[proofplan] We prove both directions by comparing the interval values of $\mu_F$ with measures defined by integration against a density. If $F$ is absolutely continuous, the fundamental theorem for absolutely continuous functions gives an $L^1$ density $F'$ for the increments $F(t)-F(s)$, and uniqueness of the Lebesgue-Stieltjes measure identifies this density with $\mu_F$. Conversely, if $\mu_F$ is absolutely continuous with respect to [Lebesgue measure](/page/Lebesgue%20Measure), the [Radon-Nikodym theorem](/theorems/1247) gives a density $g$, the interval formula reconstructs $F$ as an indefinite integral of $g$, and the integral representation theorem gives [absolute continuity](/page/Absolute%20Continuity). The final density identity follows from uniqueness in the [Radon-Nikodym theorem](/theorems/2640). [/proofplan] [step:Represent the increments of an absolutely continuous $F$ by integrating $F'$] Assume first that $F$ is absolutely continuous on $[a,b]$. By [citetheorem:10121], the derivative $F'$ exists for $\mathcal L^1$-a.e. $x\in(a,b)$, belongs to \begin{align*} L^1((a,b),\mathcal B((a,b)),\mathcal L^1), \end{align*} and satisfies, for every $x\in[a,b]$, \begin{align*} F(x)=F(a)+\int_a^x F'(u)\,d\mathcal L^1(u). \end{align*} Hence, for every $s,t\in[a,b]$ with $s<t$, \begin{align*} F(t)-F(s)=\int_s^t F'(u)\,d\mathcal L^1(u). \end{align*} Because $F$ is nondecreasing, this last integral is nonnegative for every interval $(s,t]$. In particular, the finite signed Borel measure on $[a,b]$ obtained by integrating $F'$ over Borel sets is nonnegative on the generating half-open intervals, and therefore agrees with a positive finite Borel measure. [guided] Assume that $F$ is absolutely continuous on $[a,b]$. The point of invoking the fundamental theorem for absolutely continuous functions is that absolute continuity is exactly the regularity condition that turns increments of $F$ into integrals of its derivative. By [citetheorem:10121], the derivative $F'$ exists for $\mathcal L^1$-a.e. $x\in(a,b)$, the function $F'$ belongs to \begin{align*} L^1((a,b),\mathcal B((a,b)),\mathcal L^1), \end{align*} and the reconstruction formula \begin{align*} F(x)=F(a)+\int_a^x F'(u)\,d\mathcal L^1(u) \end{align*} holds for every $x\in[a,b]$. Applying this formula once at $t$ and once at $s$, where $a\le s<t\le b$, and subtracting the two identities gives \begin{align*} F(t)-F(s)=\int_s^t F'(u)\,d\mathcal L^1(u). \end{align*} This is the exact interval identity needed to compare the Lebesgue-Stieltjes measure of $F$ with a measure having density $F'$. Since $F$ is nondecreasing, $F(t)-F(s)\ge 0$ whenever $s<t$, so these interval values are nonnegative. Thus the density $F'$ produces the same nonnegative interval masses as the Stieltjes construction. [/guided] [/step] [step:Identify $\mu_F$ with the measure having density $F'$] Define \begin{align*} \nu:\mathcal B([a,b])\to\mathbb R \end{align*} by \begin{align*} \nu(A)=\int_A F'(u)\,d\mathcal L^1(u) \end{align*} for every $A\in\mathcal B([a,b])$, where $F'$ is assigned arbitrary values at $a$ and $b$. The endpoint choices do not affect the integral because \begin{align*} \mathcal L^1(\{a,b\})=0. \end{align*} For every $s,t\in[a,b]$ with $s<t$, \begin{align*} \nu((s,t])=\int_s^t F'(u)\,d\mathcal L^1(u)=F(t)-F(s)=\mu_F((s,t]). \end{align*} Also, \begin{align*} \nu(\{a\})=0=\mu_F(\{a\}). \end{align*} By the uniqueness part of the Lebesgue-Stieltjes construction for a continuous nondecreasing function with the normalization at $a$, the two finite Borel measures agree: \begin{align*} \mu_F=\nu. \end{align*} Therefore $\mu_F\ll\mathcal L^1|_{[a,b]}$, and \begin{align*} \frac{d\mu_F}{d(\mathcal L^1|_{[a,b]})}=F' \end{align*} $\mathcal L^1$-a.e. on $(a,b)$. [/step] [step:Use Radon-Nikodym to turn absolute continuity of $\mu_F$ into an integral representation of $F$] Conversely, assume \begin{align*} \mu_F\ll\mathcal L^1|_{[a,b]}. \end{align*} The measure $\mathcal L^1|_{[a,b]}$ is finite, hence $\sigma$-finite, and $\mu_F$ is a finite positive Borel measure, hence a finite complex measure after the standard real-to-complex inclusion. By [citetheorem:10127], there exists \begin{align*} g\in L^1([a,b],\mathcal B([a,b]),\mathcal L^1|_{[a,b]}) \end{align*} such that, for every $A\in\mathcal B([a,b])$, \begin{align*} \mu_F(A)=\int_A g(u)\,d(\mathcal L^1|_{[a,b]})(u). \end{align*} Therefore, for every $s,t\in[a,b]$ with $s<t$, \begin{align*} F(t)-F(s)=\mu_F((s,t])=\int_{(s,t]} g(u)\,d(\mathcal L^1|_{[a,b]})(u)=\int_s^t g(u)\,d\mathcal L^1(u). \end{align*} Taking $s=a$, we obtain for every $x\in[a,b]$, \begin{align*} F(x)=F(a)+\int_a^x g(u)\,d\mathcal L^1(u). \end{align*} [guided] Now assume that the measure-theoretic absolute continuity holds: \begin{align*} \mu_F\ll\mathcal L^1|_{[a,b]}. \end{align*} The goal is to recover $F$ as an indefinite integral. The Radon-Nikodym theorem is the correct tool because it converts absolute [continuity of measures](/theorems/1082) into a density. We verify its hypotheses. The measure $\mathcal L^1|_{[a,b]}$ is finite because $[a,b]$ has finite Lebesgue measure, and hence it is $\sigma$-finite. The measure $\mu_F$ is finite because \begin{align*} \mu_F([a,b])=\mu_F((a,b])+\mu_F(\{a\})=F(b)-F(a)<\infty. \end{align*} Thus [citetheorem:10127] applies and gives a function \begin{align*} g\in L^1([a,b],\mathcal B([a,b]),\mathcal L^1|_{[a,b]}) \end{align*} such that every Borel set $A\in\mathcal B([a,b])$ satisfies \begin{align*} \mu_F(A)=\int_A g(u)\,d(\mathcal L^1|_{[a,b]})(u). \end{align*} Apply this identity to the half-open interval $(s,t]$, where $a\le s<t\le b$. The defining property of the Lebesgue-Stieltjes measure gives \begin{align*} \mu_F((s,t])=F(t)-F(s), \end{align*} while the Radon-Nikodym representation gives \begin{align*} \mu_F((s,t])=\int_{(s,t]} g(u)\,d(\mathcal L^1|_{[a,b]})(u). \end{align*} Since the single point $\{s\}$ has $\mathcal L^1$-measure zero, this integral is the same as the integral over $[s,t]$, so \begin{align*} F(t)-F(s)=\int_s^t g(u)\,d\mathcal L^1(u). \end{align*} Taking $s=a$ yields, for every $x\in[a,b]$, \begin{align*} F(x)=F(a)+\int_a^x g(u)\,d\mathcal L^1(u). \end{align*} This is precisely the integral representation from which absolute continuity follows. [/guided] [/step] [step:Conclude absolute continuity and identify the density with $F'$] By [citetheorem:10120], applied to the function \begin{align*} x\mapsto F(a)+\int_a^x g(u)\,d\mathcal L^1(u) \end{align*} with $g\in L^1([a,b],\mathcal B([a,b]),\mathcal L^1|_{[a,b]})$, the function $F$ is absolutely continuous on $[a,b]$. It remains to identify the density. Since we have just proved that $F$ is absolutely continuous, the first direction of the proof applies to $F$ and shows that $\mu_F$ has Radon-Nikodym density $F'$ with respect to $\mathcal L^1|_{[a,b]}$. The Radon-Nikodym density obtained above was $g$, and the uniqueness assertion in [citetheorem:10127] gives \begin{align*} g=F' \end{align*} $\mathcal L^1$-a.e. on $(a,b)$. Thus \begin{align*} \frac{d\mu_F}{d(\mathcal L^1|_{[a,b]})}=F' \end{align*} $\mathcal L^1$-a.e. on $(a,b)$. Finally, since \begin{align*} \mu_F(\{a\})=0 \end{align*} by normalization and \begin{align*} \mathcal L^1(\{a,b\})=0, \end{align*} changing the values assigned to $F'$ at $a$ or $b$ does not change the Radon-Nikodym density class. Hence the same density identity may be read on $[a,b]$ after arbitrary endpoint assignments. This proves both implications and the asserted density formula. [/step]