Assuming choice, the completion of the Borel Dirac measure at zero is defined on every subset of the real line

The singleton $\{0\}$ is Borel because its complement is the union of the open rays $(-\mathrm{\infty },0)$ and $(0,+\mathrm{\infty })$; hence $Z:=\mathbb{R}\setminus \{0\}$ is Borel and ${\delta }_0(Z)=0$.

The binary-sequence injection [L5] gives an injection $f$ from a set of cardinality $\mathfrak{c}$ into $\mathbb{R}$; the map $S\mapsto f[S]$ then injects its power set into $\mathcal{P}(\mathbb{R})$. Hence [L4] gives $\mathrm{\mid }\mathcal{P}(\mathbb{R})\mathrm{\mid }\ge 2^{\mathfrak{c}}>\mathfrak{c}=\mathrm{\mid }\mathcal{B}(\mathbb{R})\mathrm{\mid }$, so some subset $S$ of $\mathbb{R}$ is not Borel.

If $0\mathrm{\notin }E$, then $E=\mathrm{\varnothing }\cup E$ with $E\subseteq Z$ is a completed representation and has completed value $0$ by [L6]; if $0\in E$, then $E=\{0\}\cup (E\setminus \{0\})$ with the second part contained in $Z$, so [L2] and [L6] give completed value $1$.

Thus every subset of $\mathbb{R}$ lies in the completion and the displayed formula holds.

For the non-Borel set $S$ from step 1.2, the set $S\setminus \{0\}$ is still non-Borel, since otherwise adjoining the Borel singleton $\{0\}$ when necessary would make $S$ Borel; it is a subset of the Borel null set $Z$ from step 1.1. Thus the original Borel Dirac space is not complete, while step 3.1 shows that its completion is the full power set.