= Solution
Construct the densities using the <Riesz representation theorem> for a <Hilbert space>, rather than assuming the <measure>-decomposition conclusion. Put $\rho=\mu+\nu$, a finite <dominating measure>, and work in the real <L2 space> $L^2(\rho)$. The <linear functional>
$$
L(g)=\int_\Omega g\,d\nu
$$
is well-defined on its equivalence classes because $\nu\le\rho$. By the <Cauchy-Schwarz inequality>,
$$
|L(g)|\le\nu(\Omega)^{1/2}
\left(\int_\Omega |g|^2\,d\nu\right)^{1/2}
\le\nu(\Omega)^{1/2}\|g\|_{L^2(\rho)}.
$$
Thus the <Riesz representation theorem> supplies $h\in L^2(\rho)$ such that $L(g)=\int gh\,d\rho$. Since $\rho$ is finite, $h$ is also <integrable>. Testing $g=1_A$ gives
$$
\nu(A)=\int_Ah\,d\rho,\qquad
\mu(A)=\int_A(1-h)\,d\rho\qquad(A\in\Sigma).
$$
The first identity and nonnegativity of $\nu$ rule out a positive-<measure> set where $h<0$; for example, test $\{h\le-1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge1+1/n\}$. Modify a measurable representative on a $\rho$-<null set> so that $0\le h\le1$ everywhere. This is the <Hilbert-space construction of dominated measure densities>.
Set
$$
B=\{h=1\},\qquad
f=
\begin{cases}
h/(1-h),&h<1,\\
0,&h=1.
\end{cases}
$$
Then $B\in\Sigma$ and $\mu(B)=0$. The indicator identities for $\mu$ extend to integrals of nonnegative <measurable functions> by approximation with <simple functions> and the <monotone convergence theorem>. Consequently,
$$
\int_\Omega f\,d\mu
=\int_{B^c}h\,d\rho=\nu(B^c)\le\nu(\Omega)<\infty,
$$
and, for every $A$,
$$
\int_A f\,d\mu=\nu(A\setminus B).
$$
Hence
$$
\boxed{\nu(A)=\int_Af\,d\mu+\nu(A\cap B),\qquad f\ge0,\quad f\in L^1(\mu),\quad\mu(B)=0.}
$$
This proves <Lebesgue decomposition from a sum-measure density>. The argument includes the cases of zero <measures>; there is no division on $B$, where the denominator vanishes.
<Absolute continuity of measures>, written $\nu\ll\mu$, means that $\mu(A)=0$ implies $\nu(A)=0$ for every $A\in\Sigma$. If $\nu\ll\mu$, then $\nu(B)=0$ in the construction above, so the singular term vanishes. Conversely, if $\nu(A)=\int_Af\,d\mu$ with a nonnegative <integrable> $f$, that integral is zero on every $\mu$-<null set>. Thus
$$
\boxed{\nu\ll\mu\iff \nu(A)=\int_Af\,d\mu
\text{ for some nonnegative }f\in L^1(\mu).}
$$
This is the finite-<measure> <Radon-Nikodym theorem> as a consequence of the proved construction.
Finally restrict the base <measure> to the sub-<sigma-algebra> $\Sigma_0$. For a real $f\in L^1(\mu)$, write $f=f^+-f^-$, with both parts nonnegative and <integrable>. On $\Sigma_0$ define finite <positive measures>
$$
\nu_\pm(A)=\int_A f^\pm\,d\mu.
$$
They are <absolutely continuous with respect to> $\mu|_{\Sigma_0}$. The result just proved gives nonnegative, $\Sigma_0$-measurable densities $g_\pm$ satisfying
$$
\int_Ag_\pm\,d\mu=\int_Af^\pm\,d\mu\qquad(A\in\Sigma_0).
$$
Set $f_0=g_+-g_-$. Since $\int(g_++g_-)\,d\mu=\int|f|\,d\mu<\infty$, this is in $L^1(\Omega,\Sigma_0,\mu)$, and subtraction gives
$$
\boxed{\int_Af_0\,d\mu=\int_Af\,d\mu\qquad(A\in\Sigma_0).}
$$
For complex $f$, apply the real construction separately to its real and imaginary parts and combine the two results. This is <conditional expectation from finite-measure densities>. The resulting <conditional expectation> is unique almost everywhere: for two real versions, their difference has integral zero on its positive and negative level sets in $\Sigma_0$, forcing it to vanish; apply this to both components in the complex case. No probability normalization of $\mu$ is required.
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