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  "body": "\n## Proof\n\nWe reduce to the first fundamental theorem on a closed interval around each point.\n\n1. Let $I$ be an open interval, let $f \\colon I \\to \\mathbb{R}$ be continuous, and let $a \\in I$. For $x \\in I$, the restriction of $f$ to the closed interval between $a$ and $x$ is continuous.\n   Since [[continuous functions are Riemann integrable|06SESV7GQY]], the [[definite integral|021WEM680W]] $F(x) = \\int_a^x f(t) \\, dt$ is defined.\n2. Fix $x_0 \\in I$. Since $I$ is open, there are $p, q \\in I$ with $p \u003c \\min(a, x_0)$ and $q \u003e \\max(a, x_0)$. Then $[p, q] \\subseteq I$, and $f$ is continuous on $[p, q]$.\n3. Define $G(x) = \\int_p^x f(t) \\, dt$ for $x \\in [p, q]$. We claim that $F(x) = G(x) - G(a)$ for all $x \\in (p, q)$.\n4. If $x \u003e a$, the [[additivity of the integral|05RCDKHZM6]] on $[p, x]$ split at $a$ gives $G(x) = G(a) + \\int_a^x f(t) \\, dt = G(a) + F(x)$.\n5. If $x = a$, then $F(a) = 0 = G(a) - G(a)$ by the convention for the definite integral.\n6. If $x \u003c a$, additivity on $[p, a]$ split at $x$ gives $G(a) = G(x) + \\int_x^a f(t) \\, dt$. By the sign convention, $\\int_x^a f(t) \\, dt = -F(x)$. So $F(x) = G(x) - G(a)$.\n7. The function $f$ is continuous at $x_0 \\in (p, q)$. By the [[first fundamental theorem of calculus|063H5KJC69]] on $[p, q]$, $G$ is differentiable at $x_0$ with $G'(x_0) = f(x_0)$.\n8. For $0 \u003c |h|$ small enough, $x_0 + h \\in (p, q)$. Then step 3 gives $\\frac{F(x_0 + h) - F(x_0)}{h} = \\frac{G(x_0 + h) - G(x_0)}{h}$.\n9. So the two difference quotients have the same limit as $h \\to 0$. By the definition of the [[derivative|0F1WR9C83K]], $F$ is differentiable at $x_0$ with $F'(x_0) = f(x_0)$.\n10. Since $x_0 \\in I$ was arbitrary, $F$ is differentiable on $I$ with $F' = f$. So $F$ is an [[antiderivative|0BW9N5CDSB]] of $f$ on $I$.\n\n## Idea\n\nDifferentiability is a local property. Near any point $x_0$ we can place a closed interval $[p, q]$ inside $I$. There $F$ differs from $\\int_p^x f$ only by a constant, and the first fundamental theorem gives the derivative.\n",
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