= Solution
We construct the density and singular part, rather than assuming the desired <Lebesgue decomposition theorem>. Put $\rho=\mu+\nu$, a finite <dominating measure>, and use the real <Hilbert space> $L^2(\rho)$. The <linear functional>
$$
T(g)=\int_\Omega g\,d\nu
$$
is well defined on its equivalence classes, since a $\rho$-null set is $\nu$-null. The <Cauchy-Schwarz inequality> and $\nu\leq\rho$ give
$$
|T(g)|\leq\nu(\Omega)^{1/2}\left(\int g^2\,d\nu\right)^{1/2}\leq\nu(\Omega)^{1/2}\|g\|_{L^2(\rho)}.
$$
By the <Riesz representation theorem>, there is $h\in L^2(\rho)$ with $T(g)=\int gh\,d\rho$. Indicators of all measurable sets belong to this Hilbert space, so
$$
\nu(A)=\int_Ah\,d\rho,\qquad \mu(A)=\int_A(1-h)\,d\rho\quad(A\in\Sigma).
$$
The nonnegativity of these two <measures> forces $0\leq h\leq1$ almost everywhere. For instance, if $\rho(\{h\leq-1/n\})>0$, its $\nu$-<measure> would be negative; the analogous sets $\{h\geq1+1/n\}$ contradict positivity of $\mu$. Their countable union excludes every violation. Modify $h$ on its measurable $\rho$-null exceptional set so that these inequalities hold everywhere. This is the <Hilbert-space construction of dominated measure densities>.
Define
$$
B=\{h=1\},\qquad f=\begin{cases}h/(1-h),&h<1,\\0,&h=1.\end{cases}
$$
Then $B\in\Sigma$, $f$ is a nonnegative <measurable function>, and $\mu(B)=\int_B(1-h)\,d\rho=0$. The identity for integration against $\mu$ follows first for indicators, then for nonnegative simple functions and finally for all nonnegative functions by the <monotone convergence theorem>. Applying it to $f\mathbf1_A$ gives
$$
\int_Af\,d\mu=\int_{A\setminus B}h\,d\rho=\nu(A\setminus B).
$$
In particular $\int_\Omega f\,d\mu=\nu(\Omega\setminus B)<\infty$, so $f\in L^1(\mu)$. Additivity on the disjoint measurable sets $A\setminus B$ and $A\cap B$ now gives
$$
\boxed{\nu(A)=\int_A f\,d\mu+\nu(A\cap B),\qquad \mu(B)=0.}
$$
This is the <Lebesgue decomposition from a sum-measure density>: the first <measure> has a density, and the second is concentrated on a $\mu$-<null set>, so is <mutually singular> with $\mu$.
The <absolute continuity of measures> condition $\nu\ll\mu$ means that $\mu(A)=0$ implies $\nu(A)=0$ for every $A\in\Sigma$. If this holds, then $\nu(B)=0$, eliminating the singular part above. Conversely, a <measure> of the form $\nu(A)=\int_Af\,d\mu$ vanishes on every $\mu$-null set. Therefore
$$
\boxed{\nu\ll\mu\iff\nu(A)=\int_Af\,d\mu\text{ for some nonnegative }f\in L^1(\mu).}
$$
Thus absolute continuity is exactly the absence of the singular part in this construction.
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