# continuous epimorphism of compact groups preserves Haar measure

Let $G,H$ be compact Hausdorff topological groups. If $\phi:G\longrightarrow H$ is a continuous surjective homomorphism, then $\phi$ is a measure preserving transformation, in the sense that it preserves the normalized Haar measure.

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: Let $\mu$ be the Haar measure in $G$ (normalized, i.e. $\mu(G)=1$). Let $\nu$ be defined for measurable subsets $E$ of $H$ by

 $\nu(E)=\mu(\phi^{-1}(E))$

It is easy to see that $\nu$ defines a measure in $H$. Let us now see that $\nu$ is invariant under right translations. For every $s\in G$ and every measurable subset $E\subset H$ we have that

 $\displaystyle\phi^{-1}(\phi(s)E)=s\phi^{-1}(E)$ (1)

The inclusion $\supseteq$ is obvious. To prove the other inclusion notice that if $z\in\phi^{-1}(\phi(s)E)$ then $\phi(z)=\phi(s)t$ for some $t\in E$. Hence, $\phi(s^{-1}z)=t$, i.e $s^{-1}z\in\phi^{-1}(E)$. It now follows that $z=s(s^{-1}z)\in s\phi^{-1}(E)$.

Thus, equality (1) and the fact that $\mu$ is a Haar measure imply that

 $\nu(\phi(s)E)=\mu\big{(}\phi^{-1}(\phi(s)E)\big{)}=\mu(s\phi^{-1}(E))=\mu(\phi% ^{-1}(E))=\nu(E)$

Since $\phi$ is surjective it follows that $\nu$ is right invariant. It is not difficult to see that $\nu$ is regular, finite on compact sets and $\nu(H)=1$. Hence, $\nu$ is the normalized Haar measure in $H$ and, by definition, we have that

 $\nu(E)=\mu(\phi^{-1}(E))$

Thus, $\phi$ preserves the Haar measure. $\square$

Title continuous epimorphism of compact groups preserves Haar measure ContinuousEpimorphismOfCompactGroupsPreservesHaarMeasure 2013-03-22 17:59:06 2013-03-22 17:59:06 asteroid (17536) asteroid (17536) 9 asteroid (17536) Theorem msc 37A05 msc 28C10 msc 22C05