[proofplan] We prove the equivalence in the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. For $(1) \Rightarrow (2)$: absolute continuity of $F$ implies that $\mu_F$ assigns zero mass to Lebesgue-null sets. For $(2) \Rightarrow (3)$: the Radon-Nikodym Theorem provides a density $f = \frac{d\mu_F}{d\mathcal{L}^1}$, and we show $F(x) - F(a) = \int_a^x f \, d\mathcal{L}^1$, then identify $f = F'$ a.e. using the Lebesgue Differentiation Theorem. For $(3) \Rightarrow (1)$: the integral representation directly yields absolute continuity of $F$ via absolute continuity of the Lebesgue integral. [/proofplan] [step:$(1) \Rightarrow (2)$: Absolute continuity of $F$ implies $\mu_F \ll \mathcal{L}^1$] Assume $F$ is absolutely continuous on $[a,b]$. Let $E \in \mathcal{B}([a,b])$ with $\mathcal{L}^1(E) = 0$. We must show $\mu_F(E) = 0$. Since Borel measures on $[a,b]$ are regular, it suffices to show $\mu_F(U) = 0$ for every open set $U \supseteq E$ with $\mathcal{L}^1(U)$ arbitrarily small. Every open subset of $[a,b]$ is a countable disjoint union of open intervals: $U = \bigcup_{k=1}^\infty (a_k, b_k)$. Let $\varepsilon > 0$. By the absolute continuity of $F$, there exists $\delta > 0$ such that for any finite collection of non-overlapping intervals $\{(c_j, d_j)\}_{j=1}^m$ with $\sum_{j=1}^m (d_j - c_j) < \delta$, we have $\sum_{j=1}^m |F(d_j) - F(c_j)| < \varepsilon$. Since $\mathcal{L}^1(E) = 0$, choose an open cover $U \supseteq E$ with $\mathcal{L}^1(U) < \delta$. Write $U = \bigcup_{k=1}^\infty (a_k, b_k)$ as a countable union of disjoint open intervals. Then $\sum_{k=1}^\infty (b_k - a_k) = \mathcal{L}^1(U) < \delta$. For any finite $m$, the intervals $(a_1, b_1), \ldots, (a_m, b_m)$ are non-overlapping with total length $< \delta$, so: \begin{align*} \sum_{k=1}^m (F(b_k) - F(a_k)) \le \sum_{k=1}^m |F(b_k) - F(a_k)| < \varepsilon. \end{align*} Since $F$ is nondecreasing, $F(b_k) - F(a_k) \ge 0$, so taking $m \to \infty$: \begin{align*} \mu_F(U) = \sum_{k=1}^\infty \mu_F((a_k, b_k)) = \sum_{k=1}^\infty (F(b_k^-) - F(a_k^-)). \end{align*} Since $F$ is right-continuous, $F(b_k^-) \le F(b_k)$ and $F(a_k^-) \le F(a_k) = F(a_k)$ (using right-continuity at $a_k$: $F(a_k^-) \le F(a_k)$ since $F$ is nondecreasing). More precisely, for the Lebesgue-Stieltjes measure, $\mu_F((a_k, b_k]) = F(b_k) - F(a_k)$. Therefore: \begin{align*} \mu_F(U) \le \sum_{k=1}^\infty \mu_F((a_k, b_k]) = \sum_{k=1}^\infty (F(b_k) - F(a_k)) \le \varepsilon. \end{align*} Since $\varepsilon > 0$ is arbitrary, $\mu_F(E) \le \mu_F(U) \le \varepsilon \to 0$, giving $\mu_F(E) = 0$. [guided] The argument translates the $\varepsilon$-$\delta$ definition of absolute continuity of $F$ into the measure-theoretic statement $\mu_F \ll \mathcal{L}^1$. The key is that the Lebesgue-Stieltjes measure $\mu_F$ on intervals is controlled by the variation of $F$: $\mu_F((a_k, b_k]) = F(b_k) - F(a_k)$. Absolute continuity of $F$ says precisely that small total length $\sum (b_k - a_k) < \delta$ forces small total variation $\sum |F(b_k) - F(a_k)| < \varepsilon$. Since $F$ is nondecreasing, $|F(b_k) - F(a_k)| = F(b_k) - F(a_k) = \mu_F((a_k, b_k])$, so the variation equals the measure. The passage from finite to countable unions uses the nonneg of each term and the monotone convergence of partial sums. [/guided] [/step] [step:$(2) \Rightarrow (3)$: Apply Radon-Nikodym to obtain the density, then identify it as $F'$] Assume $\mu_F \ll \mathcal{L}^1$. Both $\mu_F$ and $\mathcal{L}^1$ are finite (hence $\sigma$-finite) measures on $([a,b], \mathcal{B}([a,b]))$. By the [Radon-Nikodym Theorem](/theorems/1247), there exists a Borel-measurable function: \begin{align*} f: [a,b] &\to [0, \infty), \end{align*} satisfying $\mu_F(A) = \int_A f \, d\mathcal{L}^1$ for every Borel set $A \subseteq [a,b]$. Since $\mu_F([a,b]) = F(b) - F(a) < \infty$, we have $f \in L^1([a,b], \mathcal{L}^1)$. For every $x \in [a,b]$, applying this to $A = (a, x]$: \begin{align*} F(x) - F(a) = \mu_F((a, x]) = \int_{(a,x]} f \, d\mathcal{L}^1 = \int_a^x f \, d\mathcal{L}^1(t), \end{align*} where the last equality holds because singleton sets have $\mathcal{L}^1$-measure zero. It remains to show $f = F'$ $\mathcal{L}^1$-a.e. By the [Lebesgue Differentiation Theorem](/theorems/894), for $\mathcal{L}^1$-a.e. $x \in [a,b]$: \begin{align*} f(x) = \lim_{h \to 0} \frac{1}{2h} \int_{x-h}^{x+h} f \, d\mathcal{L}^1(t) = \lim_{h \to 0} \frac{\mu_F([x-h, x+h])}{2h} = \lim_{h \to 0} \frac{F(x+h) - F(x-h)}{2h}. \end{align*} The symmetric derivative equals the ordinary derivative when the latter exists. Moreover, since $F(x) - F(a) = \int_a^x f \, d\mathcal{L}^1(t)$ and $f \in L^1$, the function $G: x \mapsto \int_a^x f \, d\mathcal{L}^1(t)$ is absolutely continuous, and $G' = f$ $\mathcal{L}^1$-a.e. (again by the Lebesgue Differentiation Theorem). Since $F(x) = F(a) + G(x)$, we conclude $F' = G' = f$ $\mathcal{L}^1$-a.e. [guided] The Radon-Nikodym Theorem provides the density $f$ representing $\mu_F$ with respect to $\mathcal{L}^1$. The function $f$ is nonneg because $\mu_F$ is a nonneg measure (since $F$ is nondecreasing). The integral representation $F(x) - F(a) = \int_a^x f \, d\mathcal{L}^1(t)$ follows from evaluating $\mu_F$ on the interval $(a, x]$. The identification $f = F'$ a.e. uses the [Lebesgue Differentiation Theorem](/theorems/894), which states that a locally integrable function is recovered a.e. as the limit of its averages over shrinking balls (or intervals in one dimension). Since the integral of $f$ over $(a, x]$ equals $F(x) - F(a)$, differentiating both sides recovers $f = F'$ a.e. This step is where the direction from $\mu_F \ll \mathcal{L}^1$ to the pointwise derivative is established. The Radon-Nikodym Theorem gives the density as a measure-theoretic object; the Lebesgue Differentiation Theorem converts it to a pointwise statement. [/guided] [/step] [step:$(3) \Rightarrow (1)$: The integral representation implies absolute continuity of $F$] Assume $F'$ exists $\mathcal{L}^1$-a.e., $F' \in L^1([a,b], \mathcal{L}^1)$, and $F(x) - F(a) = \int_a^x F'(t) \, d\mathcal{L}^1(t)$ for every $x \in [a,b]$. Let $\varepsilon > 0$. Since $F' \ge 0$ (because $F$ is nondecreasing) and $F' \in L^1$, the measure $\lambda(A) := \int_A F' \, d\mathcal{L}^1$ is absolutely continuous with respect to $\mathcal{L}^1$. Therefore there exists $\delta > 0$ such that $\mathcal{L}^1(A) < \delta$ implies $\lambda(A) < \varepsilon$ (this is the absolute continuity of the Lebesgue integral for $L^1$ functions). Now let $\{(a_k, b_k)\}_{k=1}^m$ be finitely many non-overlapping intervals with $\sum_{k=1}^m (b_k - a_k) < \delta$. Then: \begin{align*} \sum_{k=1}^m |F(b_k) - F(a_k)| &= \sum_{k=1}^m (F(b_k) - F(a_k)) = \sum_{k=1}^m \int_{a_k}^{b_k} F'(t) \, d\mathcal{L}^1(t) \\ &= \int_{\bigcup_{k=1}^m (a_k, b_k)} F' \, d\mathcal{L}^1 = \lambda\left(\bigcup_{k=1}^m (a_k, b_k)\right) < \varepsilon, \end{align*} since $\mathcal{L}^1\left(\bigcup_{k=1}^m (a_k, b_k)\right) = \sum_{k=1}^m (b_k - a_k) < \delta$. This is the $\varepsilon$-$\delta$ definition of absolute continuity of $F$. [guided] This direction is a direct consequence of the absolute continuity of the Lebesgue integral. The key fact is: if $g \in L^1(\mathcal{L}^1)$ with $g \ge 0$, then the indefinite integral $A \mapsto \int_A g \, d\mathcal{L}^1$ is an absolutely continuous measure with respect to $\mathcal{L}^1$. In the $\varepsilon$-$\delta$ formulation: for every $\varepsilon > 0$ there exists $\delta > 0$ such that $\mathcal{L}^1(A) < \delta$ implies $\int_A g \, d\mathcal{L}^1 < \varepsilon$. This can be proved by approximation: choose a bounded function $g_M := \min(g, M)$ with $\int (g - g_M) \, d\mathcal{L}^1 < \varepsilon/2$. Then $\int_A g \, d\mathcal{L}^1 \le \int_A g_M \, d\mathcal{L}^1 + \varepsilon/2 \le M \cdot \mathcal{L}^1(A) + \varepsilon/2$, so $\delta = \varepsilon/(2M)$ works. [/guided] [/step] [step:Verify the final identity $\frac{d\mu_F}{d\mathcal{L}^1} = F'$ $\mathcal{L}^1$-a.e.] Under the equivalent conditions, $\mu_F(A) = \int_A F' \, d\mathcal{L}^1$ for every Borel set $A$. This is verified for intervals $(a, x]$ by hypothesis: $\mu_F((a,x]) = F(x) - F(a) = \int_a^x F' \, d\mathcal{L}^1$. The class of intervals $(a, x]$ for $x \in [a,b]$ generates $\mathcal{B}([a,b])$ and forms a $\pi$-system. Both $A \mapsto \mu_F(A)$ and $A \mapsto \int_A F' \, d\mathcal{L}^1$ are finite measures agreeing on this $\pi$-system. By the [Uniqueness of Measure Extension](/theorems/506) (applied with the exhaustion $X_n = [a,b]$ for all $n$), the two measures agree on all of $\mathcal{B}([a,b])$. By the uniqueness clause of the [Radon-Nikodym Theorem](/theorems/1247), $\frac{d\mu_F}{d\mathcal{L}^1} = F'$ $\mathcal{L}^1$-a.e. [guided] This final identification uses a $\pi$-$\lambda$ argument. The intervals $(a, x]$ form a $\pi$-system (closed under finite intersections, since $(a, x_1] \cap (a, x_2] = (a, \min(x_1, x_2)]$). The two finite measures $\mu_F$ and $\int \cdot \, F' \, d\mathcal{L}^1$ agree on this $\pi$-system by the integral representation. By the [Uniqueness of Measure Extension](/theorems/506), they agree on $\sigma(\{(a,x] : x \in [a,b]\}) = \mathcal{B}([a,b])$. The hypothesis of Theorem 506 requires an exhaustion by sets of finite measure in the $\pi$-system; here we take the constant sequence $X_n = (a, b]$ with $\mu_F((a,b]) = F(b) - F(a) < \infty$. [/guided] [/step]