The oriented Riemann integral is linear in continuous integrands

On each partition interval, the maximum of $f+g$ is at most the sum of the separate maxima, while its minimum is at least the sum of the separate minima. Thus $U(f+g,P)\le U(f,P)+U(g,P)$ and $L(f+g,P)\ge L(f,P)+L(g,P)$. Given $\epsilon >0$, choose partitions whose upper sums for $f,g$ are within $\epsilon \mathrm{/}2$ above their infima, and refine both to a common partition $P$. By [L1], refinement cannot increase those upper sums, so ${\int }_a^b(f+g)\le U(f+g,P)<{\int }_a^{b}f+{\int }_a^{b}g+\epsilon$. Similarly choose lower sums within $\epsilon \mathrm{/}2$ below their suprema and refine to a common partition; [L1] gives ${\int }_a^b(f+g)\ge L(f+g,P)>{\int }_a^{b}f+{\int }_a^{b}g-\epsilon$. Since this holds for every $\epsilon >0$, the integral is additive in the integrand.

If $\alpha >0$, maxima and minima of $\alpha f$ on every partition piece are $\alpha$ times those of $f$, so both Darboux sums scale by $\alpha$ and their infimum/supremum do as well. If $\alpha =0$, every sum is zero. If $\alpha <0$, multiplication reverses order, giving $U(\alpha f,P)=\alpha L(f,P)$ and $L(\alpha f,P)=\alpha U(f,P)$; taking the infimum and supremum, which reverse under multiplication by negative $\alpha$, again gives ${\int }_a^b\alpha f=\alpha {\int }_a^{b}f$. Combine this homogeneity with step 1.1 to obtain the displayed formula. For reversed endpoints use ${\int }_b^a=-{\int }_a^b$; for $a=b$ use the zero-length definition.