Proof, [Exercise 6.18, Fol99]^[1]. (Finite + Positive): First suppose that $\mu$ is finite and $\nu$ is positive. Let $\lambda = \mu + \nu$, then the mapping

is a continuous linear functional on $L^{2}(X, \cm, \lambda; K)$. By the Riesz Representation Theorem, there exists a unique $g \in L^{2}(X, \cm, \lambda; K)$ such that for every $f \in L^{2}(X, \cm, \mu; K)$, $\int f d\nu = \int f g d\lambda$. By uniqueness of the Riesz representation, $\int f d\mu = \int f(1 - g)d\lambda$ for all $f \in L^{2}(X, \cm, \lambda; K)$.

Since $0 \le \int f d\nu \le \norm{f}_{L^1(X, \cm, \lambda; K)}$ for all $f \in L^{2}(X, \cm, \lambda; K) \cap L^{+}(X, \cm)$, $0 \le g \le 1$ $\lambda$-almost everywhere. Let $A = \bracs{g < 1}$, $B = \bracs{g = 1}$,

and

then

1. (1)

   Since $X = A \sqcup B$, $\nu = \nu_{a} + \nu_{s}$.

2. (2)

   Let $E \in \cm$ with $\mu(E) = 0$, then $(1 - g)|_{E} = 0$ $\lambda$-almost everywhere. Thus $E \subset B$ modulo a $\lambda$-null set, and

   \[\nu_{a}(E) = \nu(E \cap A) \le \lambda(A \cap B) = 0\]

   
   

   so $\nu_{a} \ll \mu$.

3. (3)

   Since $\mu(B) = 0$ and $\nu_{s}(A) = 0$, $\mu \perp \nu_{s}$.

Let

then for each $E \in \cm$,

so $\one_{A} g/(1 - g)$ is the desired Radon-Nikodym derivative.

($\sigma$-Finite + Positive): Now suppose that $\mu$ is $\sigma$-finite and $\nu$ is still positive. Let $\seq{E_n}\subset \cm$ be a pairwise disjoint sequence such that $\bigsqcup_{n \in \natp}E_{n} = X$ and $\mu(E_{n}) < \infty$ for all $n \in \natp$. For each $n \in \natp$ and $E \in \cm$, let $\mu_{n}(E) = \mu(E \cap E_{n})$ and $\nu_{n}(E) = \nu(E \cap E_{n})$, then there exists $g_{n} \in L^{1}(X, \cm, \mu)$ and a positive measure $\nu_{s}^{(n)}$ such that

1. (1)

   $\nu_{n}(dx) = g_{n}\mu(dx) + \nu_{s}^{(n)}(dx)$.

2. (3)

   $\nu_{s}^{(n)}$ is mutually singular with $\mu$.

Let $g = \sum_{n \in \natp}g_{n}$ and $\nu_{s} = \sum_{n \in \natp}\nu_{s}^{(n)}$, then by the Monotone Convergence Theorem, completeness of $L^{1}(X, \cm, \mu)$, and completeness of $M(X, \cm; \real)$, $g \in L^{1}(X, \cm, \mu)$ and $\nu_{s}$ is a measure with

1. (1)

   $\nu(dx) = g\mu(dx) + \nu_{s}$.

2. (3)

   $\nu_{s}$ is mutually singular with $\mu$.

(Hilbert): Finally, suppose that $\nu$ is an $H$-valued finite vector measure, then the mapping

is a continuous linear functional. By the Riesz Representation Theorem, there exists a unique $g \in L^{2}(X, \cm, |\nu|; H)$ such that for every $f \in L^{2}(X, \cm, |\nu|; H)$, $\int f d\nu = \int fg d|\nu|$. For each $E \in \cm$, let $\nu_{a}(E) = \int_{E} g d|\nu|_{a}$ and $\nu_{s}(E) = \int_{E} g d|\nu|_{s}$, then

1. (1)

   $\nu(E) = \int_{E} g d|\nu| = \int_{E} g d|\nu|_{a} + \int_{E} g d|\nu|_{s} = \nu_{a}(E) + \nu_{s}(E)$.

2. (2)

   Since $|\nu|_{a}$ is absolutely continuous with respect to $\mu$, so is $\nu_{a}$.

3. (3)

   Since $|\nu|_{s}$ is mutually singular with $\mu$, so is $\nu_{s}$.

and

so $\frac{d\nu_{a}}{d\mu}= g \cdot \frac{d|\nu|_{a}}{d\mu}$ is the desired Radon-Nikodym derivative.

(Uniqueness): For any decomposition $\nu = \rho_{a} + \rho_{s}$ satisfying the above, then $\rho_{a} - \nu_{a} = \nu_{s} - \rho_{s}$ with $\rho_{a} - \nu_{a} \perp \nu_{s} - \rho_{s}$. Therefore $\rho_{a} = \nu_{a}$, $\rho_{s} = \nu_{s}$, and the decomposition is unique.$\square$