[proofplan] We use the Radon-Nikodym representations of both $\mu$ with respect to $\nu$ and $\nu$ with respect to $\mu$, then apply the Chain Rule for Radon-Nikodym derivatives to the identity measure $\mu = \mu$. The chain $\mu \ll \nu \ll \mu$ yields $\frac{d\mu}{d\nu} \cdot \frac{d\nu}{d\mu} = 1$ $\mu$-a.e., from which positivity and the reciprocal identity follow. [/proofplan] [step:Apply the chain rule to the identity $\mu \ll \nu \ll \mu$] Since $\mu \ll \nu$ and $\nu \ll \mu$, the [Chain Rule for Radon-Nikodym Derivatives](/theorems/1208) applied to the chain $\mu \ll \nu \ll \mu$ gives: \begin{align*} \frac{d\mu}{d\mu} = \frac{d\mu}{d\nu} \cdot \frac{d\nu}{d\mu} \quad \mu\text{-a.e.} \end{align*} [guided] The Chain Rule (Theorem 1208) requires three $\sigma$-finite measures $\lambda \ll \nu \ll \mu$. Here we take $\lambda = \mu$, the first measure in the chain. The hypotheses are satisfied: $\mu \ll \nu$ by assumption, and $\nu \ll \mu$ by assumption. Both measures are $\sigma$-finite by hypothesis. The conclusion gives $\frac{d\mu}{d\mu} = \frac{d\mu}{d\nu} \cdot \frac{d\nu}{d\mu}$ $\mu$-a.e. [/guided] [/step] [step:Identify $\frac{d\mu}{d\mu} = 1$ $\mu$-a.e.] The Radon-Nikodym derivative of $\mu$ with respect to itself is the constant function $1$. To verify: for every $A \in \mathcal{A}$, \begin{align*} \int_A 1 \, d\mu = \mu(A), \end{align*} which is the defining identity for $\frac{d\mu}{d\mu}$. By the uniqueness clause of the [Radon-Nikodym Theorem](/theorems/1247), $\frac{d\mu}{d\mu} = 1$ $\mu$-a.e. [/step] [step:Deduce that $\frac{d\nu}{d\mu} > 0$ $\mu$-a.e.] Write $f := \frac{d\nu}{d\mu}$ and $g := \frac{d\mu}{d\nu}$. From the preceding two steps: \begin{align*} g(x) \cdot f(x) = 1 \quad \text{for } \mu\text{-a.e. } x \in X. \end{align*} Since $f \ge 0$ and $g \ge 0$ (both are Radon-Nikodym derivatives of nonneg measures), the product $gf = 1 > 0$ forces $f(x) > 0$ for $\mu$-a.e. $x \in X$. (If $f(x) = 0$ on a set $E$ with $\mu(E) > 0$, then $g(x) f(x) = 0$ on $E$, contradicting $gf = 1$ $\mu$-a.e.) [guided] This is where the mutual absolute continuity is consumed. If we only had $\nu \ll \mu$ (without $\mu \ll \nu$), then $f = \frac{d\nu}{d\mu}$ could vanish on a set of positive $\mu$-measure — the density $f$ would simply be zero on regions where $\nu$ assigns no mass. Mutual absolute continuity forces $f > 0$ $\mu$-a.e., because the set $\{f = 0\}$ would be a $\nu$-null set (since $\nu(\{f = 0\}) = \int_{\{f=0\}} f \, d\mu = 0$), and $\mu \ll \nu$ would then require $\mu(\{f = 0\}) = 0$. The chain rule argument provides an alternative and more direct proof of positivity: $gf = 1$ $\mu$-a.e. immediately excludes the possibility $f = 0$ on any set of positive $\mu$-measure. [/guided] [/step] [step:Conclude the reciprocal identity] Since $f(x) > 0$ for $\mu$-a.e. $x$, the reciprocal $f(x)^{-1}$ is well-defined $\mu$-a.e. The identity $g \cdot f = 1$ $\mu$-a.e. gives: \begin{align*} g = f^{-1} \quad \mu\text{-a.e.}, \end{align*} that is, \begin{align*} \frac{d\mu}{d\nu} = \left(\frac{d\nu}{d\mu}\right)^{-1} \quad \mu\text{-a.e.} \end{align*} [/step]