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
Canonical name	TheProofOfTheoremIsWrong
Date of creation	2013-03-22 19:16:05
Last modified on	2013-03-22 19:16:05
Owner	tomprimozic (26284)
Last modified by	tomprimozic (26284)
Numerical id	4
Author	tomprimozic (26284)
Entry type	Example
Classification	msc 28A12