Let $\mathcal{A}$ be an algebra of subsets of a set $E$, and let $\mu_0:\mathcal{A}\to[0,\infty]$ be a premeasure. Then there exists a measure $\mu:\sigma(\mathcal{A})\to[0,\infty]$ such that $\mu(A)=\mu_0(A)$ for every $A\in\mathcal{A}$. If there exist sets $E_1,E_2,\ldots\in\mathcal{A}$ with $E=\bigcup_{n=1}^{\infty}E_n$ and $\mu_0(E_n)<\infty$ for every $n$, then this extension is unique on $\sigma(\mathcal{A})$.