[proofplan] Part I is proved by rewriting the difference quotient of $F$ as the average value of $f$ over the interval between $x$ and $x+h$. [Continuity](/page/Continuity) of $f$ at $x$ then forces this average to converge to $f(x)$. Part II compares any antiderivative $G$ with the integral-defined antiderivative $F$; their difference has derivative zero, so the [Mean Value Theorem](/theorems/186) makes that difference constant, giving the evaluation formula. [/proofplan] [step:Rewrite the difference quotient as a local average of $f$] Fix $x \in (a,b)$. Let $h \in \mathbb{R}\setminus\{0\}$ satisfy $x+h \in [a,b]$. Define the compact interval $J_h \subset [a,b]$ by \begin{align*} J_h := \begin{cases} [x,x+h], & h > 0,\\ [x+h,x], & h < 0. \end{cases} \end{align*} By additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) on intervals, together with the convention that $\int_r^s f(t)\,d\mathcal{L}^1(t)=-\int_s^r f(t)\,d\mathcal{L}^1(t)$ when $r>s$, we have \begin{align*} \frac{F(x+h)-F(x)}{h} = \frac{1}{|h|}\int_{J_h} f(t)\,d\mathcal{L}^1(t). \end{align*} Since $\mathcal{L}^1(J_h)=|h|$, this gives \begin{align*} \frac{F(x+h)-F(x)}{h}-f(x) = \frac{1}{|h|}\int_{J_h}\bigl(f(t)-f(x)\bigr)\,d\mathcal{L}^1(t). \end{align*} [guided] Fix $x \in (a,b)$ and choose $h \in \mathbb{R}\setminus\{0\}$ with $x+h \in [a,b]$. The only point requiring care is the sign of $h$, so define the interval between $x$ and $x+h$ by \begin{align*} J_h := \begin{cases} [x,x+h], & h > 0,\\ [x+h,x], & h < 0. \end{cases} \end{align*} This interval always has one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^1(J_h)=|h|$. If $h>0$, then additivity of the integral gives \begin{align*} F(x+h)-F(x) = \int_a^{x+h} f(t)\,d\mathcal{L}^1(t) - \int_a^x f(t)\,d\mathcal{L}^1(t) = \int_x^{x+h} f(t)\,d\mathcal{L}^1(t). \end{align*} If $h<0$, then the same additivity gives \begin{align*} F(x+h)-F(x) = \int_a^{x+h} f(t)\,d\mathcal{L}^1(t) - \int_a^x f(t)\,d\mathcal{L}^1(t) = -\int_{x+h}^{x} f(t)\,d\mathcal{L}^1(t). \end{align*} Dividing by $h$ in the two cases gives the same formula: \begin{align*} \frac{F(x+h)-F(x)}{h} = \frac{1}{|h|}\int_{J_h} f(t)\,d\mathcal{L}^1(t). \end{align*} Because $\mathcal{L}^1(J_h)=|h|$, the constant value $f(x)$ can be written as its average over $J_h$: \begin{align*} f(x) = \frac{1}{|h|}\int_{J_h} f(x)\,d\mathcal{L}^1(t). \end{align*} Subtracting these two average formulas gives \begin{align*} \frac{F(x+h)-F(x)}{h}-f(x) = \frac{1}{|h|}\int_{J_h}\bigl(f(t)-f(x)\bigr)\,d\mathcal{L}^1(t). \end{align*} [/guided] [/step] [step:Use continuity at $x$ to identify the derivative of $F$] Let $\varepsilon>0$. Since $f$ is continuous at $x$, there exists $\delta_0>0$ such that $|f(t)-f(x)|<\varepsilon$ whenever $t \in [a,b]$ and $|t-x|<\delta_0$. Define \begin{align*} \delta := \min\{\delta_0, x-a, b-x\}. \end{align*} If $0<|h|<\delta$, then $J_h \subset (x-\delta_0,x+\delta_0)\cap [a,b]$. Therefore \begin{align*} \left|\frac{F(x+h)-F(x)}{h}-f(x)\right| &= \left|\frac{1}{|h|}\int_{J_h}\bigl(f(t)-f(x)\bigr)\,d\mathcal{L}^1(t)\right|\\ &\leq \frac{1}{|h|}\int_{J_h}|f(t)-f(x)|\,d\mathcal{L}^1(t)\\ &< \frac{1}{|h|}\int_{J_h}\varepsilon\,d\mathcal{L}^1(t)\\ &= \varepsilon. \end{align*} Thus \begin{align*} \lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=f(x). \end{align*} Since $x \in (a,b)$ was arbitrary, $F$ is differentiable on $(a,b)$ and $F'(x)=f(x)$ for every $x \in (a,b)$. [/step] [step:Prove the integral-defined antiderivative is continuous on the closed interval] Since $f:[a,b]\to\mathbb{R}$ is continuous, it is Borel measurable. Since $[a,b]$ is compact, the [Extreme Value Theorem](/theorems/???) gives a constant $M\geq 0$ such that $|f(t)|\leq M$ for every $t\in [a,b]$. Because $\mathcal{L}^1([a,b])=b-a<\infty$, this bound implies $f\in L^1([a,b])$. Let $y,z\in [a,b]$ with $y<z$. By additivity of the integral on intervals and the preceding bound, \begin{align*} |F(z)-F(y)| &= \left|\int_y^z f(t)\,d\mathcal{L}^1(t)\right|\\ &\leq \int_y^z |f(t)|\,d\mathcal{L}^1(t)\\ &\leq \int_y^z M\,d\mathcal{L}^1(t)\\ &=M(z-y). \end{align*} The same estimate is immediate when $y=z$, and when $z<y$ it follows by interchanging $y$ and $z$. Therefore \begin{align*} |F(z)-F(y)|\leq M|z-y| \end{align*} for all $y,z\in [a,b]$. Hence $F$ is Lipschitz continuous, and in particular continuous, on $[a,b]$. [/step] [step:Compare an arbitrary antiderivative with the integral-defined antiderivative] Let $G:[a,b]\to\mathbb{R}$ be an antiderivative of $f$ in the sense of the statement. Define the [function](/page/Function) \begin{align*} H:[a,b] &\to \mathbb{R}\\ y &\mapsto G(y)-F(y). \end{align*} The functions $G$ and $F$ are continuous on $[a,b]$ and differentiable on $(a,b)$; for $F$, differentiability on $(a,b)$ follows from Part I. Hence $H$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with \begin{align*} H'(y)=G'(y)-F'(y)=f(y)-f(y)=0 \end{align*} for every $y \in (a,b)$. Apply the Mean Value Theorem to $H$ on any interval $[u,v]\subset [a,b]$ with $u<v$. Its hypotheses hold because $H$ is continuous on $[u,v]$ and differentiable on $(u,v)$. There exists $c \in (u,v)$ such that \begin{align*} H(v)-H(u)=H'(c)(v-u)=0. \end{align*} Thus $H$ is constant on $[a,b]$. Therefore there exists $C \in \mathbb{R}$ such that \begin{align*} G(y)-F(y)=C \end{align*} for every $y \in [a,b]$. [/step] [step:Evaluate the constant difference at the endpoints] Since $F(a)=\int_a^a f(t)\,d\mathcal{L}^1(t)=0$ and $F(b)=\int_a^b f(t)\,d\mathcal{L}^1(t)$, the identity $G(y)-F(y)=C$ gives \begin{align*} G(b)-F(b) &= C,\\ G(a)-F(a) &= C. \end{align*} Subtracting the second equality from the first yields \begin{align*} G(b)-G(a)-F(b)+F(a)=0. \end{align*} Using $F(a)=0$ and the definition of $F(b)$, we obtain \begin{align*} \int_a^b f(t)\,d\mathcal{L}^1(t) = F(b) = G(b)-G(a). \end{align*} This proves the evaluation formula and completes the proof. [/step]