Every finite Borel measure on R has a unique absolutely continuous, discrete, and singular-continuous decomposition

By [L1], write $\mu ={\mu }_{\mathrm{d}}+\eta$, where ${\mu }_{\mathrm{d}}$ is discrete and $\eta$ is atomless.

Choose an increasing exhaustion $(X_n)$ of $\mathbb{R}$ with $\lambda (X_n)<\mathrm{\infty }$. Since $\eta$ is finite, also $\eta (X_n)<\mathrm{\infty }$, so [L3] applies to $\eta$ relative to Lebesgue measure $\lambda$ from [L2]. This gives $\eta ={\mu }_{\mathrm{a}\mathrm{c}}+{\mu }_{\mathrm{s}\mathrm{c}}$ with ${\mu }_{\mathrm{a}\mathrm{c}}\ll \lambda$ and ${\mu }_{\mathrm{s}\mathrm{c}}\perp \lambda$. Because $\eta$ is atomless and both summands are positive, ${\mu }_{\mathrm{a}\mathrm{c}}(\{x\})+{\mu }_{\mathrm{s}\mathrm{c}}(\{x\})=\eta (\{x\})=0$ forces ${\mu }_{\mathrm{a}\mathrm{c}}(\{x\})={\mu }_{\mathrm{s}\mathrm{c}}(\{x\})=0$ for every $x$, so ${\mu }_{\mathrm{s}\mathrm{c}}$ is atomless as well.

Combining steps 1.1 and 2.1 gives the required decomposition $\mu ={\mu }_{\mathrm{a}\mathrm{c}}+{\mu }_{\mathrm{d}}+{\mu }_{\mathrm{s}\mathrm{c}}$. Uniqueness follows because [L1] uniquely determines the discrete part and atomless remainder, while [L3] uniquely decomposes that atomless remainder into its absolutely continuous and singular pieces. The singular-continuous part is nonvacuous by [L4], which provides an atomless singular finite Borel measure.