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 $\mu(U)=\mu(f(U))$?

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    $\begingroup$ 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 $\nu(U)=\mu(f(U))$. $\endgroup$ – drhab Nov 8 '15 at 13:58
  • $\begingroup$ @drhab: worth adding that there's nothing "special" about $\nu$ since first of all it may not be unique. $\endgroup$ – Ilya Nov 10 '15 at 7:49

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