_{i=1}^{n} (M_{i}-m_{i})(x_{i}-x_{i-1}) \\ &\leq \frac{\varepsilon}{b-a} \sum_{i=1}^{n}(x_{i}-x_{i-1}) = \frac{\varepsilon}{b-a} (x_{n}-x_{0}) = \varepsilon. \end{align} $