A non-sigma-finite premeasure has distinct Borel extensions

The interval family is an algebra containing all finite half-open intervals, so [L2] gives $\mathcal{B}(\mathbb{R})\subseteq \sigma ({\mathcal{A}}_0)$. Every extended-endpoint half-open interval is Borel, and finite unions of Borel sets are Borel, so ${\mathcal{A}}_0\subseteq \mathcal{B}(\mathbb{R})$ and the reverse inclusion follows. The zero-or-infinity function satisfies [F1] because any nonempty disjoint union has a nonempty term. Every nonempty algebra member contains an interval component and an infinite midpoint-bisection sequence, so only $\mathrm{\varnothing }$ has finite premeasure and no sigma-finite cover exists.

Counting measure from [L1] restricts to a Borel extension because every nonempty algebra member is infinite. The Borel function $\nu (B)=0$ for $B=\mathrm{\varnothing }$ and $+\mathrm{\infty }$ otherwise is also countably additive by the same empty-versus-nonempty argument, and it too extends ${\mu }_0$.

For every $x\in \mathbb{R}$, counting measure gives $\mathrm{\#}(\{x\})=1$ while $\nu (\{x\})=+\mathrm{\infty }$, so the extensions are distinct.