 Solution

Let $\mu$ have the stated properties. Translation invariance and the disjoint partition of $(0,1]$ into $n$ intervals of length $1/n$ giveFinite additivity then gives $\mu ((0,r])=r$ for every positive rational $r$. Continuity from below and above extends this to every real $r>0$, and translation invariance yieldsHence $\mu$ and $\lambda$ agree on the half-open intervals. Both are $\sigma$-finite, so uniqueness of extension gives $\mu =\lambda$ on the Borel $\sigma$-algebra.