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  "body": "\n## Proof\n\nWe work directly with Riemann sums.\n\n1. Let $a \u003c b$, let $f, g \\colon [a, b] \\to \\mathbb{R}$ be [[Riemann integrable|0N1JQXZXHC]], and let $\\alpha, \\beta \\in \\mathbb{R}$. Let $I_f$ and $I_g$ be their [[integrals|0QM4TXT3DR]], and let $h = \\alpha f + \\beta g$.\n2. For each tagged partition $(P, t)$ with $P = (x_0, \\dots, x_n)$, the [[Riemann sums|0HBY48G8G1]] satisfy\n   $$S(h, P, t) = \\sum_{i=1}^{n} \\big(\\alpha f(t_i) + \\beta g(t_i)\\big)(x_i - x_{i-1}) = \\alpha S(f, P, t) + \\beta S(g, P, t).$$\n3. Let $\\varepsilon \u003e 0$. Take $\\delta_f \u003e 0$ with $|S(f, P, t) - I_f| \u003c \\frac{\\varepsilon}{2(|\\alpha| + 1)}$ for all tagged partitions of [[mesh|04GM4NYXJB]] less than $\\delta_f$.\n   Take $\\delta_g \u003e 0$ in the same way for $g$, with $\\frac{\\varepsilon}{2(|\\beta| + 1)}$. Let $\\delta = \\min(\\delta_f, \\delta_g)$.\n4. Let $(P, t)$ be a tagged partition of mesh less than $\\delta$. By step 2, the [[triangle inequality|03HF5PMY5T]] and [[multiplicativity of the absolute value|0GMV7GP8W0]],\n   $$\\big|S(h, P, t) - (\\alpha I_f + \\beta I_g)\\big| \\le |\\alpha| \\, |S(f, P, t) - I_f| + |\\beta| \\, |S(g, P, t) - I_g|.$$\n5. By step 3, the right side is at most $\\frac{|\\alpha| \\varepsilon}{2(|\\alpha| + 1)} + \\frac{|\\beta| \\varepsilon}{2(|\\beta| + 1)} \u003c \\frac{\\varepsilon}{2} + \\frac{\\varepsilon}{2} = \\varepsilon$.\n6. 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 $\\int_a^b h(x) \\, dx = \\alpha \\int_a^b f(x) \\, dx + \\beta \\int_a^b g(x) \\, dx$.\n\n## Idea\n\nA Riemann sum is a finite sum, so it is linear in the function. If the sums of $f$ and of $g$ are close to their integrals for fine partitions, then the sums of $\\alpha f + \\beta g$ are close to $\\alpha I_f + \\beta I_g$ for the same partitions.\n",
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