# Defining measure on homeomorphic spaces

Say you have two homeomorphic spaces $X\cong Y$, and a measure $\mu$ defined on the measurable space $(X,\scr{B}$$(X)), where \scr B$$(X)$ is the Borel sigma-algebra of $X$. Since $\scr B$$(X) is generate by the open sets of X, and U\subseteq X is open in X \Leftrightarrow f(U)\subseteq Y is open, isn't it true that \mu is also a measure on (Y,\scr B$$(X))$, with $\mu(U)=\mu(f(U))$?

• If $\mu$ is allready a measure on one of them and the spaces are not the same then $\mu$ is not a measur on the other. This in spite of being homeomorphic. However, it has indeed a corresponding measure $\nu$ on the other defined by $\nu(U)=\mu(f(U))$. – drhab Nov 8 '15 at 13:58
• @drhab: worth adding that there's nothing "special" about $\nu$ since first of all it may not be unique. – Ilya Nov 10 '15 at 7:49