Understanding the Lebesgue's Decomposition Theorem In his book Bauer proves the Lebesgue's decomposition theorem. Actually he proves it only in the case where $\mu$ and $\nu$ are finite, leaving the $\sigma$-finite case as an exercise. Looking at the proof however I don't see anywhere where the finiteness of $\mu$ is used. For the $\sigma$-finite case I did the following:
Since $\nu$ is $\sigma$-finite we can find a partition $(A_n)$ of $\Omega$ into sets of finite $\nu$ measure.  For each $n$, let $\nu_n$ denote the finite measure defined by $\nu_n(A):=\nu(A\cap A_n)$ for measurable $A$. From the finite measure case there is a unique decomposition
$$\nu_n=\nu_n^c+\nu_n^s \hspace{0.7cm}\nu_n^c\ll\mu \hspace{0.7cm} \nu_n^s\perp\mu.$$
For each $n$, let $N_n$ be such that $\mu(N_n)=0=\nu_n^s(N_n^c)$  and let $N=\bigcup_{n\in\mathbb{N}}N_n$. Define the measures $\nu_c$ and $\nu_s$ by $\nu_c(A):=\nu(A\cap N^c)$ and $\nu_s(A):=\nu(A\cap N)$ for measurable $A$. We see that $\nu=\nu_c+\nu_s$ and $\nu_s \perp\mu$. Also $\nu_c\ll\mu $, since $\mu(A)=0$ implies
$$\nu_c(A)=\nu(A \cap N^c)=\sum_{n=1}^\infty \nu_n(A\cap N^c)=\sum_{n=1}^\infty \nu^c_n(A\cap N^c)+\sum_{n=1}^\infty \nu^s_n(A\cap N^c)$$$$\leq \sum_{n=1}^\infty \nu^c_n(A)+\sum_{n=1}^\infty \nu^s_n( N_n^c)=0$$
This shows existence as well as uniqueness, because if $\nu=\nu'_c+\nu'_s$ is any such decomposition, then defining the measures $\nu_n^{'c}(A):=\nu'_c(A\cap A_n)$ and $\nu_n^{'s}(A):=\nu'_s(A\cap A_n)$ for measurable $A$  and $n\in\mathbb{N}$ we get
$$\nu_n=\nu_n^c+\nu_n^s=\nu_n^{'c}+\nu_n^{'s} \hspace{0.7cm}\nu_n^{'c}\ll\mu \hspace{0.7cm} \nu_n^{'s}\perp\mu $$
which implies $\nu_n^c=\nu_n^{'c}$ and $\nu_n^s=\nu_n^{'s}$ for each $n$ by uniqueness. Hence $\nu'_c=\sum_{n=1}^\infty \nu_n^{'c}$ and $\nu'_s=\sum_{n=1}^\infty \nu_n^{'s}$ are completely determined.
Again I don't see where the $\sigma$-finiteness of $\mu$ is needed in the argument. I also get confused by the hint Bauer gives for the exercise:
Hint. For the existence proof use 17.6. For the uniqueness proof choose a sequence $(A_n)$ of measurable sets with $A_n \uparrow \Omega$ and $\mu(A_n),\nu(A_n)$ finite for each $n$, and consider the measures $\nu_n(A):=\nu(A\cap A_n)$ for mesurable $A$ and $n\in\mathbb{N}$.
EDIT: I think the requirement that $\mu$ be $\sigma$-finite is due to the fact that Lebesgue's decomposition theorem is usually proven in conjunction with the Radon-Nikodym theorem. See my answer below. Still I don't get how to use the hint provided to prove existence and uniqueness. Any help is appreciated.
 A: It seems like the reason we require $\mu$ to be also $\sigma$-finite is because Lebesgue's decomposition theorem is usually proven in conjunction with the Radon-Nikodym theorem, and presented under the name Radon-Nikodym-Lebesgue theorem. Indeed in the decomposition
$$v_n=v_n^c+v_n^s \hspace{0.7cm}v_n^c\ll\mu \hspace{0.7cm} v_n^s\perp\mu$$
we need $\mu$ to be $\sigma$-finite in order for $v_n^c$ to have a density with respect to $\mu$ using the Radon-Nikodym theorem. In fact, the Radon-Nikodym theorem can be used to prove the existence of the decomposition when $\mu$ and $v$ are assumed to be $\sigma$-finite:
Let $\lambda$ be the measure defined by $\lambda:=\mu+v$. Since $\mu$ and $v$ are both  $\sigma$-finite, so is $\lambda$. We see also that $\mu \ll \lambda$ and $v \ll \lambda$. Hence the  Radon-Nikodym theorem implies that there exist densities $f,g$ such that
$$v(A)=\int_A f d\lambda$$
$$\mu(A)=\int_A g d\lambda$$
for $A$ measurable. Let $N=\{\omega\in\Omega:g(\omega)=0\}$. Define the measures $v_c$ and $v_s$ by
$$v_c(A):=v(A\cap N^c)$$
$$v_s(A):=v(A\cap N)$$
for $A$ measurable. Since $\mu(N)=0$ we see that $ v_s \perp \mu$. Also $v_c \ll \mu$, since $\mu(A)=0$ implies $1_A g=0$ $\lambda$-a.e., i.e. $\lambda(A\cap N^c)=0$, so that from $v \ll \lambda$ we get $v_c(A)=v(A\cap N^c)=0$. Clearly $v=v_c+v_s$ so we have found a decomposition. Uniqueness follows as before.
