Let $P$ and $Q$ denote finite measures. Let $\mu$ denote a finite measure that dominates them both, and $p, q$ the Radon-Nikodym derivatives of $P, Q$, respectively, with respect to $\mu$.
The Lebesgue decomposition of $Q$ is that $Q = Q_a + Q_s$ where $$ Q_a(E) = Q(E \cap \{p > 0\}) \quad \mbox{and} \quad Q_s(E) = Q(E \cap \{p = 0\}). $$
It is then claimed that $Q_a \ll P$.
My reasoning for this is as follows. Let $E$ be such that $P(E) = 0$. Fix $\delta > 0$ and define the sequence of sets $$ E_0 = E \cap \{p > \delta\}, \quad \mbox{and} \quad E_j = E \cap \{2^{-j} \delta < p \leq 2^{-j + 1}\delta\}, $$ for $j \geq 1$. Evidently $E \cap \{p > 0\} = \cup_{j \geq 0} E_j$, and this is a disjoint union. Therefore if $P(E) = 0$ then $P(E_j) = 0$ for all $j$, so $$ 0 = P(E_j) = \int_{E_j} p \, d \mu \geq 2^{-j} \delta \,\mu(E_j). $$ Thus $\mu(E_j) = 0$ and hence $Q_a(E_j) = 0$. Since $Q_a(E) = \sum_{j \geq 0} Q_a(E_j)$, we have $Q_a(E) = 0$, which proves the claim.
My question: Is it necessary to pass through this disjoint union argument (or something like it)? Is there a more direct proof that $Q_a \ll P$?
I think there should be actually. Let
$r(x) = \frac{q(x)}{p(x)}$ whenever $p(x)>0$ and $0$ otherwise.
Then, $1_{p>0}\cdot q = r\cdot p$. Recall that $dP = p d\mu$ and therefore, $$r\cdot dP = rp~d\mu = 1_{p>0}\cdot q~d\mu = 1_{p>0} dQ = dQ_a.$$
We deduce that $Q_a$ is absolutely continuous with respect to $P$.
To see that $r$ is measurable, note that $\{x : p(x)=0\}$ is measurable, which deals with the $p=0$ case, and when $p(x)>0$ note that $\frac{1}{x}$ is continuous on $\mathbb{R}/\{0\}$ so the composition with $p$ is measurable.