Lebesgue decomposition theorem

For sigma-finite measures $\nu$ and $\mu$, there is a unique decomposition $\nu ={\nu }_{\mathrm{ac}}+{\nu }_{\mathrm{s}}$ in which ${\nu }_{\mathrm{ac}}\ll \mu$ and ${\nu }_{\mathrm{s}}\perp \mu$.

For a finite positive measure on $[0,1)$, average its mass over each half-open dyadic interval. The resulting nonnegative martingale has constant expectation $\mu ([0,1))$: a parent interval's mass is the sum of its children's masses. Its almost-sure limit is integrable but the martingale need not be uniformly integrable. The limit recovers the density of the absolutely continuous component in the Lebesgue decomposition theorem.

The Conditional Fatou lemma gives $X_n\ge \mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ for the dyadic density martingale. Integration first on dyadic sets, then extension by the Monotone class theorem, makes $\nu =\mu -X_{\infty }\lambda$ a nonnegative measure. Its finite-level density is $Y_n=X_n-\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$, which tends to zero almost everywhere for Lebesgue measure. The density level sets separate a singular measure argument shows that this zero-limit set has $\nu$-measure zero. Thus $\nu$ and $\lambda$ are mutually singular measures, obtaining the decomposition without assuming it to identify the martingale limit.

Use the Hilbert-space construction of dominated measure densities with $\rho =\mu +\nu$. The set $B=\{h=1\}$ is a null set for $\mu$. Put $f=0$ on $B$ and $f=h/(1-h)$ elsewhere. Then $\int fd\mu =\nu (B^c)<\infty$ and $\nu (A)={\int }_{A}fd\mu +\nu (A\cap B)$. This is the Lebesgue decomposition into a density part and a part concentrated on a $\mu$-null set.