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?

  • $\begingroup$ For countable set $X$, the subspace with property D is exactly those absolutely continuous with respect to some measure. Are there counterexamples in the uncountable case? $\endgroup$ – John Jiang Aug 15 '10 at 1:02
  • $\begingroup$ Regarding the last question by Vaughn, notice that you could take the singular part to be the $\mathcal{S}$ as well, or you could mix the parts. For example take a subspace $$\mathcal{S} = \{\mu \in \mathcal{M}(\mathbb{R}) \mid \mu|_{(-\infty,0]} \ll \mathcal{L}|_{(-\infty,0]} \text{ and } \mu|_{(0,\infty)} \perp \mathcal{L}|_{(0,\infty)} \}.$$ Then $\mathcal{S}$ has the property D but neither $\mathcal{S}$ nor $\mathcal{S}^\perp$ satisfies a condition of the type $$\nu \ll \mu \in \mathcal{S} \Longrightarrow \nu \in \mathcal{S}.$$ $\endgroup$ – Tapio Rajala Dec 28 '10 at 19:16

A little further searching turned up a simple proof of Lebesgue's decomposition theorem in "The Lebesgue Decomposition Theorem for Measures", J. K. Brooks, The American Mathematical Monthly, 78 (1971), pp. 660-662. Without much extra work, it admits the following generalisation.

Theorem. Let $\mathcal{N} \subset \Omega$ be a collection of subsets such that

  1. if $E\in \mathcal{N}$ and $F\in \Omega$, $F\subset E$, then $F\in \mathcal{N}$;
  2. if $E_n \in \mathcal{N}$ is a countable collection, then $\bigcup_{n} E_n \in \mathcal{N}$ as well.

Consider the subspace $\mathcal{S} = \lbrace \mu \mid \mu(E) = 0 \text{ for all } E\in \mathcal{N} \rbrace$. Then $\mathcal{M} = \mathcal{S} \oplus \mathcal{S}^\perp$.

Proof. Fix $\nu\in \mathcal{M}$, and consider the following collection of subsets: $$ \mathcal{R} = \lbrace E \in \mathcal{N} \mid \nu(E) > 0 \rbrace. $$ Let $\alpha = \sup \lbrace \nu(E) \mid E\in \mathcal{R} \rbrace$, and let $E_n\in \mathcal{R}$ be a sequence of sets such that $\nu(E_n) \to \alpha$. Let $A = \bigcup_n E_n$. Then $\nu(A) = \alpha$ and $A \in \mathcal{N}$.

Furthermore, given any $E\in \mathcal{R}$, we have $\nu(E\setminus A) = 0$. Indeed, if $\nu(E\setminus A)>0$, then $\nu(E) = \nu(A) + \nu(E\setminus A) > \alpha$, contradicting the definition of $\alpha$. Similarly, $\nu(E\setminus A) = 0$ for every $E\in \mathcal{N}$.

Thus we may take $\nu_1 = \nu|_{X\setminus A}$ and $\nu_2 = \nu|_A$. It follows that $\nu_2 \in \mathcal{S}^\perp$, since $\nu_2(A)=1$ and $A\in \mathcal{N}$, and $\nu_1\in \mathcal{S}$, since $\nu(E\setminus A) = 0$ for every $E\in \mathcal{N}$.

Finally, uniqueness follows since $\mathcal{S} \cap \mathcal{S}^\perp = \lbrace 0\rbrace$.

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