$\textbf{Theorem 4.3}$ Suppose $\mu$ is a $\sigma$-finite positive measure on the measure space $(X, \mathcal{M})$ and $\nu$ a $\sigma$-finite signed measure on $\mathcal{M}$. Then there exist unique signed measures $\nu_a$ and $\nu_s$ on $\mathcal{M}$ such that $\nu_a \ll \mu$, $\nu_s \perp \mu$, and $\nu = \nu_a + \nu_s$. In addition, the measure $\nu_a$ takes the form $d\nu_a = f \, d\mu$; that is,
$ \nu_a(E) = \int_E f(x) \, d\mu(x) $
for some extended $\mu$-integrable function $f$.
The proof is easy for the case of finite $\mu$,$\nu$ and it can be proven that f is in $L^{1}(u)$.(Although the proof is extensive to be copied here).If μ and ν are σ-finite and positive, we may clearly find sets $ E_j \in \mathcal{M} $ such that $ X = \bigcup E_j $ and $ \mu(E_j) < \infty, \nu(E_j) < \infty $ for all $ j $. We may define positive and finite measures on $ \mathcal{M} $ by $ \mu_j(E) = \mu(E \cap E_j) \quad \text{and} \quad \nu_j(E) = \nu(E \cap E_j), $ and then we can write for each $ j $, $ \nu_j = \nu_{j,a} + \nu_{j,s} $ where $ \nu_{j,s} \perp \mu_j $ and $ \nu_{j,a} = f_j \, d\mu_j. $ Then it suffices to set $ f = \sum f_j, \quad \nu_s = \sum \nu_{j,s}, \quad \text{and} \quad \nu_a = \sum \nu_{j,a}. $ but the book does not mention that f is in $L^{1}(\mu)$
