I am asked to proof the following Lebesgue Decomposition Theorem for the case of $\sigma$-finite positive measure:
Lebesgue Decomposition Theorem$\quad$ Let $(X,\mathscr{A})$ be a measurable space, let $\mu$ be a positive measure on $(X,\mathscr{A})$, and let $\nu$ be a $\sigma$-finite positive measure on $(X,\mathscr{A})$. Then there are unique positive measures $\nu_a$ and $\nu_s$ on $(X,\mathscr{A})$ such that
(a) $\nu_a$ is absolutely continuous with respect to $\mu$,
(b) $\nu_s$ is singular with respect to $\mu$, and
(c) $\nu=\nu_a+\nu_s$.
The decomposition $\nu=\nu_a+\nu_s$ is called the Lebesgue decomposition of $\nu$.
For this post, let us forget the uniqueness. I tried myself to prove the existence of such $\nu_a$ anhd $\nu_s$, but I am not sure whether my attempt is correct or not, especially the final part of showing the absolute continuity of $\nu_a$ with respect to $\mu$. I would really appreciate it if someone could help me check my work (especially for the last part of Step 3)! Thank you very much!
Here is my attempt:
Proof of Existence$\quad$ Since $\nu$ is a $\sigma$-finite positive measure on $(X,\mathscr{A})$, we can write$$X = \bigcup_{i=1}^{\infty}D_i,$$where $D_i\in\mathscr{A}$ and $\nu(D_i)<+\infty$ for all $i$.
Step 1$\quad$ Consider the set $\mathscr{A}_k$ defined by $\mathscr{A}_k = \{A\in\mathscr{A}:A\subseteq D_k\}$. We show that $\mathscr{A_k}$ is a $\sigma$-algebra on $D_k$.
Clearly, $D_k\in\mathscr{A}_k$.
Let $A\in\mathscr{A}_k$, then $A\in\mathscr{A}$ and $A\subseteq D_k$, and thus $D_k\backslash A = A^c\bigcap D_k$, so that $D_k\backslash A\in\mathscr{A}_k$.
Let $\{A_i\}$ be a sequence of sets that belong to $\mathscr{A}_k$, then $A_i\in\mathscr{A}$ and $A_i\subseteq D_k$ for all $i$, and thus $\bigcup_{i=1}^{\infty}A_i\in\mathscr{A}$ and $\bigcup_{i=1}^{\infty}A_i\subseteq D_k$, so that $\bigcup_{i=1}^{\infty}A_i\in\mathscr{A}_k$.
Step 2$\quad$ Now let $k$ be a positive integer and consider the restrictions of $\mu$ and $\nu$ to the spaces $(D_k,\mathscr{A}_k)$. Then $\mu|_{(D_k,\mathscr{A}_k)}$ is a positive measure and $\nu|_{(D_k,\mathscr{A}_k)}$ is a finite positive measure on $(D_k,\mathscr{A}_k)$. Define $\mathscr{N}_{\mu_k}$ by$$\mathscr{N}_{\mu_k} = \{B_k\in\mathscr{A}_k:\mu(B_k)=0\},$$and choose a sequence $\{B_{k,j}\}_{j=1}^{\infty}$ of sets in $\mathscr{N}_{\mu_k}$ such that$$\lim_{j\to\infty}\nu(B_{k,j}) = \sup\{\nu(B_k):B_k\in\mathscr{N}_{\mu_k}\}.$$Let $N_k=\bigcup_jB_{k,j}$ and let $N=\bigcup_kN_k$. Define $\nu_a$ and $\nu_s$ by $\nu_a(A)=\nu(A\bigcap N^c)$ and $\nu_s(A) = \nu(A\bigcap N)$.
Step 3$\quad$ We prove that the $\nu_a$ and $\nu_s$ defined in Step 2 form a Lebesgue decomposition of $\nu$.
$\nu=\nu_a+\nu_s$: $\nu(A) = \nu(A\bigcap N^c) + \nu(A\bigcap N) = \nu_a(A)+\nu_s(A)$.
$\nu_s$ is singular with respect to $\mu$: The countable subadditivity of $\mu$ implies$$0\leq\mu(N)=\mu\left(\bigcup_{k=1}^{\infty}N_k\right)\leq\sum_{k=1}^{\infty}\mu(N_k)=\sum_{k=1}^{\infty}\mu\left(\bigcup_{j=1}^{\infty}B_{k,j}\right)\leq\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}\mu(B_{k,j})=0,$$and so $\mu(N)=0$. We also have $\nu_s(N^c)=\nu(\emptyset)=0$. So $\nu_s$ is singular with respect to $\mu$.
$\nu_a$ is absolutely continuous with respect to $\mu$: Note that$$\nu(N_k) = \sup\{\nu(B_k):B_k\in\mathscr{N}_{\mu_k}\},$$so each $\mathscr{A}_k$-measurable subset $B_k$ of $D_k\backslash N_k$ that satisfies $\mu(B_k)=0$ also satisfies $\nu(B_k)=0$, because otherwise $N_k\bigcup B_k$ would belong to $\mathscr{N}_{\mu_k}$ and satisfy $\nu(N_k\bigcup B)>\nu(N_k)$. Now, if $B$ is an $\mathscr{A}$-measurable subset of $N^c = \left(\bigcup_{k=1}^{\infty}N_k\right)^c = \bigcap_{k=1}^{\infty}N_k^c$, then it must be an $\mathscr{A}$-measurable subset of $D_k\backslash N_k$ for some $k$, and so $B$ is an $\mathscr{A}_k$-measurable subset of $D_k\backslash N_k$. So if this set $B$ satisfies $\mu(B)=0$ then it must also satisfy $\nu(B)=0$. Define$$\mathscr{N}_{\mu} = \{B\in\mathscr{A}:\mu(B)=0\}.$$Note that $N=\bigcup_k\bigcup_jB_{k,j}\in\mathscr{A}$ so that $N^c\in\mathscr{A}$. Let $B\in\mathscr{N}_{\mu}$, then $\mu(B)=0$. Also, $B\bigcap N^c\in\mathscr{A}$, $B\bigcap N^c\subseteq N^c$, and $0\leq\mu(B\bigcap N^c)\leq\mu(B)=0$ so that $\mu(B\bigcap N^c)=0$. Therefore, $\nu_a(B)=\nu(B\bigcap N^c)=0$. Hence, $\nu_a$ is absolutely continuous with respect to $\mu$.