Every finite Borel measure on $\mathbb{R}$ splits as an atomic part plus an atomless part

$A:=\{x\in \mathbb{R}:\mu (\{x\})>0\}$

is at most countable. If $A=\mathrm{\varnothing }$, then $\mu (\{x\})=0$ for every $x\in \mathbb{R}$, so $\mu$ is already atomless; taking $\nu :=\mu$ proves the theorem. Otherwise choose an injective enumeration $A=\{a_i:i\in I\}$, where $I=\{1,\dots ,m\}$ if $A$ is finite and $I=\mathbb{N}$ if $A$ is infinite.

${\mu }_{\mathrm{a}\mathrm{t}}:={\sum }_{i\in I}\mu (\{a_i\}){\delta }_{a_i}$

is a Borel measure. For every Borel set $E$,

${\mu }_{\mathrm{a}\mathrm{t}}(E)={\sum }_{a_i\in E}\mu (\{a_i\})=\mu (E\cap A),$

because the singletons $\{a_i\}$ are pairwise disjoint and countable additivity of $\mu$ applies to their union. [step 1.1, L3, algebra]

$\nu (E):=\mu (E\setminus A)(E\in \mathcal{B}(\mathbb{R}))\mathrm{.}$

Because $(E_n\setminus A)$ remains pairwise disjoint whenever $(E_n)$ is, the same countable additivity as for $\mu$ shows that $\nu$ is a finite Borel measure. [step 2.1, given, algebra]

$\mu (E)=\mu (E\cap A)+\mu (E\setminus A)={\mu }_{\mathrm{a}\mathrm{t}}(E)+\nu (E)\mathrm{.}$

If $x\in A$, then $\nu (\{x\})=\mu (\mathrm{\varnothing })=0$; if $x\mathrm{\notin }A$, then $\mu (\{x\})=0$ by definition of $A$, so also $\nu (\{x\})=0$. Hence $\nu$ is atomless. [step 2.1, step 3.1, algebra]