# The proof of theorem is wrong

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

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 $\mathcal{A}$.

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.

Title The proof of theorem is wrong TheProofOfTheoremIsWrong 2013-03-22 19:16:05 2013-03-22 19:16:05 tomprimozic (26284) tomprimozic (26284) 4 tomprimozic (26284) Example msc 28A12