= Solution
Let $\mu$ have the stated properties. Translation invariance and the disjoint partition of $(0,1]$ into $n$ intervals of length $1/n$ give
$$
\mu((0,1/n])=\frac1n.
$$
Finite 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 yields
$$
\mu((a,b])=b-a.
$$
\b[Hence $\mu$ and $\lambda$ agree on the half-open intervals. Both are $\sigma$-finite, so uniqueness of extension gives $\mu=\lambda$ on the Borel $\sigma$-algebra.]
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