= Solution
The <uniqueness theorem for measures> says that two measures agreeing on a generating <pi-system> agree on its generated <sigma-algebra>, provided they are finite with the same total mass, or there is a common increasing exhaustion by pi-system sets of finite measure.
For the finite case let $\mathcal D=\{A:\mu(A)=\nu(A)\}$. Equality of total masses gives closure under complements, and <countable additivity> gives closure under disjoint countable unions. Thus $\mathcal D$ is a <Dynkin system> containing the generating <pi-system>. The <pi-lambda theorem> puts the entire generated sigma-algebra inside $\mathcal D$.
For the exhaustion version, let $E_n\uparrow E$ with $E_n$ in the pi-system and $\mu(E_n)=\nu(E_n)<\infty$. The finite measures $A\mapsto\mu(A\cap E_n)$ and $A\mapsto\nu(A\cap E_n)$ agree on the generating pi-system, since it is intersection-closed. The finite proof applies, and <continuity from below of a measure> as $n\to\infty$ gives $\mu(A)=\nu(A)$ for every measurable $A$. A finiteness hypothesis is essential to this standard extension-uniqueness statement.
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