> [!NOTE] Lemma (upper integral is never smaller than lower integral) > Let $[a,b]$ be a [[Closed Real Interval|closed real interval]]. > > Let $f:[a,b]\to \mathbb{R}$ be a [[Bounded Real Function|bounded]] [[Real Function|real function]]. > >Let $\overline{\int }f$ and $\underline{\int }f$ denote the [[Upper Darboux Integral|upper Riemann integral]] and [[Lower Darboux Integral|lower Riemann integral]] of $f$ respectively. Then $\underline{\int } f\leq \overline{\int } f$ **Proof**: Let $I=\inf_{P}U(f,P)$ and $S=\sup_{P} L(f,P).$ Suppose $S>I$ and let $\varepsilon=\frac{S-I}{2}>0.$ By [[Characteristic Property of Infimum of Subset of Real Numbers]], there exist a [[Finite Partition of Closed Real Interval|finite partition]] of $[a,b],$ denote $P,$ such that $U(f,P)<I+\frac{\varepsilon}{2}$and again by [[Characteristic Property of Supremum of Subset of Real Numbers]], there exists a finite partition of $[a,b],$ denote $Q,$ such that $S-\frac{\varepsilon}{2}< L(f,Q).$Combining the above inequalities gives $U(f,P)<I+\frac{\varepsilon}{2}=S-\frac{\varepsilon}{2}<L(f,Q)$contradicting [[Upper Darboux Sum is Never Smaller than Lower Riemann Sum]]. Therefore $I\geq S$ $\square$