> [!NOTE] Theorem > Let $[a,b]$ be a [[Closed Real Interval|closed real interval]]. Let $f,g:[a,b]\to \mathbb{R}$ be [[Darboux Integrable Function|Darboux integrable functions]]. Let $\lambda,\mu\in\mathbb{R}.$ Then $\lambda f+\mu g$ is also Darboux integrable and its [[Darboux Integral|Darboux integral]] is given by $\int_{a}^{b} (\lambda f(x)+ \mu g(x)) \, dx = \lambda \int_{a}^{b} f(x) \, dx + \mu \int_{a}^{b} g(x) \, dx \tag{\dagger}$ **Proof**: Note that $({\dagger})$ is equivalent to $\int_{a}^{b} \lambda f(x) \, dx= \lambda \int_{a}^{b} f(x) \, dx\tag{1}$and $\int_{a}^{b} (f(x)+g(x)) \, dx =\int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \tag{2}$ ?It is clear that $(1)$ is true since $U(\lambda f,P_{n})=\lambda U(f,P_{n})$ so by [[Algebra of Limits of Convergent Sequences]] $\lim_{ n \to \infty }U(\lambda f,P_{n})=\lambda \lim_{ n \to \infty }U(f,P_{n})$. Let $P$ be a [[Finite Partition of Closed Real Interval|finite partition]] of $[a,b].$ By [[Triangle Inequality for Supremum of Function on Interval]], the [[Upper Darboux Sum|upper Darboux sums]] with respect to $P$ satisfy $U(f+g,P)\leq U(f,P)+U(g,P)$and the reverse holds for [[Lower Darboux Sum|lower Darboux sums]]. STS that the [[Upper Darboux Integral|upper]] and [[Lower Darboux Integral|lower Darboux integrals]] satisfy $\overline{\int} (f+g) \leq \overline{\int} f + \overline{\int} g = \underline{\int} f + \underline{\int} g \leq \underline{\int}(f+g)$since this together with [[Upper Darboux Integral is Never Smaller than Lower Riemann Integral]], would force the upper and lower integrals of $f+g$ to be equal, and equal to $\int f +\int g.$ Let $\varepsilon>0.$ By [[Characteristic Property of Infimum of Subset of Real Numbers]], there exists $P,Q$ finite partitions of $[a,b]$ so that $U(f,P)\leq \overline{\int}f+ \frac{\varepsilon}{2} \quad \land \quad U(f,Q)\leq \overline{\int} g+ \frac{\varepsilon}{2}$Let $R$ be a [[Common Refinement of Finite Partition of Closed Real Interval|common refinement]] of $P$ and $Q.$ By triangle inequality again and [[Upper & Lower Darboux Sums of Refinement]], $ \begin{aligned} \overline{\int}(f+g) \leq U(f+g, R) & \leq U(f, R)+U(g, R) \\ & \leq U(f, P)+U(g, Q) \\ & \leq \overline{\int} f+\overline{\int} g+\varepsilon \end{aligned} $Since $\varepsilon>0,$ we have $\overline{\int}(f+g)\leq \overline{\int}f + \overline{\int}g.$