FALSE: the extension of a premeasure is always unique

The family ${\mathcal{A}}_0$ is an algebra containing the 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. A disjoint sequence in ${\mathcal{A}}_0$ with empty union has all terms empty, while one with nonempty union has a nonempty term, so [F1] gives countable additivity of ${\mu }_0$. Every nonempty member contains a nonempty interval component and hence the distinct midpoint-bisection sequence $a+(b-a)\mathrm{/}{2}^n$ after restricting to finite endpoints; therefore it is infinite, and the only finite-${\mu }_0$ member is $\mathrm{\varnothing }$, so ${\mu }_0$ is not sigma-finite.

By [L1], counting measure restricted to $\mathcal{B}(\mathbb{R})$ is a measure and agrees with ${\mu }_0$ because every nonempty source-algebra member is infinite. The function $\nu (B)=0$ for $B=\mathrm{\varnothing }$ and $\nu (B)=+\mathrm{\infty }$ otherwise is also a Borel measure: a disjoint union is empty exactly when every term is empty, and otherwise one term is nonempty. Thus both are extensions.

On a singleton $\{x\}$, counting measure is $1$ while $\nu (\{x\})=+\mathrm{\infty }$, so the extensions are distinct.