[proofplan] Both assertions follow directly from the definition of absolute continuity and the uniqueness clause of the Radon-Nikodym Theorem. We verify that the measure $\alpha \nu_1 + \beta \nu_2$ is absolutely continuous with respect to $\mu$, exhibit the candidate density $\alpha \frac{d\nu_1}{d\mu} + \beta \frac{d\nu_2}{d\mu}$, verify that it represents the measure, and invoke uniqueness. [/proofplan] [step:Verify that $\alpha \nu_1 + \beta \nu_2 \ll \mu$] Let $A \in \mathcal{A}$ satisfy $\mu(A) = 0$. Since $\nu_1 \ll \mu$, we have $\nu_1(A) = 0$. Since $\nu_2 \ll \mu$, we have $\nu_2(A) = 0$. Therefore: \begin{align*} (\alpha \nu_1 + \beta \nu_2)(A) = \alpha \nu_1(A) + \beta \nu_2(A) = 0. \end{align*} This holds for every $\mu$-null set $A$, so $\alpha \nu_1 + \beta \nu_2 \ll \mu$. [/step] [step:Write the Radon-Nikodym representations of $\nu_1$ and $\nu_2$] Since $\nu_1 \ll \mu$ and both measures are $\sigma$-finite, the [Radon-Nikodym Theorem](/theorems/1247) provides an $\mathcal{A}$-measurable function $f_1: X \to [0,\infty)$ with: \begin{align*} \nu_1(A) = \int_A f_1 \, d\mu \quad \text{for every } A \in \mathcal{A}. \end{align*} Similarly, there exists an $\mathcal{A}$-measurable function $f_2: X \to [0,\infty)$ with: \begin{align*} \nu_2(A) = \int_A f_2 \, d\mu \quad \text{for every } A \in \mathcal{A}. \end{align*} We write $f_1 = \frac{d\nu_1}{d\mu}$ and $f_2 = \frac{d\nu_2}{d\mu}$. [/step] [step:Show that $\alpha f_1 + \beta f_2$ is a Radon-Nikodym derivative of $\alpha \nu_1 + \beta \nu_2$] Define $h: X \to [0,\infty)$ by $h := \alpha f_1 + \beta f_2$. Since $\alpha, \beta \ge 0$ and $f_1, f_2 \ge 0$ are $\mathcal{A}$-measurable, the function $h$ is $\mathcal{A}$-measurable and nonnegative. For every $A \in \mathcal{A}$, linearity of the integral gives: \begin{align*} \int_A h \, d\mu &= \int_A (\alpha f_1 + \beta f_2) \, d\mu = \alpha \int_A f_1 \, d\mu + \beta \int_A f_2 \, d\mu \\ &= \alpha \nu_1(A) + \beta \nu_2(A) = (\alpha \nu_1 + \beta \nu_2)(A). \end{align*} [guided] The linearity of the Lebesgue integral is the key tool here. For nonneg functions and nonneg scalars, we have $\int_A (\alpha f_1 + \beta f_2) \, d\mu = \alpha \int_A f_1 \, d\mu + \beta \int_A f_2 \, d\mu$. This holds without any integrability restriction when $\alpha, \beta \ge 0$ and $f_1, f_2 \ge 0$, since all terms are nonneg (possibly $+\infty$). In particular, we do not need to verify that $h$ is integrable — the identity holds in $[0, \infty]$. However, since $\alpha \nu_1 + \beta \nu_2$ is a ($\sigma$-finite) measure, the integral $\int_A h \, d\mu$ must be finite for sets $A$ of finite $\mu$-measure within an exhaustion, which is automatic from the $\sigma$-finiteness of $\nu_1$ and $\nu_2$. [/guided] [/step] [step:Conclude by uniqueness of the Radon-Nikodym derivative] The function $h = \alpha f_1 + \beta f_2$ satisfies $(\alpha \nu_1 + \beta \nu_2)(A) = \int_A h \, d\mu$ for every $A \in \mathcal{A}$. By the uniqueness clause of the [Radon-Nikodym Theorem](/theorems/1247), any two densities for $\alpha \nu_1 + \beta \nu_2$ with respect to $\mu$ agree $\mu$-a.e. Therefore: \begin{align*} \frac{d(\alpha \nu_1 + \beta \nu_2)}{d\mu} = \alpha \frac{d\nu_1}{d\mu} + \beta \frac{d\nu_2}{d\mu} \quad \mu\text{-a.e.} \end{align*} [/step]