Theorem 23.6.2 (Lebesgue-Radon-Nikodym (Localisable)).label Let $(X, \cm)$ be a measurable space, $\mu, \nu: \cm \to [0, \infty]$ be positive measures, and $\cf \subset \cm$ be an ideal such that:
- (a)
$(X, \cm, \mu + \nu)$ and $(X, \cm, \mu)$ are localisable.
- (b)
$\cf$ is a scaffold for both $\mu$ and $\nu$.
then there exists a unique pair of positive measures $\nu_{a}, \nu_{s}: \cm \to [0, \infty]$ such that:
- (1)
$\nu = \nu_{a} + \nu_{s}$.
- (2)
$\nu_{a}$ is absolutely continuous with respect to $\mu$.
- (3)
$\nu_{s}$ is mutually singular with $\mu$.
- (4)
$\cf$ is a scaffold for $\nu_{a}$ and $\nu_{s}$.
The pair $(\nu_{a}, \nu_{s})$ is the Lebesgue decomposition of $\nu$ with respect to $\mu$. Moreover, there exists a unique $\frac{d\nu_{a}}{d\mu}\in L^{+}(X, \cm, \mu)$ such that for every $A \in \cm$,
If $\nu \ll \mu$, then $\nu_{s} = 0$ and $\nu(dx) = \frac{d\nu_{a}}{d\mu}\mu(dx) = \frac{d\nu_{a}}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the Radon-Nikodym derivative of $\nu$ with respect to $\mu$.
Proof. For each $A \in \cf$ and $E \in \cm$, let
then $\mu^{A}$ and $\nu^{A}$ are finite measures on $A$. By the finite case, there exists an a.e. unique $f_{A} \in L^{+}(A, \mu)$ and $\nu_{s}^{A}: \cm \to [0, \infty]$ such that:
- (1)
$d\nu^{A} = f_{A}\mu^{A} + \nu_{s}^{A} = f_{A}\mu + \nu_{s}^{A}$.
- (2)
$\nu_{s}^{A}$ is mutually singular with $\mu$.
The uniqueness given by the finite case implies that for any $A, B \in \cf$, $f_{A}|_{A \cap B}= f_{B}|_{A \cap B}$ $\mu$-almost everywhere. By the gluing lemma for measurable functions, there exists $f \in L^{+}(X, \mu)$ such that $f|_{A} = f_{A}$ $\mu$-almost everywhere for all $A \in \cf$.
By the gluing lemma for measures,
is a measure.
(4): By definition, $\cf$ is a scaffold for $\nu_{s}$. By Lemma 22.4.5, it is also a scaffold for $f d\mu$.
(1): Let $E \in \cm$, then since $\cf$ is a scaffold for $\nu$,
As $\cf$ is an ideal, for any $A, B \in \cf$, $A \cup B \in \cf$, and
Thus the sum and the supremum may be interchanged, so
Now, since $\cf$ is a scaffold for $f\mu$,
By definition of $\nu_{s}$,
Therefore $\nu(E) = \int_{E} f d\mu + \nu_{s}(E)$, so $\nu(dx) = f(x)\mu(dx) + \nu_{s}(dx)$.
(2): $\nu_{a}(dx) = f(x)\mu(dx)$.
(3): For each $A \in \cf$, $f_{A} d\mu \perp \nu_{s}^{A}$, so there exists $E_{A}, F_{A} \in \cm$ such that:
- (i)
$A = E_{A} \sqcup F_{A}$.
- (ii)
$\mu(F_{A}) = 0$ and $\nu_{s}^{A}(E_{A}) = 0$.
Let $F$ be an essential supremum of $\bracs{F_A}_{A \in \cf}$ with respect to $(\mu + \nu)$. By Lemma 22.6.6, for each $A \in \cf$,
Since $\cf$ is a scaffold for $\mu$, $\mu(F) = 0$. On the other hand, for each $A \in \cf$,
By definition of $\nu_{s}$, $\nu(F^{c}) = 0$. Therefore $\mu \perp \nu_{s}$.
(Uniqueness): By uniqueness of the finite case, the restriction of $(\nu_{a}, \nu_{s})$ to each set in $\cf$ is unique. Since $\cf$ is a scaffold for both measures, they are uniquely determined by their restrictions to each set in $\cf$.$\square$
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