Proof of the linearity of the integral

Proof

We work directly with Riemann sums.

1. Let $a < b$, let $f, g \colon [a, b] \to \mathbb{R}$ be Riemann integrable, and let $\alpha, \beta \in \mathbb{R}$. Let $I_f$ and $I_g$ be their integrals, and let $h = \alpha f + \beta g$.
2. For each tagged partition $(P, t)$ with $P = (x_0, \dots, x_n)$, the Riemann sums satisfy $$S(h, P, t) = \sum_{i=1}^{n} \big(\alpha f(t_i) + \beta g(t_i)\big)(x_i - x_{i-1}) = \alpha S(f, P, t) + \beta S(g, P, t).$$
3. Let $\varepsilon > 0$. Take $\delta_f > 0$ with $|S(f, P, t) - I_f| < \frac{\varepsilon}{2(|\alpha| + 1)}$ for all tagged partitions of mesh less than $\delta_f$. Take $\delta_g > 0$ in the same way for $g$, with $\frac{\varepsilon}{2(|\beta| + 1)}$. Let $\delta = \min(\delta_f, \delta_g)$.
4. Let $(P, t)$ be a tagged partition of mesh less than $\delta$. By step 2, the triangle inequality and multiplicativity of the absolute value, $$\big|S(h, P, t) - (\alpha I_f + \beta I_g)\big| \le |\alpha| \, |S(f, P, t) - I_f| + |\beta| \, |S(g, P, t) - I_g|.$$
5. By step 3, the right side is at most $\frac{|\alpha| \varepsilon}{2(|\alpha| + 1)} + \frac{|\beta| \varepsilon}{2(|\beta| + 1)} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$.
6. So the number $\alpha I_f + \beta I_g$ has the property in the definition of integrability for $h$. Hence $h$ is Riemann integrable, and $\int_a^b h(x) \, dx = \alpha \int_a^b f(x) \, dx + \beta \int_a^b g(x) \, dx$.