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  "body": "\n## Proof\n\nWe use Riemann's criterion and split partitions at $c$.\nWrite $U(g, P)$ and $L(g, P)$ for the upper and lower [[Darboux sums|0QRQVH0CRW]] of a bounded function $g$ for a partition $P$.\n\n1. First a lemma: a function $g$ that is [[Riemann integrable|0N1JQXZXHC]] on an interval $[r, s]$ is bounded there. Let $I$ be its integral. Take $\\delta \u003e 0$ such that each tagged partition of mesh less than $\\delta$ has [[Riemann sum|0HBY48G8G1]] within $1$ of $I$.\n2. By the [[Archimedean property|0A3GWZ86AY]], fix a partition $P = (x_0, \\dots, x_n)$ of $[r, s]$ with all subintervals shorter than $\\delta$, and fix tags $t$.\n3. Fix $i$ and $x \\in [x_{i-1}, x_i]$. Let $t'$ be the tags $t$ with $t_i$ replaced by $x$. Both sums are within $1$ of $I$, so $|S(g, P, t') - S(g, P, t)| \u003c 2$.\n   This difference is $(g(x) - g(t_i))(x_i - x_{i-1})$. So $|g(x)| \u003c |g(t_i)| + \\frac{2}{x_i - x_{i-1}}$.\n4. So $g$ is bounded on each of the $n$ subintervals. The largest of these $n$ bounds bounds $g$ on $[r, s]$. This proves the lemma.\n5. Now let $a \u003c c \u003c b$ and $f \\colon [a, b] \\to \\mathbb{R}$. For a [[partition|069P20KYQT]] $P_1$ of $[a, c]$ and a partition $P_2$ of $[c, b]$, let $P_1 \\oplus P_2$ be the partition of $[a, b]$ with the points of both.\n   Its subintervals are those of $P_1$ and those of $P_2$. So $U(f, P_1 \\oplus P_2) = U(f, P_1) + U(f, P_2)$, and the same holds for $L$.\n6. Each partition of $[a, b]$ that contains $c$ has this form $P_1 \\oplus P_2$.\n7. Let $P$ be a partition of $[a, b]$ and let $P'$ be $P$ with the point $c$ added. We claim $U(f, P') \\le U(f, P)$ and $L(f, P') \\ge L(f, P)$.\n   If $c$ is a point of $P$, then $P' = P$. Otherwise $c$ lies inside one subinterval $[x_{j-1}, x_j]$. Its term $M_j (x_j - x_{j-1})$ is replaced by $M' (c - x_{j-1}) + M'' (x_j - c)$.\n8. Here $M'$ and $M''$ are [[suprema|0BTMQ04556]] over subsets of $[x_{j-1}, x_j]$, so $M', M'' \\le M_j$. This gives $U(f, P') \\le U(f, P)$. The same argument with [[infima|04KFTA34C5]] gives $L(f, P') \\ge L(f, P)$.\n9. Assume that $f$ is integrable on $[a, b]$. By the lemma, $f$ is bounded. Let $\\varepsilon \u003e 0$. By [[Riemann's criterion|017DH8WXTE]], there is $P$ with $U(f, P) - L(f, P) \u003c \\varepsilon$.\n10. Write $P' = P_1 \\oplus P_2$ as in steps 6 and 7. Since $U(f, P_2) \\ge L(f, P_2)$, steps 5 and 8 give\n    $$U(f, P_1) - L(f, P_1) \\le U(f, P') - L(f, P') \\le U(f, P) - L(f, P) \u003c \\varepsilon.$$\n11. By Riemann's criterion, $f$ is integrable on $[a, c]$. The same argument with $P_1$ and $P_2$ exchanged shows that $f$ is integrable on $[c, b]$.\n12. Conversely, assume that $f$ is integrable on $[a, c]$ and on $[c, b]$. By the lemma, $f$ is bounded on each of them, so on $[a, b]$.\n13. Let $\\varepsilon \u003e 0$. By Riemann's criterion, there are partitions $P_1$ of $[a, c]$ and $P_2$ of $[c, b]$ with $U(f, P_i) - L(f, P_i) \u003c \\varepsilon/2$ for $i = 1, 2$.\n14. Let $P = P_1 \\oplus P_2$. By step 5, $U(f, P) - L(f, P) \u003c \\varepsilon$. By Riemann's criterion, $f$ is integrable on $[a, b]$.\n15. Now let $f$ be integrable on all three intervals. Let $I$, $I_1$ and $I_2$ be its integrals over $[a, b]$, $[a, c]$ and $[c, b]$.\n16. Take a bounded integrable function and any partition $Q$ of its interval. Then $L(Q)$ is at most the lower [[Darboux integral|09FQT057T7]] and $U(Q)$ is at least the upper one.\n    Both equal the integral, since [[Darboux and Riemann integrals agree|021XJAQVDC]]. So $L(Q)$ is at most the integral, and $U(Q)$ is at least the integral.\n17. Let $\\varepsilon \u003e 0$, and take $P_1$, $P_2$ and $P = P_1 \\oplus P_2$ as in steps 13 and 14. By step 16 on each piece, $L(f, P_1) + L(f, P_2) \\le I_1 + I_2 \\le U(f, P_1) + U(f, P_2)$.\n18. By step 5, this says $L(f, P) \\le I_1 + I_2 \\le U(f, P)$. By step 16 on $[a, b]$, also $L(f, P) \\le I \\le U(f, P)$.\n19. So $|I - (I_1 + I_2)| \\le U(f, P) - L(f, P) \u003c \\varepsilon$. This holds for each $\\varepsilon \u003e 0$, so $\\int_a^b f(x) \\, dx = \\int_a^c f(x) \\, dx + \\int_c^b f(x) \\, dx$.\n\n## Idea\n\nAdding the point $c$ to a partition only brings the upper and lower sums closer. A partition containing $c$ is two partitions side by side, and its Darboux sums are the sums of the two pieces. So small gaps on $[a, b]$ give small gaps on both halves, and conversely, and the integrals are trapped in the same small interval.\n",
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