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  "body": "\n## Proof\n\nWe show that $b \\mapsto \\int_a^b f$ is increasing and bounded, so it has a limit by the completeness of $\\R$.\n\n1. For $b \u003e a$, let $F(b) = \\int_a^b f(x) \\, dx$ and $G(b) = \\int_a^b g(x) \\, dx$. They exist by hypothesis.\n2. If $h \\colon [c, d] \\to \\R$ is Riemann integrable with $h \\ge 0$, then $\\int_c^d h \\ge 0$. Indeed, the [[linearity of the integral|080RWVVSB6]] gives $\\int_c^d 0 \\, dx = 0 \\cdot \\int_c^d h = 0$. Then the [[monotonicity of the integral|0MBWQKM8QR]] gives $\\int_c^d h \\ge 0$.\n3. $F$ is increasing. Let $a \u003c b \u003c b'$. By the [[additivity over intervals|05RCDKHZM6]], $F(b') = F(b) + \\int_b^{b'} f$. By step 2 and $f \\ge 0$, $F(b') \\ge F(b)$. In the same way, $G$ is increasing.\n4. For $b \u003e a$, $F(b) \\le G(b)$, by the monotonicity of the integral and $f \\le g$ on $[a, b]$.\n5. Now assume that $\\int_a^\\infty g$ converges. By the definition of the [[improper integral|0VCTBJMEJF]], $G(b) \\to L$ as $b \\to \\infty$, where $L = \\int_a^\\infty g(x) \\, dx$.\n6. Claim: $G(b) \\le L$ for all $b \u003e a$. Suppose that $G(b_0) \u003e L$ for some $b_0 \u003e a$, and let $\\varepsilon = G(b_0) - L \u003e 0$.\n7. By the definition of the [[limit at infinity|0Y12QYTJK3]], there is $M$ with $\\abs{G(b) - L} \u003c \\varepsilon$ for all $b \u003e M$.\n8. Take $b \u003e \\max(M, b_0)$. By step 3, $G(b) \\ge G(b_0) = L + \\varepsilon$. So $G(b) - L \\ge \\varepsilon$, which contradicts step 7. This proves the claim.\n9. By steps 4 and 6, $F(b) \\le L$ for all $b \u003e a$. So the set $S = \\set{F(b) : b \u003e a}$ is nonempty and bounded above by $L$.\n10. Let $s$ be the [[supremum|0BTMQ04556]] of $S$. It exists by the [[completeness axiom|0AXWET80YW]]. Since $L$ is an upper bound of $S$, $s \\le L$.\n11. Claim: $F(b) \\to s$ as $b \\to \\infty$. Let $\\varepsilon \u003e 0$. By the [[approximation property of the supremum|0M8TGGP9JP]], there is $b_1 \u003e a$ with $s - \\varepsilon \u003c F(b_1) \\le s$.\n12. For $b \u003e b_1$, step 3 and the definition of $s$ give $s - \\varepsilon \u003c F(b_1) \\le F(b) \\le s$. So $\\abs{F(b) - s} \u003c \\varepsilon$ for all $b \u003e b_1$, which proves the claim.\n13. By step 11, $\\int_a^\\infty f(x) \\, dx$ converges and equals $s$. By step 10, $\\int_a^\\infty f(x) \\, dx = s \\le L = \\int_a^\\infty g(x) \\, dx$.\n14. Finally, assume that $\\int_a^\\infty f$ diverges. If $\\int_a^\\infty g$ converged, then steps 5 to 13 would show that $\\int_a^\\infty f$ converges. So $\\int_a^\\infty g$ diverges.\n\n## Idea\n\nFor a nonnegative function, the integral up to $b$ can only grow as $b$ grows. If it stays below a fixed bound, it must settle at its supremum, by the completeness of the real numbers. The convergent integral of the larger function provides this bound.\n",
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