[proofplan] Uniqueness follows from the standard fact that two integrable functions with equal integrals over every measurable set agree almost everywhere. For existence, we reduce to the case $\nu, \mu$ both finite positive measures (using $\sigma$-finiteness for $\mu$ and Jordan decomposition / linearity for $\nu$). In the reduced setting, we maximise $\int_\Omega h \, d\mu$ over the family $\mathcal{F}$ of measurable $h: \Omega \to [0, \infty]$ with $\int_A h \, d\mu \le \nu(A)$ for all $A$, achieving the supremum at some $f \in \mathcal{F}$. The defect measure $\rho(A) := \nu(A) - \int_A f \, d\mu$ is shown to be zero via Hahn decompositions of the auxiliary signed measures $\rho - n^{-1} \mu$: a positive part of any of these would let us improve $f$, contradicting maximality. [/proofplan]

[step:Establish uniqueness from agreement of set integrals] Suppose $f, g \in L^1(\mu)$ both represent $\nu$, so $\int_A f \, d\mu = \nu(A) = \int_A g \, d\mu$ for all $A \in \Sigma$. Then $\int_A (f - g) \, d\mu = 0$ for all $A \in \Sigma$. Set $h := \operatorname{Re}(f - g)$ and decompose $h = h^+ - h^-$ with $h^+ = \max(h, 0)$, $h^- = \max(-h, 0)$, both non-negative measurable. Apply the equality on $A_+ := \{h > 0\} = \{h^+ > 0\}$ to get $0 = \int_{A_+} (f - g) \, d\mu$, whose real part is $\int_{A_+} h \, d\mu = \int_{A_+} h^+ \, d\mu = \int_\Omega h^+ \, d\mu$. Since $h^+ \ge 0$ and $\int_\Omega h^+ \, d\mu = 0$, we conclude $h^+ = 0$ $\mu$-a.e. Symmetrically, $h^- = 0$ $\mu$-a.e. on $A_- := \{h < 0\}$. Hence $h = 0$ $\mu$-a.e. The same argument applied to the imaginary part of $f - g$ gives $\operatorname{Im}(f - g) = 0$ $\mu$-a.e. Therefore $f = g$ $\mu$-a.e. [/step]

[step:Reduce existence to the case of finite positive $\nu$ and $\mu$]Assume the theorem is proved when both $\nu$ and $\mu$ are finite positive measures. We deduce the general case as follows. **Reduction from complex $\nu$ to positive $\nu$.** Decompose $\nu = \nu_1 + i \nu_2$ into real and imaginary parts (finite real signed measures). Since $\nu$ is a complex measure, it is finite by definition, so each $\nu_j$ is a finite signed measure on the $\sigma$-algebra $\Sigma$. Hence the **Jordan Decomposition Theorem** for finite signed measures applies and gives $\nu_j = \nu_j^+ - \nu_j^-$ as a difference of finite positive measures. (The complex case is identical: it reduces to two finite real signed measures via $\operatorname{Re}$ and $\operatorname{Im}$.) Since $|\nu_j(A)| \le |\nu(A)|$ and $\nu \ll \mu$, each $\nu_j^\pm \ll \mu$ as well: if $\mu(A) = 0$, then $|\nu(A')| = 0$ for all $A' \subset A$ measurable, hence $\nu_j(A') = 0$ for all such $A'$, forcing $\nu_j^+(A) = \nu_j^-(A) = 0$. If we find $f_j^\pm \in L^1(\mu)$ representing $\nu_j^\pm$, then $f := f_1^+ - f_1^- + i(f_2^+ - f_2^-) \in L^1(\mu)$ represents $\nu$ by linearity of the integral. **Reduction from $\sigma$-finite $\mu$ to finite $\mu$ (for positive $\nu$).** Now suppose $\nu$ is finite positive and $\mu$ is $\sigma$-finite. Write $\Omega = \bigsqcup_{n=1}^\infty B_n$ with $B_n \in \Sigma$, $\mu(B_n) < \infty$. Define on each $B_n$ the restricted measures $\mu_n(A) := \mu(A \cap B_n)$ and $\nu_n(A) := \nu(A \cap B_n)$, both finite, with $\nu_n \ll \mu_n$. The finite-finite case yields $f_n \in L^1(\mu_n)$ representing $\nu_n$ on $\Sigma$, with $f_n \ge 0$ supported on $B_n$. Set $f := \sum_n f_n \mathbb{1}_{B_n}$ (well-defined and measurable since the $B_n$ are disjoint and each $f_n \mathbb{1}_{B_n}$ is non-negative measurable). Then for $A \in \Sigma$, applying the **Monotone Convergence Theorem** to the partial sums $S_N := \sum_{n=1}^N f_n \mathbb{1}_{A \cap B_n}$ — these are non-negative, measurable, and increase pointwise to $f \mathbb{1}_A$ — we obtain \begin{align*} \int_A f \, d\mu = \sum_n \int_{A \cap B_n} f_n \, d\mu_n = \sum_n \nu_n(A) = \sum_n \nu(A \cap B_n) = \nu(A), \end{align*} where the last equality uses countable additivity of $\nu$. Similarly $\int_\Omega |f| \, d\mu = \sum_n \int_{B_n} |f_n| \, d\mu_n \le \sum_n \nu_n(\Omega) = \nu(\Omega) < \infty$, so $f \in L^1(\mu)$. It therefore suffices to prove the theorem when $\nu$ and $\mu$ are both finite positive measures. We carry this out in the remaining steps.[/step]

[guided]Assume the theorem is established for finite positive $\nu$ and finite positive $\mu$. We deduce the general case in two reductions: complex $\nu \to$ positive $\nu$, then $\sigma$-finite $\mu \to$ finite $\mu$. **Reduction from complex $\nu$ to positive $\nu$.** A complex measure $\nu : \Sigma \to \mathbb{C}$ takes values in $\mathbb{C}$, but our supremum construction will require non-negative measurable functions. We split $\nu$ into pieces that are positive measures. Decompose $\nu = \nu_1 + i \nu_2$ where $\nu_1(A) := \operatorname{Re}\nu(A)$ and $\nu_2(A) := \operatorname{Im}\nu(A)$. Since $\nu$ is finite (every complex measure is finite by definition), each $\nu_j : \Sigma \to \mathbb{R}$ is a finite real signed measure on the $\sigma$-algebra $\Sigma$. The hypotheses of the **Jordan Decomposition Theorem** — finite signed measure on a $\sigma$-algebra — are therefore satisfied for each $\nu_j$. We apply Jordan decomposition to obtain $\nu_j = \nu_j^+ - \nu_j^-$ where $\nu_j^\pm$ are finite positive measures on $\Sigma$. (The complex case is identical: we have just reduced it to two real signed measures via the real and imaginary part maps.) We verify each $\nu_j^\pm \ll \mu$. Suppose $\mu(A) = 0$. By absolute continuity $\nu \ll \mu$, every measurable $A' \subseteq A$ satisfies $\mu(A') = 0$, hence $\nu(A') = 0$, hence $\nu_j(A') = \operatorname{Re}\nu(A') = 0$ (similarly for $\nu_2$). The Jordan decomposition expresses $\nu_j^+(A)$ and $\nu_j^-(A)$ as variations over measurable subsets of $A$, all of which give value $0$. Hence $\nu_j^\pm(A) = 0$, so $\nu_j^\pm \ll \mu$. The finite-finite case applied to each $\nu_j^\pm$ produces $f_j^\pm \in L^1(\mu)$ with $\nu_j^\pm(A) = \int_A f_j^\pm \, d\mu$ for all $A \in \Sigma$. Define \begin{align*} f := f_1^+ - f_1^- + i(f_2^+ - f_2^-) \in L^1(\mu). \end{align*} By linearity of the integral, $\int_A f \, d\mu = (\nu_1^+ - \nu_1^-)(A) + i(\nu_2^+ - \nu_2^-)(A) = \nu_1(A) + i \nu_2(A) = \nu(A)$ for all $A \in \Sigma$. So $f$ represents $\nu$. **Reduction from $\sigma$-finite $\mu$ to finite $\mu$.** Now assume $\nu$ is finite positive and $\mu$ is $\sigma$-finite. We need to handle the case $\mu(\Omega) = \infty$, which is incompatible with the bound $\alpha \le \nu(\Omega)$ that drives the supremum construction once we factor through $\mu$. The remedy is to partition $\Omega$ into pieces of finite $\mu$-measure. By $\sigma$-finiteness, write $\Omega = \bigsqcup_{n=1}^\infty B_n$ with $B_n \in \Sigma$ pairwise disjoint and $\mu(B_n) < \infty$ (a standard disjointification of any countable cover by finite-measure sets). Define \begin{align*} \mu_n &: \Sigma \to [0, \infty), & \mu_n(A) &:= \mu(A \cap B_n), \\ \nu_n &: \Sigma \to [0, \infty), & \nu_n(A) &:= \nu(A \cap B_n). \end{align*} Both are finite positive measures: $\mu_n(\Omega) = \mu(B_n) < \infty$ and $\nu_n(\Omega) = \nu(B_n) \le \nu(\Omega) < \infty$. We verify $\nu_n \ll \mu_n$: if $\mu_n(A) = \mu(A \cap B_n) = 0$, then $\nu(A \cap B_n) = 0$ by $\nu \ll \mu$, i.e. $\nu_n(A) = 0$. Apply the finite-finite case to each pair $(\mu_n, \nu_n)$ to obtain $f_n \in L^1(\mu_n)$ with $f_n \ge 0$ representing $\nu_n$. Without loss of generality $f_n$ is supported in $B_n$ (replacing $f_n$ by $f_n \mathbb{1}_{B_n}$ does not change any integral $\int_A f_n \, d\mu_n$ because $\mu_n$ is concentrated on $B_n$). Define \begin{align*} f : \Omega &\to [0, \infty), & f(\omega) &:= \sum_{n=1}^\infty f_n(\omega)\, \mathbb{1}_{B_n}(\omega). \end{align*} Since the $B_n$ are disjoint, at each $\omega \in \Omega$ exactly one term is nonzero, so the sum is finite-valued and $f$ is measurable as a pointwise limit of partial sums of non-negative measurable functions. We compute $\int_A f \, d\mu$ via the **Monotone Convergence Theorem**. Define partial sums \begin{align*} S_N : \Omega \to [0, \infty), \quad S_N := \sum_{n=1}^N f_n \mathbb{1}_{A \cap B_n}. \end{align*} Each $S_N$ is non-negative and measurable; $S_N \le S_{N+1}$ pointwise (each new term is non-negative); and $S_N \uparrow f \mathbb{1}_A$ pointwise as $N \to \infty$. The hypotheses of MCT (non-negativity, measurability, monotone increase) are verified, so \begin{align*} \int_\Omega f \mathbb{1}_A \, d\mu = \lim_{N \to \infty} \int_\Omega S_N \, d\mu = \sum_{n=1}^\infty \int_\Omega f_n \mathbb{1}_{A \cap B_n} \, d\mu. \end{align*} Since $f_n$ is supported in $B_n$ and $\mu$ restricted to $B_n$ equals $\mu_n$ on subsets of $B_n$, $\int_\Omega f_n \mathbb{1}_{A \cap B_n} \, d\mu = \int_{A \cap B_n} f_n \, d\mu_n = \nu_n(A) = \nu(A \cap B_n)$. Summing and using countable additivity of $\nu$: \begin{align*} \int_A f \, d\mu = \sum_{n=1}^\infty \nu(A \cap B_n) = \nu\Big(\bigsqcup_n (A \cap B_n)\Big) = \nu(A). \end{align*} Finally, $f \in L^1(\mu)$: by the same MCT argument with $A = \Omega$, \begin{align*} \int_\Omega f \, d\mu = \sum_{n=1}^\infty \int_{B_n} f_n \, d\mu_n = \sum_{n=1}^\infty \nu_n(\Omega) = \sum_{n=1}^\infty \nu(B_n) = \nu(\Omega) < \infty. \end{align*} Both reductions complete, the remaining work is the finite-finite case, treated in the next steps.[/guided]

[step:Set up the family $\mathcal{F}$ and the supremum $\alpha$] Henceforth assume $\nu$ and $\mu$ are finite positive measures with $\nu \ll \mu$. Define \begin{align*} \mathcal{F} := \Big\{h: \Omega \to [0, \infty] \text{ measurable} : \int_A h \, d\mu \le \nu(A) \text{ for all } A \in \Sigma\Big\}, \end{align*} and \begin{align*} \alpha := \sup_{h \in \mathcal{F}} \int_\Omega h \, d\mu. \end{align*} Note $0 \in \mathcal{F}$, so $\mathcal{F} \ne \varnothing$. Taking $A = \Omega$ in the defining inequality gives $\int_\Omega h \, d\mu \le \nu(\Omega) < \infty$, so $\alpha \le \nu(\Omega) < \infty$. **$\mathcal{F}$ is closed under finite maxima.** Given $h_1, h_2 \in \mathcal{F}$ and $A \in \Sigma$, partition $A = A_1 \sqcup A_2$ where $A_1 := A \cap \{h_1 \ge h_2\}$, $A_2 := A \cap \{h_1 < h_2\}$. Then $\max(h_1, h_2) = h_1 \mathbb{1}_{\{h_1 \ge h_2\}} + h_2 \mathbb{1}_{\{h_1 < h_2\}}$, and \begin{align*} \int_A \max(h_1, h_2) \, d\mu = \int_{A_1} h_1 \, d\mu + \int_{A_2} h_2 \, d\mu \le \nu(A_1) + \nu(A_2) = \nu(A). \end{align*} So $\max(h_1, h_2) \in \mathcal{F}$. **$\mathcal{F}$ is closed under monotone increasing limits.** If $g_n \in \mathcal{F}$ and $g_n \uparrow g$ pointwise, then by the monotone convergence theorem applied to $g_n \mathbb{1}_A \uparrow g \mathbb{1}_A$, for every $A \in \Sigma$ \begin{align*} \int_A g \, d\mu = \lim_n \int_A g_n \, d\mu \le \nu(A), \end{align*} so $g \in \mathcal{F}$. [/step]

[step:Construct $f \in \mathcal{F}$ achieving the supremum $\alpha$] By definition of $\alpha$, choose $h_n \in \mathcal{F}$ with $\int_\Omega h_n \, d\mu > \alpha - 1/n$. Set $g_n := \max(h_1, \ldots, h_n)$. Iterating closure of $\mathcal{F}$ under finite maxima, $g_n \in \mathcal{F}$. The sequence $g_n$ is monotone increasing, so let $f := \lim_n g_n = \sup_n g_n$, a measurable $[0, \infty]$-valued function. By closure under monotone increasing limits, $f \in \mathcal{F}$. By monotone convergence, \begin{align*} \int_\Omega f \, d\mu = \lim_n \int_\Omega g_n \, d\mu \ge \lim_n \int_\Omega h_n \, d\mu \ge \lim_n (\alpha - 1/n) = \alpha, \end{align*} where the first inequality uses $g_n \ge h_n$. Combined with $\int_\Omega f \, d\mu \le \alpha$ (from $f \in \mathcal{F}$), we get $\int_\Omega f \, d\mu = \alpha$. Since $\alpha < \infty$, the set $\{f = \infty\}$ has $\mu$-measure $0$ (else $\int f \, d\mu = \infty$). Modify $f$ on this $\mu$-null set to set $f := 0$ there; this preserves $f \in \mathcal{F}$ (changing $f$ on a $\mu$-null set does not affect any integral $\int_A f \, d\mu$) and gives $f: \Omega \to [0, \infty)$ finite-valued. We have $\int_\Omega f \, d\mu = \alpha \le \nu(\Omega) < \infty$, so $f \in L^1(\mu)$. [/step]

[step:Show $f$ represents $\nu$ via Hahn decomposition of auxiliary signed measures]Define the defect measure $\rho: \Sigma \to [0, \infty)$ by \begin{align*} \rho(A) := \nu(A) - \int_A f \, d\mu. \end{align*} This is a finite positive measure (it is positive because $f \in \mathcal{F}$, finite because $\nu(\Omega) < \infty$). To prove $\nu(A) = \int_A f \, d\mu$ it suffices to show $\rho \equiv 0$. Suppose for contradiction $\rho(\Omega) > 0$. Since $\mu(\Omega) < \infty$, choose $n \in \mathbb{N}$ large enough that $\rho(\Omega) > \mu(\Omega)/n$, i.e. $\rho(\Omega) - n^{-1} \mu(\Omega) > 0$. Consider \begin{align*} \sigma_n : \Sigma &\to \mathbb{R}, & \sigma_n(A) &:= \rho(A) - n^{-1} \mu(A) = \nu(A) - \int_A f \, d\mu - n^{-1} \mu(A). \end{align*} We verify $\sigma_n$ is a finite signed measure on $\Sigma$: $\rho$ is a finite positive measure (verified above) and $n^{-1} \mu$ is a finite positive measure (since $\mu(\Omega) < \infty$), so their difference is a real-valued countably additive set function on $\Sigma$ with $|\sigma_n(A)| \le \rho(\Omega) + n^{-1} \mu(\Omega) < \infty$ for all $A$. The hypotheses of the **Hahn Decomposition Theorem** — finite signed measure on a $\sigma$-algebra — are therefore satisfied. The theorem yields $P_n, N_n \in \Sigma$ with $\Omega = P_n \sqcup N_n$, $\sigma_n(A) \ge 0$ for all measurable $A \subseteq P_n$, and $\sigma_n(A) \le 0$ for all measurable $A \subseteq N_n$. By the choice of $n$, \begin{align*} \sigma_n(\Omega) = \sigma_n(P_n) + \sigma_n(N_n) > 0, \end{align*} and $\sigma_n(N_n) \le 0$, so $\sigma_n(P_n) > 0$. Now $\sigma_n(P_n) > 0$ together with $\nu \ll \mu$ implies $\mu(P_n) > 0$: if $\mu(P_n) = 0$, then $\nu(P_n) = 0$ (by absolute continuity), so $\rho(P_n) = -\int_{P_n} f \, d\mu \le 0$ (in fact $= 0$ since $\rho \ge 0$), giving $\sigma_n(P_n) = \rho(P_n) - n^{-1} \mu(P_n) = 0$, a contradiction.[/step]

[guided]We construct a candidate density $f$ in the previous step; we now show $f$ represents $\nu$. Define the **defect measure** \begin{align*} \rho : \Sigma &\to [0, \infty), & \rho(A) &:= \nu(A) - \int_A f \, d\mu. \end{align*} We verify $\rho$ is a finite positive measure. Positivity: $\rho(A) \ge 0$ for all $A \in \Sigma$ because $f \in \mathcal{F}$, which by definition gives $\int_A f \, d\mu \le \nu(A)$. Countable additivity: both $\nu(\cdot)$ and $A \mapsto \int_A f \, d\mu$ are countably additive on $\Sigma$ (the latter by countable additivity of the Lebesgue integral with respect to non-negative integrand $f$), and the difference of two countably additive set functions taking finite values is countably additive. Finiteness: $\rho(\Omega) = \nu(\Omega) - \alpha \le \nu(\Omega) < \infty$. The identity $\nu(A) = \int_A f \, d\mu$ for all $A \in \Sigma$ is equivalent to $\rho \equiv 0$. We argue by contradiction: assume $\rho(\Omega) > 0$. The strategy is to find a "boost set" $P$ on which we can increase $f$ by a small constant $\varepsilon > 0$ — replacing $f$ by $h := f + \varepsilon \mathbb{1}_P$ — while keeping $h \in \mathcal{F}$ and producing $\int_\Omega h \, d\mu > \alpha$, contradicting maximality of $\alpha = \int_\Omega f \, d\mu$. To make this work, we need $\int_A h \, d\mu \le \nu(A)$ on subsets $A \subseteq P$, i.e. $\int_A f \, d\mu + \varepsilon \mu(A) \le \nu(A)$, i.e. $\rho(A) \ge \varepsilon \mu(A)$ on $A \subseteq P$. This says exactly that the signed measure $\rho - \varepsilon \mu$ is non-negative on $P$. For $\varepsilon = 1/n$, the **Hahn Decomposition Theorem** produces precisely such a $P$. Set \begin{align*} \sigma_n : \Sigma &\to \mathbb{R}, & \sigma_n(A) &:= \rho(A) - n^{-1} \mu(A). \end{align*} We verify that $\sigma_n$ is a finite signed measure on the $\sigma$-algebra $\Sigma$. Both $\rho$ and $n^{-1} \mu$ are finite positive measures (with $\rho(\Omega) < \infty$ verified above and $n^{-1} \mu(\Omega) = \mu(\Omega)/n < \infty$ from finiteness of $\mu$ in the reduced setting). Their difference is real-valued, countably additive (both summands are), and bounded: $|\sigma_n(A)| \le \rho(\Omega) + n^{-1} \mu(\Omega) < \infty$ for all $A \in \Sigma$. The hypotheses of the Hahn Decomposition Theorem are met, so it produces $P_n, N_n \in \Sigma$ with \begin{align*} \Omega = P_n \sqcup N_n, \quad \sigma_n(A) \ge 0 \text{ for all measurable } A \subseteq P_n, \quad \sigma_n(A) \le 0 \text{ for all measurable } A \subseteq N_n. \end{align*} We choose $n$ to make $\sigma_n(P_n) > 0$. Since $\rho(\Omega) > 0$ and $\mu(\Omega) < \infty$, pick $n \in \mathbb{N}$ with $n > \mu(\Omega)/\rho(\Omega)$. Then $n^{-1} \mu(\Omega) < \rho(\Omega)$, so \begin{align*} \sigma_n(\Omega) = \rho(\Omega) - n^{-1} \mu(\Omega) > 0. \end{align*} Splitting via the Hahn decomposition: $\sigma_n(\Omega) = \sigma_n(P_n) + \sigma_n(N_n)$. Since $N_n \subseteq N_n$, we have $\sigma_n(N_n) \le 0$. Therefore $\sigma_n(P_n) \ge \sigma_n(\Omega) > 0$. We now show $\mu(P_n) > 0$, which is the crucial place where absolute continuity $\nu \ll \mu$ enters. Suppose for contradiction $\mu(P_n) = 0$. By $\nu \ll \mu$, $\nu(P_n) = 0$. Then \begin{align*} \rho(P_n) = \nu(P_n) - \int_{P_n} f \, d\mu = 0 - \int_{P_n} f \, d\mu = -\int_{P_n} f \, d\mu. \end{align*} Since $\rho \ge 0$ (verified) and $f \ge 0$, both $\rho(P_n) \ge 0$ and $-\int_{P_n} f \, d\mu \le 0$. The only number that is simultaneously $\ge 0$ and $\le 0$ is $0$, so $\rho(P_n) = 0$. Then $\sigma_n(P_n) = \rho(P_n) - n^{-1} \mu(P_n) = 0 - 0 = 0$, contradicting $\sigma_n(P_n) > 0$. Hence $\mu(P_n) > 0$, and $P_n$ is a genuine boost set with strictly positive $\mu$-mass on which the boost $\varepsilon = 1/n$ is admissible. The next step uses this to manufacture the contradiction with maximality of $\alpha$.[/guided]

[step:Derive a contradiction with the maximality of $\alpha$] On the set $P_n$ from the previous step, we have $\sigma_n(A) \ge 0$ for every measurable $A \subseteq P_n$, i.e. \begin{align*} \nu(A) - \int_A f \, d\mu - n^{-1} \mu(A) \ge 0, \end{align*} which rearranges to $\int_A (f + n^{-1} \mathbb{1}_{P_n}) \, d\mu \le \nu(A)$ for $A \subseteq P_n$. Define $h := f + n^{-1} \mathbb{1}_{P_n}: \Omega \to [0, \infty)$. We claim $h \in \mathcal{F}$. For any $A \in \Sigma$, write $A = (A \cap P_n) \sqcup (A \cap N_n)$. On $A \cap N_n$ the indicator $\mathbb{1}_{P_n}$ vanishes, so \begin{align*} \int_{A \cap N_n} h \, d\mu = \int_{A \cap N_n} f \, d\mu \le \nu(A \cap N_n), \end{align*} using $f \in \mathcal{F}$. On $A \cap P_n$, the inequality $\int_{A \cap P_n} h \, d\mu \le \nu(A \cap P_n)$ is the rearrangement of $\sigma_n(A \cap P_n) \ge 0$ from the Hahn decomposition (applicable because $A \cap P_n \subseteq P_n$). Adding, \begin{align*} \int_A h \, d\mu \le \nu(A), \end{align*} so $h \in \mathcal{F}$. But \begin{align*} \int_\Omega h \, d\mu = \int_\Omega f \, d\mu + n^{-1} \mu(P_n) = \alpha + n^{-1} \mu(P_n) > \alpha \end{align*} since $\mu(P_n) > 0$, contradicting the maximality $\alpha = \sup_{g \in \mathcal{F}} \int_\Omega g \, d\mu$. This contradiction shows $\rho(\Omega) = 0$, hence $\rho \equiv 0$ (as $\rho$ is a positive measure with total mass zero), so $\nu(A) = \int_A f \, d\mu$ for all $A \in \Sigma$. [/step]

[step:Verify the value-type assertions for signed and positive $\nu$] Throughout the existence proof for finite positive $\nu$, we constructed $f \ge 0$ pointwise, so $f$ takes non-negative real values when $\nu$ is positive. For signed real $\nu$, the Jordan decomposition gives $\nu = \nu^+ - \nu^-$ with finite positive parts, and the densities $f^+ \ge 0$ representing $\nu^+$ and $f^- \ge 0$ representing $\nu^-$ are real, so $f = f^+ - f^-$ is real-valued. This completes the proof of the Radon-Nikodym theorem. [/step]