measure theory - Generalisation of Lebesgue decomposition theorem

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 $Ein mathcal{N}$ and $Fin Omega$, $Fsubset E$, then $Fin mathcal{N}$;


2. if $E_n in mathcal{N}$ is a countable collection, then $bigcup_{n} E_n in mathcal{N}$ as well.

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

Proof. Fix $nuin mathcal{M}$, and consider the following collection of subsets:
$$
mathcal{R} = lbrace E in mathcal{N} mid nu(E) > 0 rbrace.
$$
Let $alpha = sup lbrace nu(E) mid Ein mathcal{R} rbrace$, and let $E_nin mathcal{R}$ 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 $Ein mathcal{R}$, we have $nu(Esetminus A) = 0$. Indeed, if $nu(Esetminus A)>0$, then $nu(E) = nu(A) + nu(Esetminus A) > alpha$, contradicting the definition of $alpha$. Similarly, $nu(Esetminus A) = 0$ for every $Ein mathcal{N}$.

Thus we may take $nu_1 = nu|_{Xsetminus A}$ and $nu_2 = nu|_A$. It follows that $nu_2 in mathcal{S}^perp$, since $nu_2(A)=1$ and $Ain mathcal{N}$, and $nu_1in mathcal{S}$, since $nu(Esetminus A) = 0$ for every $Ein mathcal{N}$.

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