Lebesgue decomposition theorem

Remarks

One proof applies the Radon–Nikodym theorem to $\mu$ and $\mu + \nu$. With $f = d\mu / d(\mu + \nu)$, the set $S = \{f = 0\}$ is $\mu$-null, and one puts $\nu_s(A) = \nu(A \cap S)$ and $\nu_a(A) = \nu(A \setminus S)$. Uniqueness holds because a measure that is both absolutely continuous and singular with respect to $\mu$ is zero.

Example: let $\mu$ be Lebesgue measure on $\R$ and $\nu = \mu|_{[0,1]} + \delta_2$, where $\delta_2$ is the point mass at $2$. Then $\nu_a = \mu|_{[0,1]}$ and $\nu_s = \delta_2$.

On the real line with $\mu$ Lebesgue measure, the singular part splits further into a part made of point masses and a continuous singular part. The distribution of the Cantor function is an example of a continuous singular measure.

Going deeper

For a student

Combined with the Radon–Nikodym theorem, it says $\nu = f\mu + \nu_s$ with $f = d\nu_a / d\mu$ and $\nu_s \perp \mu$. In probability this splits a distribution on $\R$ into a part with a density, a discrete part and a singular continuous part. The same decomposition holds for finite signed and complex measures, applied to the total variation.