The Lebesgue Decomposition Theorem I am using is: Let $\mu$ be a $\sigma$-finite signed measure and let $\nu$ be finite positive measure. Then, there exist positive measures $\lambda$ and $\rho$ such that:
- $\nu = \lambda + \rho$
- $\lambda \perp \mu$
- $\rho \ll \mu$
This makes sense to me. An exercise I'm given uses a slightly different notation, and I'm not really sure how to interpret it. It says that the Lebesgue decomposition of $\nu$ can be written: \begin{align} d\nu = d\lambda + f d\mu & &(1) \end{align} This is confusing to me, as I have learned that, given $\eta$ $\sigma$-finite, $\nu$ finite, and $\nu \ll \mu$:
\begin{align} d\nu &= f d\mu & &\Leftrightarrow & \nu(A) = \int_A f d\mu \quad \forall A \in \mathcal{A} & &(2) \end{align}
How do I reconcile this notion of the Radon-Nikodym derivative in (2) with the expression in (1)? The expression in (1) seems sort of nonsensical to me (what does $d \lambda$ even mean here?), so clearly I'm missing something.
Since integrals often are written $\int f \, d\mu,$ it is common to write for example $d\nu = d\lambda + f \, d\mu$ for $\nu(A) = \lambda(A) + \int_A f \, d\mu.$ The symbolic foundation for this is $$\int_A d\nu = \nu(A) = \lambda(A) + \int_A f \, d\mu = \int_A d\lambda + \int_A f \, d\mu = \int_A (d\lambda + f \, d\mu).$$ Remove $\int_A$ and you get $d\nu = d\lambda + f \, d\mu.$