[proofplan] We translate the statement from random variables to measures by introducing the law $\mu_X=\mathbb P\circ X^{-1}$ on $(\mathbb R,\mathcal B(\mathbb R))$. The forward implication is exactly the Radon-Nikodym theorem applied to the absolute continuity relation $\mu_X\ll\mathcal L^1$. The reverse implication follows because an integral of a nonnegative measurable function over a Lebesgue-null set is zero. Finally, uniqueness follows by applying the density identity to the sets where two candidate densities differ. [/proofplan]

[step:Express the distribution of $X$ as a measure on the Borel line] Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space on which $X$ is defined. Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, and let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $(\mathbb R,\mathcal B(\mathbb R))$. Define the law of $X$ as the pushforward measure \begin{align*} \mu_X:\mathcal B(\mathbb R)&\to[0,1] \\ A&\mapsto \mathbb P(X^{-1}(A)). \end{align*} Since $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable, $X^{-1}(A)\in\mathcal F$ for every $A\in\mathcal B(\mathbb R)$, so $\mu_X$ is a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$. For every $A\in\mathcal B(\mathbb R)$, the event $\{X\in A\}$ is $X^{-1}(A)$, hence \begin{align*} \mu_X(A)=\mathbb P(X\in A). \end{align*} By the definition of absolute continuity for a real-valued random variable, $X$ is absolutely continuous precisely when $\mu_X\ll\mathcal L^1$. [/step]

[step:Apply Radon-Nikodym to obtain a density from absolute continuity]Assume that $X$ is absolutely continuous. Then $\mu_X\ll\mathcal L^1$. The measure $\mu_X$ is a finite measure because $\mu_X(\mathbb R)=1$, and $\mathcal L^1$ is $\sigma$-finite on $\mathbb R$ because \begin{align*} \mathbb R=\bigcup_{m\in\mathbb N}[-m,m], \qquad \mathbb N:=\{1,2,3,\ldots\}, \qquad \mathcal L^1([-m,m])=2m<\infty. \end{align*} Therefore the hypotheses of the [Radon-Nikodym Theorem](/theorems/???) apply to the measures $\mu_X$ and $\mathcal L^1$ on $(\mathbb R,\mathcal B(\mathbb R))$. Hence there exists a nonnegative Borel measurable function \begin{align*} f_X:(\mathbb R,\mathcal B(\mathbb R))&\to([0,\infty],\mathcal B([0,\infty])) \end{align*} such that, for every $A\in\mathcal B(\mathbb R)$, \begin{align*} \mu_X(A)=\int_A f_X(x)\,d\mathcal L^1(x). \end{align*} Using $\mu_X(A)=\mathbb P(X\in A)$ gives \begin{align*} \mathbb P(X\in A)=\int_A f_X(x)\,d\mathcal L^1(x) \end{align*} for every $A\in\mathcal B(\mathbb R)$.[/step]

[guided]Assume that $X$ is absolutely continuous. By definition, this means that the law $\mu_X$ of $X$ is absolutely continuous with respect to Lebesgue measure: \begin{align*} \mu_X\ll\mathcal L^1. \end{align*} We want to convert this measure-theoretic absolute continuity into integration against a function. The precise tool for this conversion is the [Radon-Nikodym Theorem](/theorems/???). We verify its hypotheses. The measure $\mu_X$ is finite because it is a probability measure: \begin{align*} \mu_X(\mathbb R)=\mathbb P(X\in\mathbb R)=1. \end{align*} The reference measure $\mathcal L^1$ is $\sigma$-finite on $\mathbb R$. Indeed, with $\mathbb N:=\{1,2,3,\ldots\}$, \begin{align*} \mathbb R=\bigcup_{m\in\mathbb N}[-m,m], \qquad \mathcal L^1([-m,m])=2m<\infty. \end{align*} Thus the Radon-Nikodym theorem applies to the measure space $(\mathbb R,\mathcal B(\mathbb R))$, the finite measure $\mu_X$, and the $\sigma$-finite measure $\mathcal L^1$. The theorem gives a nonnegative Borel measurable function \begin{align*} f_X:(\mathbb R,\mathcal B(\mathbb R))&\to([0,\infty],\mathcal B([0,\infty])) \end{align*} such that \begin{align*} \mu_X(A)=\int_A f_X(x)\,d\mathcal L^1(x) \end{align*} for every $A\in\mathcal B(\mathbb R)$. Since $\mu_X(A)=\mathbb P(X\in A)$ by the definition of the law of $X$, this becomes \begin{align*} \mathbb P(X\in A)=\int_A f_X(x)\,d\mathcal L^1(x) \end{align*} for every $A\in\mathcal B(\mathbb R)$. This is exactly the density representation required in the theorem.[/guided]

[step:Use the density formula to recover absolute continuity] Conversely, suppose there exists a nonnegative Borel measurable function \begin{align*} f_X:(\mathbb R,\mathcal B(\mathbb R))&\to([0,\infty],\mathcal B([0,\infty])) \end{align*} such that \begin{align*} \mathbb P(X\in A)=\int_A f_X(x)\,d\mathcal L^1(x) \end{align*} for every $A\in\mathcal B(\mathbb R)$. Let $N\in\mathcal B(\mathbb R)$ satisfy $\mathcal L^1(N)=0$. Since $f_X$ is nonnegative and Borel measurable, the defining property of the Lebesgue integral over a null set gives \begin{align*} \int_N f_X(x)\,d\mathcal L^1(x)=0. \end{align*} Therefore \begin{align*} \mu_X(N)=\mathbb P(X\in N)=\int_N f_X(x)\,d\mathcal L^1(x)=0. \end{align*} Thus $\mathcal L^1(N)=0$ implies $\mu_X(N)=0$, so $\mu_X\ll\mathcal L^1$. Hence $X$ is absolutely continuous. [/step]

[step:Compare two candidate densities on the sets where they differ] Suppose \begin{align*} f:(\mathbb R,\mathcal B(\mathbb R))&\to([0,\infty],\mathcal B([0,\infty])),\\ g:(\mathbb R,\mathcal B(\mathbb R))&\to([0,\infty],\mathcal B([0,\infty])) \end{align*} are nonnegative Borel measurable functions such that, for every $A\in\mathcal B(\mathbb R)$, \begin{align*} \mu_X(A)=\int_A f(x)\,d\mathcal L^1(x) =\int_A g(x)\,d\mathcal L^1(x). \end{align*} Taking $A=\mathbb R$ gives \begin{align*} \int_{\mathbb R} f(x)\,d\mathcal L^1(x) = \int_{\mathbb R} g(x)\,d\mathcal L^1(x) = \mu_X(\mathbb R)=1, \end{align*} so both integrals are finite. Define the Borel sets \begin{align*} E_+&:=\{x\in\mathbb R:f(x)>g(x)\},\\ E_-&:=\{x\in\mathbb R:g(x)>f(x)\}. \end{align*} We prove that $\mathcal L^1(E_+)=0$. For each $m\in\mathbb N$, define \begin{align*} E_{+,m}:=\{x\in\mathbb R:f(x)\ge g(x)+m^{-1}\}. \end{align*} Then $E_{+,m}\in\mathcal B(\mathbb R)$ and \begin{align*} E_+=\bigcup_{m\in\mathbb N}E_{+,m}. \end{align*} For each $m\in\mathbb N$, the pointwise inequality on $E_{+,m}$ gives \begin{align*} \int_{E_{+,m}} f(x)\,d\mathcal L^1(x) \ge \int_{E_{+,m}} g(x)\,d\mathcal L^1(x) + m^{-1}\mathcal L^1(E_{+,m}). \end{align*} But the density identities applied to $A=E_{+,m}$ give \begin{align*} \int_{E_{+,m}} f(x)\,d\mathcal L^1(x) = \int_{E_{+,m}} g(x)\,d\mathcal L^1(x). \end{align*} Since both integrals are finite, subtraction yields \begin{align*} m^{-1}\mathcal L^1(E_{+,m})=0, \end{align*} and therefore $\mathcal L^1(E_{+,m})=0$. Countable subadditivity of $\mathcal L^1$ gives \begin{align*} \mathcal L^1(E_+) \le \sum_{m\in\mathbb N}\mathcal L^1(E_{+,m}) =0. \end{align*} The same argument with $f$ and $g$ interchanged gives $\mathcal L^1(E_-)=0$. Hence \begin{align*} \{x\in\mathbb R:f(x)\ne g(x)\}=E_+\cup E_- \end{align*} has $\mathcal L^1$-measure zero. Thus $f=g$ $\mathcal L^1$-a.e., proving uniqueness of the density up to equality $\mathcal L^1$-a.e. [/step]