[proofplan] We reduce the complex-valued statement to the real-valued theorem by writing $f=u+iv$, where $u=\operatorname{Re}f$ and $v=\operatorname{Im}f$. [Absolute continuity](/page/Absolute%20Continuity) passes to the real and imaginary parts because each component increment is bounded by the complex increment. Applying the real-valued fundamental theorem for absolutely continuous functions gives almost-everywhere differentiability, integrability of $u'$ and $v'$, and the integral formula componentwise. Combining the two componentwise formulas produces a complex-valued $L^1$ representative of the derivative and the desired identity for every $x\in[a,b]$. [/proofplan] [step:Pass absolute continuity to the real and imaginary parts] Define $u:[a,b]\to\mathbb R$ by $u(x)=\operatorname{Re}f(x)$, and define $v:[a,b]\to\mathbb R$ by $v(x)=\operatorname{Im}f(x)$. We prove that $u$ and $v$ are absolutely continuous on $[a,b]$. Let $\varepsilon>0$. Since $f$ is absolutely continuous, there exists $\delta>0$ such that for every finite family of pairwise disjoint intervals $(\alpha_j,\beta_j)\subset[a,b]$ with \begin{align*} \sum_{j=1}^N(\beta_j-\alpha_j)<\delta, \end{align*} one has \begin{align*} \sum_{j=1}^N |f(\beta_j)-f(\alpha_j)|<\varepsilon. \end{align*} For the same family, \begin{align*} \sum_{j=1}^N |u(\beta_j)-u(\alpha_j)| \leq \sum_{j=1}^N |f(\beta_j)-f(\alpha_j)| < \varepsilon, \end{align*} because the absolute value of the real part of a complex number is bounded by its complex modulus. For each index $j$, the imaginary increment is controlled by the complex increment: \begin{align*} |v(\beta_j)-v(\alpha_j)| = |\operatorname{Im}(f(\beta_j)-f(\alpha_j))| \leq |f(\beta_j)-f(\alpha_j)|. \end{align*} Summing over $j$ gives \begin{align*} \sum_{j=1}^N |v(\beta_j)-v(\alpha_j)| \leq \sum_{j=1}^N |f(\beta_j)-f(\alpha_j)| < \varepsilon. \end{align*} Thus $u$ and $v$ are absolutely continuous on $[a,b]$. [guided] The point of this reduction is that the standard real-valued theorem is available for absolutely continuous functions with values in $\mathbb R$, while our function takes values in $\mathbb C$. We therefore split $f$ into its two coordinate functions. Define $u:[a,b]\to\mathbb R$ by $u(x)=\operatorname{Re}f(x)$, and define $v:[a,b]\to\mathbb R$ by $v(x)=\operatorname{Im}f(x)$. To prove that $u$ is absolutely continuous, we must verify the same interval condition as for $f$. Let $\varepsilon>0$. Since $f$ is absolutely continuous, there is a number $\delta>0$ such that every finite pairwise disjoint family of intervals $(\alpha_j,\beta_j)\subset[a,b]$ satisfying \begin{align*} \sum_{j=1}^N(\beta_j-\alpha_j)<\delta \end{align*} also satisfies \begin{align*} \sum_{j=1}^N |f(\beta_j)-f(\alpha_j)|<\varepsilon. \end{align*} For each index $j$, the real increment of $f$ is controlled by the complex increment: \begin{align*} |u(\beta_j)-u(\alpha_j)| = |\operatorname{Re}(f(\beta_j)-f(\alpha_j))| \leq |f(\beta_j)-f(\alpha_j)|. \end{align*} Summing this inequality over $j$ gives \begin{align*} \sum_{j=1}^N |u(\beta_j)-u(\alpha_j)| \leq \sum_{j=1}^N |f(\beta_j)-f(\alpha_j)| < \varepsilon. \end{align*} This is exactly the absolute continuity condition for $u$. We now verify the absolute continuity condition for $v$ explicitly. For each index $j$, \begin{align*} |v(\beta_j)-v(\alpha_j)| = |\operatorname{Im}(f(\beta_j)-f(\alpha_j))| \leq |f(\beta_j)-f(\alpha_j)|. \end{align*} Summing this inequality over $j$ gives \begin{align*} \sum_{j=1}^N |v(\beta_j)-v(\alpha_j)| \leq \sum_{j=1}^N |f(\beta_j)-f(\alpha_j)| < \varepsilon. \end{align*} Hence both $u$ and $v$ are absolutely continuous real-valued functions on $[a,b]$. [/guided] [/step] [step:Apply the real-valued theorem to obtain componentwise derivatives] By the real-valued [fundamental theorem of calculus](/theorems/632) for absolutely continuous functions, applied to the absolutely continuous maps $u:[a,b]\to\mathbb R$ and $v:[a,b]\to\mathbb R$, there exist functions $p:(a,b)\to\mathbb R$ and $q:(a,b)\to\mathbb R$ with $p,q\in L^1((a,b),\mathcal B((a,b)),\mathcal L^1)$ such that $u$ and $v$ are differentiable at $\mathcal L^1$-almost every point of $(a,b)$, $p(x)=u'(x)$ for $\mathcal L^1$-almost every point $x$ where $u$ is differentiable, $q(x)=v'(x)$ for $\mathcal L^1$-almost every point $x$ where $v$ is differentiable, and for every $x\in[a,b]$, \begin{align*} u(x)=u(a)+\int_{(a,x)} p(t)\,d\mathcal L^1(t), \end{align*} and \begin{align*} v(x)=v(a)+\int_{(a,x)} q(t)\,d\mathcal L^1(t). \end{align*} Here the case $x=a$ uses the convention that the integral over the empty interval $(a,a)$ is $0$. [/step] [step:Assemble the complex derivative as an $L^1$ function] Define $g:(a,b)\to\mathbb C$ by \begin{align*} g(t)=p(t)+iq(t). \end{align*} Since $p$ and $q$ are measurable real-valued functions, $g$ is a measurable complex-valued function. Also, for every $t\in(a,b)$, \begin{align*} |g(t)|\leq |p(t)|+|q(t)|. \end{align*} Integrating this inequality with respect to $\mathcal L^1$ on $(a,b)$ gives \begin{align*} \int_{(a,b)} |g(t)|\,d\mathcal L^1(t) \leq \int_{(a,b)} |p(t)|\,d\mathcal L^1(t) + \int_{(a,b)} |q(t)|\,d\mathcal L^1(t) < \infty. \end{align*} Thus $g\in L^1((a,b),\mathcal B((a,b)),\mathcal L^1)$. Let $E\subset(a,b)$ be the set of points at which both $u$ and $v$ are differentiable and at which $p=u'$ and $q=v'$. The complements of the corresponding real-valued differentiability and equality sets have $\mathcal L^1$-measure zero, so $\mathcal L^1((a,b)\setminus E)=0$. For $x\in E$, \begin{align*} \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{u(x+h)-u(x)}{h} + i\lim_{h\to 0}\frac{v(x+h)-v(x)}{h} = p(x)+iq(x) = g(x), \end{align*} where the limit is taken over real $h$ with $x+h\in[a,b]$. Hence $f$ is differentiable at every $x\in E$, and $f'(x)=g(x)$ there. Since $\mathcal L^1((a,b)\setminus E)=0$, this proves that $f$ is differentiable $\mathcal L^1$-almost everywhere on $(a,b)$ and that $g(x)=f'(x)$ for $\mathcal L^1$-almost every differentiability point of $f$. Thus $g$ represents the almost-everywhere derivative $f'$ in $L^1$. [/step] [step:Combine the componentwise integral identities] For every $x\in[a,b]$, the componentwise identities give \begin{align*} f(x)=u(x)+iv(x). \end{align*} Substituting the two real-valued integral formulas yields \begin{align*} f(x)=u(a)+\int_{(a,x)}p(t)\,d\mathcal L^1(t)+i v(a)+i\int_{(a,x)}q(t)\,d\mathcal L^1(t). \end{align*} By linearity of the complex [Lebesgue integral](/page/Lebesgue%20Integral) for integrable real and imaginary parts, \begin{align*} f(x)=f(a)+\int_{(a,x)}(p(t)+iq(t))\,d\mathcal L^1(t). \end{align*} Since $g=p+iq$, this becomes \begin{align*} f(x)=f(a)+\int_{(a,x)}g(t)\,d\mathcal L^1(t). \end{align*} Because $g=f'$ at $\mathcal L^1$-almost every point of $(a,b)$, the same identity may be written with $f'$ denoting this $L^1$ representative: \begin{align*} f(x)=f(a)+\int_{(a,x)}f'(t)\,d\mathcal L^1(t). \end{align*} This proves the asserted differentiability, integrability, and reconstruction formula for all $x\in[a,b]$. [/step]