[proofplan] We prove the two implications separately. **From $W^{1,1}$ to absolutely continuous:** given $u \in W^{1,1}(a,b)$ with weak derivative $v$, we define a candidate representative $\tilde{u}$ as an indefinite integral of $v$. Absolute continuity of $\tilde{u}$ follows from absolute continuity of the Lebesgue integral. To identify $u = \tilde{u}$ a.e., we approximate $u$ by mollifications $u_\varepsilon = u * \eta_\varepsilon$, write the (smooth, classical) fundamental theorem of calculus for each $u_\varepsilon$, and pass to the $L^1$ limit using that mollifications converge to $u$ in $L^1$ and to $v$ in $L^1$ for the derivatives. **From absolutely continuous to $W^{1,1}$:** the a.e. classical derivative $v = \tilde{u}'$ exists and lies in $L^1$ by the fundamental theorem of calculus for Lebesgue integrals; we verify it is the weak derivative by integration by parts against test functions, with all boundary terms vanishing because the test function has compact support inside $(a,b)$. [/proofplan]

[step:Construct the candidate absolutely continuous representative] Assume $u \in W^{1,1}(a,b)$ with weak derivative $v := Du \in L^1(a,b)$. Since $v \in L^1(a,b)$, the function \begin{align*} F: [a,b] &\to \mathbb{R} \\ x &\mapsto \int_a^x v(t)\, d\mathcal{L}^1(t) \end{align*} is well-defined, and by absolute continuity of the Lebesgue integral the function $F$ is absolutely continuous on $[a,b]$ with $F'= v$ $\mathcal{L}^1$-a.e. on $(a,b)$. We will show that there exists a constant $c \in \mathbb{R}$ such that $u(x) = F(x) + c$ for $\mathcal{L}^1$-a.e. $x \in (a,b)$. With $c$ determined, the representative is \begin{align*} \tilde{u}: [a,b] &\to \mathbb{R} \\ x &\mapsto F(x) + c. \end{align*} [/step]

[step:Mollify $u$ and apply the classical fundamental theorem of calculus]Extend $u$ and $v$ by zero outside $(a,b)$ to functions on $\mathbb{R}$ (still denoted $u$, $v$). For $\varepsilon > 0$, let $\eta_\varepsilon: \mathbb{R} \to \mathbb{R}$ be the standard mollifier (so $\eta_\varepsilon \in C_c^\infty(\mathbb{R})$, $\operatorname{supp} \eta_\varepsilon \subseteq [-\varepsilon, \varepsilon]$, $\eta_\varepsilon \geq 0$, and $\int_{\mathbb{R}} \eta_\varepsilon\, d\mathcal{L}^1 = 1$). Define \begin{align*} u_\varepsilon: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto (u * \eta_\varepsilon)(x) = \int_{\mathbb{R}} u(y)\, \eta_\varepsilon(x - y)\, d\mathcal{L}^1(y). \end{align*} Fix any compact subinterval $[\alpha, \beta] \subset (a,b)$, and consider $\varepsilon < \min\{\alpha - a, b - \beta\}$ so that for every $x \in [\alpha, \beta]$ the support of $\eta_\varepsilon(x - \cdot)$ lies in $(a,b)$. By the [Differentiation Passes Through Convolution](/theorems/3096) result applied on the open interval $V = (\alpha, \beta) \subset\subset (a,b)$, we have $u_\varepsilon \in C^\infty(\alpha, \beta)$ with \begin{align*} u_\varepsilon'(x) = (v * \eta_\varepsilon)(x) \quad \text{for all } x \in (\alpha, \beta). \end{align*} Since $u_\varepsilon$ is smooth (in particular $C^1$) on $(\alpha, \beta)$, the classical fundamental theorem of calculus gives, for any $\alpha \leq y < x \leq \beta$, \begin{align*} u_\varepsilon(x) - u_\varepsilon(y) = \int_y^x u_\varepsilon'(t)\, d\mathcal{L}^1(t) = \int_y^x (v * \eta_\varepsilon)(t)\, d\mathcal{L}^1(t). \end{align*}[/step]

[guided]The strategy is to write the smooth fundamental theorem of calculus for $u_\varepsilon$, where everything is justified classically, and then send $\varepsilon \to 0$ to recover an identity about $u$ itself. The issue is matching the domains: $u$ is defined on $(a,b)$, but $u_\varepsilon = u * \eta_\varepsilon$ requires evaluating $u$ on a $\varepsilon$-neighbourhood of each point. To avoid boundary issues we extend $u$ and $v$ by zero outside $(a,b)$ — both extensions remain in $L^1(\mathbb{R})$ — and restrict attention to a compact subinterval $[\alpha, \beta] \subset (a,b)$. Choosing $\varepsilon < \min\{\alpha - a, b - \beta\}$ ensures the smoothing involves only values of $u$ from inside $(a,b)$, so the extension by zero does not contaminate the calculation on $[\alpha, \beta]$. The key analytic input is [Differentiation Passes Through Convolution](/theorems/3096), which we apply with $V = (\alpha, \beta)$. Its hypotheses are: $u \in W^{1,p}$ for some $1 \leq p < \infty$ (we have $p = 1$); $V \subset\subset \Omega$ open with $\overline{V} \subset \Omega$ (we have $\overline{(\alpha, \beta)} = [\alpha, \beta] \subset (a,b)$); and $\varepsilon < \operatorname{dist}(V, \partial\Omega) = \min\{\alpha - a, b - \beta\}$. All hypotheses verified, the theorem yields $u_\varepsilon \in C^\infty(\alpha, \beta)$ and the identity $u_\varepsilon' = v * \eta_\varepsilon$ on $(\alpha, \beta)$. Now we write the classical fundamental theorem of calculus for $u_\varepsilon$ on $[\alpha, \beta]$. Since $u_\varepsilon$ is $C^1$ on $(\alpha, \beta)$ and continuous on $[\alpha, \beta]$ (continuous on a slightly larger open set, in fact), the fundamental theorem of calculus for $C^1$ functions applies — its only hypothesis is that the integrand $u_\varepsilon'$ be continuous on $[y, x]$, and continuity of $u_\varepsilon'$ on $(\alpha, \beta) \supseteq [y, x]$ is granted by smoothness. We obtain \begin{align*} u_\varepsilon(x) - u_\varepsilon(y) = \int_y^x u_\varepsilon'(t)\, d\mathcal{L}^1(t) = \int_y^x (v * \eta_\varepsilon)(t)\, d\mathcal{L}^1(t). \end{align*} This is the identity we will pass to the $\varepsilon \to 0$ limit.[/guided]

[step:Pass to the $L^1$ limit and identify $u$ with $F$ up to a constant]By the standard convergence of mollifications in $L^p$, as $\varepsilon \to 0$, \begin{align*} u_\varepsilon \to u \quad \text{in } L^1_{\mathrm{loc}}(\mathbb{R}), \qquad v * \eta_\varepsilon \to v \quad \text{in } L^1_{\mathrm{loc}}(\mathbb{R}). \end{align*} In particular, on $[\alpha, \beta]$, \begin{align*} \int_\alpha^\beta |u_\varepsilon - u|\, d\mathcal{L}^1 \to 0, \qquad \int_\alpha^\beta |v * \eta_\varepsilon - v|\, d\mathcal{L}^1 \to 0. \end{align*} Fix any $y \in (\alpha, \beta)$ and integrate the identity from Step 2 in $x$ over $[\alpha, \beta]$: \begin{align*} \int_\alpha^\beta u_\varepsilon(x)\, d\mathcal{L}^1(x) - (\beta - \alpha)\, u_\varepsilon(y) = \int_\alpha^\beta \int_y^x (v * \eta_\varepsilon)(t)\, d\mathcal{L}^1(t)\, d\mathcal{L}^1(x). \end{align*} The left-hand side does not directly give the desired identity for $u$; instead we proceed pointwise using the Lebesgue differentiation theorem. Since $u \in L^1(a,b)$ implies $u \in L^1_{\mathrm{loc}}(a,b)$ (any $L^1$ function on a bounded interval is locally integrable), $\mathcal{L}^1$-a.e. point of $(a,b)$ is a Lebesgue point of $u$, and similarly for $v$. The standard mollifier $\eta_\varepsilon$ is non-negative, radial, and supported in $[-\varepsilon,\varepsilon]$ with unit mass, so at every Lebesgue point $x \in (\alpha, \beta)$ of $u$, \begin{align*} u_\varepsilon(x) = \int_{\mathbb{R}} u(x - z)\, \eta_\varepsilon(z)\, d\mathcal{L}^1(z) \xrightarrow{\varepsilon \to 0} u(x), \end{align*} by mollifier convergence at Lebesgue points (the average of $u$ against the radial weight $\eta_\varepsilon$ on $B(x,\varepsilon)$ converges to $u(x)$), and similarly $(v * \eta_\varepsilon)(x) \to v(x)$ at Lebesgue points of $v$. Moreover, for any $y < x$ in $(\alpha, \beta)$, \begin{align*} \int_y^x (v * \eta_\varepsilon)(t)\, d\mathcal{L}^1(t) \xrightarrow{\varepsilon \to 0} \int_y^x v(t)\, d\mathcal{L}^1(t) = F(x) - F(y), \end{align*} since $v * \eta_\varepsilon \to v$ in $L^1([\alpha, \beta])$ and integration over $[y, x]$ is a continuous functional on $L^1$. Take $y \in (\alpha, \beta)$ to be a common Lebesgue point of $u$, and fix it. For every Lebesgue point $x \in (\alpha, \beta)$ of $u$ with $x > y$ (the case $x < y$ is symmetric), Step 2 gives \begin{align*} u_\varepsilon(x) - u_\varepsilon(y) = \int_y^x (v * \eta_\varepsilon)(t)\, d\mathcal{L}^1(t). \end{align*} Sending $\varepsilon \to 0$, \begin{align*} u(x) - u(y) = F(x) - F(y). \end{align*} Setting $c := u(y) - F(y)$, this gives $u(x) = F(x) + c$ for $\mathcal{L}^1$-a.e. $x \in (\alpha, \beta)$. Since the Lebesgue points of $u$ have full measure in $(a,b)$ and $[\alpha, \beta] \subset (a,b)$ was arbitrary, $u(x) = F(x) + c$ for $\mathcal{L}^1$-a.e. $x \in (a,b)$. Define $\tilde{u}(x) := F(x) + c$ on $[a,b]$. Then $\tilde{u}$ is absolutely continuous on $[a,b]$ (adding a constant preserves the $\varepsilon$-$\delta$ definition of absolute continuity verbatim, since differences $|\tilde{u}(x_i) - \tilde{u}(y_i)| = |F(x_i) - F(y_i)|$ are unaffected), $\tilde{u}' = F' = v \in L^1(a,b)$ $\mathcal{L}^1$-a.e., and $\tilde{u} = u$ $\mathcal{L}^1$-a.e. on $(a,b)$.[/step]

[guided]We now have, for each fixed $\varepsilon > 0$, the identity \begin{align*} u_\varepsilon(x) - u_\varepsilon(y) = \int_y^x (v * \eta_\varepsilon)(t)\, d\mathcal{L}^1(t). \end{align*} We want to send $\varepsilon \to 0$ and recover an identity about $u$. The left-hand side involves the values of $u_\varepsilon$ at *individual* points $x, y$, not integrals — so $L^1$ convergence alone is not enough. We need pointwise convergence at $x$ and $y$. The standard tool is the Lebesgue differentiation theorem: an $L^1_{\mathrm{loc}}$ function is the limit of its averages at $\mathcal{L}^1$-a.e. point (Lebesgue points), and at Lebesgue points the mollifications converge pointwise (since $\eta_\varepsilon$ is radial and supported in $[-\varepsilon,\varepsilon]$ with unit mass): \begin{align*} u_\varepsilon(x) = \int_{\mathbb{R}} u(x - z)\, \eta_\varepsilon(z)\, d\mathcal{L}^1(z) \to u(x). \end{align*} The argument is: $u_\varepsilon(x)$ is a weighted average of $u$ over the ball $B(x, \varepsilon)$ with weight $\eta_\varepsilon$; the Lebesgue point property says these averages converge to $u(x)$. So we choose $y$ to be a Lebesgue point of $u$ (such points have full measure, so they exist), fix it, and for every Lebesgue point $x \in (\alpha, \beta)$ pass to the limit in the identity. The right-hand side is an integral $\int_y^x (v * \eta_\varepsilon)\, d\mathcal{L}^1$, and integration over the bounded interval $[y, x]$ is a continuous linear functional on $L^1([\alpha, \beta])$ — so $L^1$ convergence $v * \eta_\varepsilon \to v$ implies \begin{align*} \int_y^x (v * \eta_\varepsilon)\, d\mathcal{L}^1 \to \int_y^x v\, d\mathcal{L}^1 = F(x) - F(y). \end{align*} The result is $u(x) - u(y) = F(x) - F(y)$ at $\mathcal{L}^1$-a.e. $x \in (\alpha, \beta)$, with $y$ fixed. Setting $c = u(y) - F(y)$, this is $u(x) = F(x) + c$ a.e. The compact subinterval $[\alpha, \beta]$ was arbitrary, and the union of an exhausting sequence of such subintervals covers $(a,b)$, so $u = F + c$ a.e. on $(a,b)$. To finish, define $\tilde{u}(x) = F(x) + c$ on $[a,b]$. This is the desired absolutely continuous representative: absolutely continuous because $F$ is, agreeing with $u$ a.e., with derivative $\tilde{u}' = F' = v$ a.e. by the differentiation theorem for indefinite integrals.[/guided]

[step:Prove the converse: absolutely continuous $\implies u \in W^{1,1}$] Conversely, suppose $\tilde{u}: [a,b] \to \mathbb{R}$ is absolutely continuous with $\tilde{u}' \in L^1(a,b)$. By the fundamental theorem of calculus for Lebesgue integrals, the classical derivative $\tilde{u}'$ exists $\mathcal{L}^1$-a.e. in $(a,b)$, lies in $L^1(a,b)$, and satisfies \begin{align*} \tilde{u}(x) - \tilde{u}(y) = \int_y^x \tilde{u}'(t)\, d\mathcal{L}^1(t) \quad \text{for all } a \leq y \leq x \leq b. \end{align*} Define $u := \tilde{u}|_{(a,b)} \in L^1(a,b)$ and $v := \tilde{u}' \in L^1(a,b)$. We verify that $v$ is the weak derivative of $u$. For any $\phi \in C_c^\infty(a,b)$, both $\tilde{u}$ and $\phi$ are absolutely continuous on $[a,b]$ (the AC function $\tilde{u}$ is continuous on the compact interval $[a,b]$, hence bounded; $\phi \in C_c^\infty(a,b)$ is smooth with compact support, hence bounded), and the product of two bounded AC functions on a compact interval is again AC, so $\tilde{u}\phi$ is absolutely continuous on $[a,b]$. Hence by integration by parts for absolutely continuous functions, \begin{align*} \int_a^b \tilde{u}(t)\, \phi'(t)\, d\mathcal{L}^1(t) + \int_a^b \tilde{u}'(t)\, \phi(t)\, d\mathcal{L}^1(t) = \tilde{u}(b)\phi(b) - \tilde{u}(a)\phi(a) = 0, \end{align*} where the boundary terms vanish because $\phi$ has compact support in $(a,b)$, hence $\phi(a) = \phi(b) = 0$. Rearranging, \begin{align*} \int_a^b u(t)\, \phi'(t)\, d\mathcal{L}^1(t) = -\int_a^b v(t)\, \phi(t)\, d\mathcal{L}^1(t). \end{align*} This is precisely the integration-by-parts identity defining $v = Du$ as the weak derivative of $u$. Since both $u, v \in L^1(a,b)$, we conclude $u \in W^{1,1}(a,b)$. [/step]