 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 functionalis well defined on its equivalence classes, since a $\rho$-null set is $\nu$-null. The Cauchy-Schwarz inequality and $\nu \le \rho$ giveBy the Riesz representation theorem, there is $h\in L^2(\rho )$ with $T(g)=\int ghd\rho$. Indicators of all measurable sets belong to this Hilbert space, soThe nonnegativity of these two measures forces $0\le h\le 1$ almost everywhere. For instance, if $\rho (\{h\le -1/n\})>0$, its $\nu$-measure would be negative; the analogous sets $\{h\ge 1+1/n\}$ 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.

DefineThen $B\in \Sigma$, $f$ 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 $f{1}_A$ givesIn particular ${\int }_{\Omega }fd\mu =\nu (\Omega \setminus B)<\infty$, so $f\in L^1(\mu )$. Additivity on the disjoint measurable sets $A\setminus B$ and $A\cap B$ now givesThis 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 $\nu (A)={\int }_{A}fd\mu$ vanishes on every $\mu$-null set. ThereforeThus absolute continuity is exactly the absence of the singular part in this construction.