**Background.** The Lebesgue decomposition theorem states that if $(X,\Omega)$ is a measurable space and $\mu$ is a finite measure on $X$, then for every measure $\nu$, there is a unique decomposition $\nu = \nu_1 + \nu_2$ such that $\nu_1 \ll \mu$ and $\nu_2 \perp \mu$.

Let us denote the space of all finite measures on $(X,\Omega)$ by $\mathcal{M}$. Then the above is equivalent to the statement that $\mathcal{M} = \mathcal{S} \oplus\mathcal{T}$, where $\mathcal{S}$ is the space of all measures that are absolutely continuous with respect to $\mu$, while $\mathcal{T}$ is the space of all measures that are singular with respect to $\mu$.

We can characterise $\mathcal{T}$ in terms of $\mathcal{S}$ as $$ \mathcal{T} = \mathcal{S}^\perp = \lbrace \nu\in \mathcal{M} \mid \nu \perp m \text{ for all } m\in \mathcal{S} \rbrace. $$ Let us say that a subspace $\mathcal{S} \subset \mathcal{M}$ has property D (for decomposition) if $\mathcal{M} = \mathcal{S} \oplus \mathcal{S}^\perp$. Then the Lebesgue decomposition theorem says that $\lbrace \nu \mid \nu \ll \mu\rbrace$ has property D for any fixed $\mu$.

**Question.** Which subspaces have property D? Are there conditions on $\mathcal{S}$ that are equivalent to property D, or at least imply it? Presumably $\mathcal{S}$ should have the property that if $\nu \ll \mu \in \mathcal{S}$, then $\nu\in \mathcal{S}$ as well; is this sufficient, or are there other requirements?

`$$\mathcal{S} = \{\mu \in \mathcal{M}(\mathbb{R}) \mid \mu|_{(-\infty,0]} \ll \mathcal{L}|_{(-\infty,0]} \text{ and } \mu|_{(0,\infty)} \perp \mathcal{L}|_{(0,\infty)} \}.$$`

Then $\mathcal{S}$ has the property D but neither $\mathcal{S}$ nor $\mathcal{S}^\perp$ satisfies a condition of the type $$\nu \ll \mu \in \mathcal{S} \Longrightarrow \nu \in \mathcal{S}.$$ – Tapio Rajala Dec 28 '10 at 19:16