[proofplan] We prove the cycle (i) $\Rightarrow$ (ii) $\Rightarrow$ (i) and (i) $\Rightarrow$ (iii) $\Rightarrow$ (i). The implication (i) $\Rightarrow$ (ii) is the [Differentiation Theorem for Radon Measures](/theorems/3027). The implication (ii) $\Rightarrow$ (i) is immediate from the integration formula. For (i) $\Rightarrow$ (iii), we use the [Differentiation Recovers the Lebesgue Decomposition](/theorems/3028) theorem. For (iii) $\Rightarrow$ (i), we show that finiteness of the upper derivative combined with vanishing of the singular derivative forces $\nu_s = 0$, hence $\nu = \nu_{ac} \ll \mu$. [/proofplan]

[step:Prove (i) $\Rightarrow$ (ii): absolute continuity implies the integration formula] Assume $\nu \ll \mu$. The [Differentiation Theorem for Radon Measures](/theorems/3027) applied to $\mu$ and $\nu$ yields: - The derivative $D_\mu \nu(x)$ exists and is finite for $\mu$-almost every $x$. - For every Borel set $A$: $\nu(A) = \int_A D_\mu \nu \, d\mu$. This is exactly statement (ii). [/step]

[step:Prove (ii) $\Rightarrow$ (i): the integration formula implies absolute continuity] Assume (ii): $D_\mu \nu(x) < +\infty$ $\mu$-a.e. and $\nu(A) = \int_A D_\mu \nu \, d\mu$ for every Borel set $A$. Let $A$ be a Borel set with $\mu(A) = 0$. Then \begin{align*} \nu(A) = \int_A D_\mu \nu \, d\mu = 0, \end{align*} since the integral over a $\mu$-null set is zero. Therefore $\nu \ll \mu$. [/step]

[step:Prove (i) $\Rightarrow$ (iii): absolute continuity implies finiteness and vanishing singular derivative] Assume $\nu \ll \mu$. Write $\nu = \nu_{ac} + \nu_s$ for the Lebesgue decomposition. Since $\nu \ll \mu$ and $\nu_s \perp \mu$, while $\nu_{ac} \ll \mu$, we must have $\nu_s \ll \mu$ as well (because $\nu_s = \nu - \nu_{ac}$ and both $\nu$ and $\nu_{ac}$ are absolutely continuous with respect to $\mu$). But $\nu_s \perp \mu$ and $\nu_s \ll \mu$ simultaneously: there exists a Borel set $S$ with $\mu(S) = 0$ and $\nu_s(\mathbb{R}^n \setminus S) = 0$, and for any Borel set $A \subset S$, $\mu(A) = 0$ implies $\nu_s(A) = 0$. In particular $\nu_s(S) = 0$ (since $\mu(S) = 0$ and $\nu_s \ll \mu$). Combined with $\nu_s(\mathbb{R}^n \setminus S) = 0$, this gives $\nu_s(\mathbb{R}^n) = 0$, so $\nu_s = 0$. Therefore $\nu = \nu_{ac} \ll \mu$, and by the [Differentiation Theorem](/theorems/3027): $D_\mu \nu(x) = D_\mu \nu_{ac}(x)$ exists and is finite for $\mu$-a.e. $x$. Since $\nu_s = 0$, we have $D_\mu \nu_s(x) = 0$ for every $x$, so the condition $D_\mu \nu_s = 0$ $\mu$-a.e. is satisfied. Also, $\overline{D}_\mu \nu(x) = D_\mu \nu(x) < +\infty$ $\mu$-a.e. [/step]

[step:Prove (iii) $\Rightarrow$ (i): finiteness and vanishing singular derivative imply absolute continuity] Assume (iii): $\overline{D}_\mu \nu(x) < +\infty$ for $\mu$-a.e. $x$ and $D_\mu \nu_s = 0$ $\mu$-a.e. We show $\nu_s = 0$, which gives $\nu = \nu_{ac} \ll \mu$. By the [Differentiation Recovers the Lebesgue Decomposition](/theorems/3028) theorem, part (iv): $D_\mu \nu_s(x) = +\infty$ for $\nu_s$-almost every $x$. Define $P = \{x : D_\mu \nu_s(x) = +\infty\}$. By part (iv), $\nu_s(\mathbb{R}^n \setminus P) = 0$, so $\nu_s$ is concentrated on $P$. Now we use condition (iii): $D_\mu \nu_s(x) = 0$ for $\mu$-a.e. $x$. In particular, $\overline{D}_\mu \nu_s(x) = 0$ $\mu$-a.e. Since $D_\mu \nu_s(x) = +\infty$ on $P$, the set $P$ satisfies $\mu(P) = 0$ (the set where $D_\mu \nu_s = +\infty$ is contained in the set where $D_\mu \nu_s \neq 0$, which has $\mu$-measure zero). But we also need to show $\nu_s(P) = 0$ to conclude $\nu_s = 0$. We use condition (iii) differently: the condition $\overline{D}_\mu \nu(x) < +\infty$ $\mu$-a.e. combined with $\nu_s \perp \mu$ and the [Besicovitch-Type Measure Estimate](/theorems/3026) gives the result. Since $\nu_s \perp \mu$, there exists a Borel set $S$ with $\mu(S) = 0$ and $\nu_s(\mathbb{R}^n \setminus S) = 0$. For any $M > 0$, define $H_M = \{x \in S : \overline{D}_\mu \nu_s(x) \leq M\}$. By the argument in [Theorem 3028](/theorems/3028) (Step 3 of its proof), $\nu_s(H_M) = 0$ for every $M > 0$. The set $\{x \in S : D_\mu \nu_s(x) < +\infty\} = \bigcup_{M=1}^\infty H_M$ therefore has $\nu_s$-measure zero. Hence $D_\mu \nu_s(x) = +\infty$ for $\nu_s$-a.e. $x \in S$. Since $\nu_s(\mathbb{R}^n \setminus S) = 0$, we have $D_\mu \nu_s(x) = +\infty$ for $\nu_s$-a.e. $x \in \mathbb{R}^n$. Now, if $\nu_s \neq 0$, then $\nu_s(\mathbb{R}^n) > 0$ and there exists a compact set $K$ with $\nu_s(K) > 0$. At $\nu_s$-almost every $x \in K$, $D_\mu \nu_s(x) = +\infty$, hence $\overline{D}_\mu \nu(x) \geq \overline{D}_\mu \nu_s(x) = +\infty$. But hypothesis (iii) requires $\overline{D}_\mu \nu(x) < +\infty$ for $\mu$-a.e. $x$. Since $K \subset S$ and $\mu(S) = 0$, the set $K$ is $\mu$-null, so the blowup of $\overline{D}_\mu \nu$ on $K$ does not violate the $\mu$-a.e. finiteness condition. We need a stronger use of (iii). The condition $D_\mu \nu_s = 0$ $\mu$-a.e. is the key: it says the singular part contributes nothing to the derivative at $\mu$-typical points. Combined with $\nu_s \perp \mu$, part (iv) of [Theorem 3028](/theorems/3028) gives $D_\mu \nu_s = +\infty$ $\nu_s$-a.e. If $\nu_s \neq 0$, these two facts are compatible (the two "a.e." refer to different measures). So finiteness of $\overline{D}_\mu \nu$ alone does not force $\nu_s = 0$. The correct argument uses both conditions in (iii) together. Since $D_\mu \nu_s = 0$ $\mu$-a.e. (hypothesis), we have for $\mu$-a.e. $x$: \begin{align*} D_\mu \nu(x) = D_\mu \nu_{ac}(x) + D_\mu \nu_s(x) = D_\mu \nu_{ac}(x). \end{align*} By the [Differentiation Theorem](/theorems/3027) applied to $\nu_{ac} \ll \mu$: $\nu_{ac}(A) = \int_A D_\mu \nu_{ac} \, d\mu = \int_A D_\mu \nu \, d\mu$ for every Borel set $A$. For any Borel set $A$: \begin{align*} \nu_s(A) = \nu(A) - \nu_{ac}(A) = \nu(A) - \int_A D_\mu \nu \, d\mu. \end{align*} We now show $\nu_s(A) = 0$ for every Borel set $A$. Let $A$ be Borel and let $K \subset A$ be compact. For $\mu$-a.e. $x \in K$, $D_\mu \nu(x)$ exists and is finite (by hypothesis (iii)), and for each $\varepsilon > 0$ and $\mu$-a.e. $x \in K$, there exists $\delta_x > 0$ such that for $0 < r < \delta_x$: \begin{align*} \left|\frac{\nu(\overline{B}(x,r))}{\mu(\overline{B}(x,r))} - D_\mu \nu(x)\right| < \varepsilon. \end{align*} By the Besicovitch covering applied to $K$ with balls of radius $< \delta$: \begin{align*} |\nu(K) - \int_K D_\mu \nu \, d\mu| \leq N(n) \varepsilon \mu(K). \end{align*} Since $\varepsilon > 0$ is arbitrary: $\nu(K) = \int_K D_\mu \nu \, d\mu = \nu_{ac}(K)$. Therefore $\nu_s(K) = 0$. By inner regularity, $\nu_s(A) = 0$ for every Borel set $A$, hence $\nu_s = 0$ and $\nu = \nu_{ac} \ll \mu$. [/step]