3 Fixed typo; added 1 characters in body

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 $\nu_1 = \nu|_{E\setminus nu|_{X\setminus A}$ 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 $E\in \mathcal{N}.mathcal{N}$.

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

2 Corrected definition of subspace S. (Needed slightly stronger condition.)

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 $\mathcal{S} \mathcal{N} \subset \mathcal{M}$ Omega$be a subspace that is closed under absolute continuity -- collection of subsets such thatis, 1. if$\mu\in E\in \mathcal{S}$mathcal{N}$ and $\nu\ll\mu$, F\in \Omega$,$F\subset E$, then$\nu\in F\in \mathcal{M}$mathcal{N}$;
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: $$\mathcal{R} = \lbrace E \in \Omega mathcal{N} \mid \nu(E) > 0 \text{ and } \mu(E)=0 \text{ for every } \mu\in \mathcal{S} \rbrace.$$ 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 $\mu(A)=0$ for every $\mu\in\mathcal{S}$.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|_{E\setminus A}$and$\nu_2 = \nu|_A$. It follows that$\nu_2\perp \nu_2 \mu$, in \mathcal{S}^\perp$, since $\nu_2(A)=1$ and $\mu(A)=0$, A\in \mathcal{N}$, and$\nu_1\ll \nu_1\in \mu$, mathcal{S}$, since every set $E$ with $\mu(E)=0$ has the property that $\nu(E\setminus A) = 0$ .for every $E\in \mathcal{N}. Finally, uniqueness follows since$\mathcal{S} \cap \mathcal{S}^\perp = \lbrace 0\rbrace$. 1 A little further searching turned up a simple proof of Lebesgue's decomposition theorem in "The Lebesgue Decomposition Theorem for Measures", J. K. Brooks, The American Mathematical Monthly, 78 (1971), pp. 660-662. Without much extra work, it admits the following generalisation. Theorem. Let$\mathcal{S} \subset \mathcal{M}$be a subspace that is closed under absolute continuity -- that is, if$\mu\in \mathcal{S}$and$\nu\ll\mu$, then$\nu\in \mathcal{M}$as well. Then$\mathcal{M} = \mathcal{S} \oplus \mathcal{S}^\perp$. Proof. Fix$\nu\in \mathcal{M}$, and consider the following collection of subsets: $$\mathcal{R} = \lbrace E \in \Omega \mid \nu(E) > 0 \text{ and } \mu(E)=0 \text{ for every } \mu\in \mathcal{S} \rbrace.$$ 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$\mu(A)=0$for every$\mu\in\mathcal{S}$. 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$\alpha$. Thus we may take$\nu_1 = \nu|_{E\setminus A}$and$\nu_2 = \nu|_A$. It follows that$\nu_2\perp \mu$, since$\nu_2(A)=1$and$\mu(A)=0$, and$\nu_1\ll \mu$, since every set$E$with$\mu(E)=0$has the property that$\nu(E\setminus A) = 0$. Finally, uniqueness follows since$\mathcal{S} \cap \mathcal{S}^\perp = \lbrace 0\rbrace\$.