The proof of theorem is wrong

Let’s create a very simple measurable space: $X=\{a,b\}$, $𝒜=\{\mathrm{\varnothing },\{a\},\{b\},X\}$.

Let’s take the $\pi$-system $P=\{\{a\}\}$ containing only one subset of $X$.

Let’s create two measures $\mu ={\delta }_a+{\delta }_b$ and $\nu ={\delta }_a+2{\delta }_b$. Then obviously $\mu$ and $\nu$ agree on $P$ and are finite, but they obviously are not equal on $𝒜$.

The proof, however, claims that it is sufficient if $\mu$ and $\nu$ are finite. I believe that $\mu (X)=\nu (X)$ is a necessary condition.