[proofplan] We first verify that $\lambda \ll \mu$ follows from the transitivity of absolute continuity. We then show that the product $\frac{d\lambda}{d\nu} \cdot \frac{d\nu}{d\mu}$ serves as a Radon-Nikodym derivative of $\lambda$ with respect to $\mu$ by verifying the integral identity on all measurable sets, using the change-of-density formula for integration. Uniqueness of the Radon-Nikodym derivative completes the argument. [/proofplan]

[step:Verify that $\lambda \ll \mu$]Let $A \in \mathcal{A}$ with $\mu(A) = 0$. Since $\nu \ll \mu$, we have $\nu(A) = 0$. Since $\lambda \ll \nu$, we have $\lambda(A) = 0$. Therefore $\lambda \ll \mu$.[/step]

[guided]Absolute continuity is transitive: if every $\mu$-null set is $\nu$-null, and every $\nu$-null set is $\lambda$-null, then every $\mu$-null set is $\lambda$-null. This is a purely set-theoretic observation about the containment of null-set families, requiring no measure-theoretic machinery.[/guided]

[step:Write the Radon-Nikodym representations and define the candidate density] Since $\lambda \ll \nu$ and both are $\sigma$-finite, the [Radon-Nikodym Theorem](/theorems/1247) provides an $\mathcal{A}$-measurable function: \begin{align*} f: X &\to [0, \infty), \quad f = \frac{d\lambda}{d\nu}, \end{align*} satisfying $\lambda(A) = \int_A f \, d\nu$ for every $A \in \mathcal{A}$. Similarly, since $\nu \ll \mu$ and both are $\sigma$-finite, there exists an $\mathcal{A}$-measurable function: \begin{align*} g: X &\to [0, \infty), \quad g = \frac{d\nu}{d\mu}, \end{align*} satisfying $\nu(A) = \int_A g \, d\mu$ for every $A \in \mathcal{A}$. We claim that the product $h := f \cdot g: X \to [0, \infty)$ is a Radon-Nikodym derivative of $\lambda$ with respect to $\mu$. [/step]

[step:Verify the integral identity $\lambda(A) = \int_A fg \, d\mu$ for indicator functions, then extend]We first establish the change-of-density formula: for any nonneg $\mathcal{A}$-measurable function $\varphi: X \to [0, \infty]$, \begin{align*} \int_X \varphi \, d\nu = \int_X \varphi \, g \, d\mu. \end{align*} [claim:Change-of-density formula] If $\nu(A) = \int_A g \, d\mu$ for every $A \in \mathcal{A}$ and $\varphi: X \to [0, \infty]$ is $\mathcal{A}$-measurable, then $\int_X \varphi \, d\nu = \int_X \varphi \, g \, d\mu$. [/claim] [proof] For an indicator function $\varphi = \mathbb{1}_A$ with $A \in \mathcal{A}$, the identity holds by definition: \begin{align*} \int_X \mathbb{1}_A \, d\nu = \nu(A) = \int_A g \, d\mu = \int_X \mathbb{1}_A \, g \, d\mu. \end{align*} By linearity, the identity extends to nonneg simple functions: if $\varphi = \sum_{k=1}^m c_k \mathbb{1}_{A_k}$ with $c_k \ge 0$ and $A_k \in \mathcal{A}$, then: \begin{align*} \int_X \varphi \, d\nu = \sum_{k=1}^m c_k \nu(A_k) = \sum_{k=1}^m c_k \int_{A_k} g \, d\mu = \int_X \varphi \, g \, d\mu. \end{align*} For a general nonneg measurable $\varphi$, choose a sequence $(\varphi_k)_{k=1}^\infty$ of nonneg simple functions with $\varphi_k \uparrow \varphi$ pointwise. By the [Monotone Convergence Theorem](/theorems/509) applied to the measure $\nu$: \begin{align*} \int_X \varphi \, d\nu = \lim_{k \to \infty} \int_X \varphi_k \, d\nu = \lim_{k \to \infty} \int_X \varphi_k \, g \, d\mu. \end{align*} Since $\varphi_k g \uparrow \varphi g$ pointwise (as $g \ge 0$), the [Monotone Convergence Theorem](/theorems/509) applied to the measure $\mu$ gives: \begin{align*} \lim_{k \to \infty} \int_X \varphi_k \, g \, d\mu = \int_X \varphi \, g \, d\mu. \end{align*} [/proof] Now apply the change-of-density formula with $\varphi = f \cdot \mathbb{1}_A$ for any $A \in \mathcal{A}$. Since $f \cdot \mathbb{1}_A \ge 0$ is $\mathcal{A}$-measurable: \begin{align*} \lambda(A) = \int_A f \, d\nu = \int_X f \cdot \mathbb{1}_A \, d\nu = \int_X f \cdot \mathbb{1}_A \cdot g \, d\mu = \int_A f \, g \, d\mu. \end{align*}[/step]

[guided]The heart of the proof is the change-of-density formula. The idea is that integrating against the measure $\nu$ is the same as integrating against the weighted measure $g \, d\mu$, since $g = \frac{d\nu}{d\mu}$. This is immediate for indicator functions (by definition of the Radon-Nikodym derivative), extends to simple functions by linearity, and passes to general nonneg measurable functions by monotone approximation. The [Monotone Convergence Theorem](/theorems/509) is invoked twice: once for the measure $\nu$ (to pass the limit through $\int \cdot \, d\nu$) and once for the measure $\mu$ (to pass the limit through $\int \cdot \, g \, d\mu$). In both cases, the hypotheses are satisfied because $\varphi_k$ is a nondecreasing sequence of nonneg measurable functions converging pointwise to $\varphi$, and $g \ge 0$ ensures that $\varphi_k g$ is also nondecreasing. Once the change-of-density formula is established, the chain rule follows by a single application: $\lambda(A) = \int_A f \, d\nu = \int_A fg \, d\mu$, which says that $fg$ represents $\lambda$ with respect to $\mu$.[/guided]

[step:Conclude by uniqueness] The function $h = fg = \frac{d\lambda}{d\nu} \cdot \frac{d\nu}{d\mu}$ is $\mathcal{A}$-measurable, nonneg, and satisfies $\lambda(A) = \int_A h \, d\mu$ for every $A \in \mathcal{A}$. By the uniqueness clause of the [Radon-Nikodym Theorem](/theorems/1247), the Radon-Nikodym derivative $\frac{d\lambda}{d\mu}$ is $\mu$-a.e. unique, so: \begin{align*} \frac{d\lambda}{d\mu} = \frac{d\lambda}{d\nu} \cdot \frac{d\nu}{d\mu} \quad \mu\text{-a.e.} \end{align*} [/step]