Surjective endomorphism preserves Haar measure How to prove the following statement:
Let $G$ be a compact topological group and let $m$ be the Haar measure on it. Let $\varphi$ be a continuous endomorphism of $G$ onto $G$, i.e., the map $\varphi$ is surjective. Then $\varphi$ preserves $m$.
Is compact necessary, or is it still true for locally compact groups?
 A: For $G=\langle \mathbb{R},+\rangle$ let $\varphi$ be the surjective endomorphism defined by  $x\stackrel{\varphi}{\mapsto}\frac{1}{2}x$, then 
$$m(\varphi^{-1}([0,1])=m([0,2])=m([0,1])+m([1,2])=2m([0,1])$$
since $m$ is an additive Haar measure.
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Added: The fact is that a Haar measure on a compact set is finite (since it is regular), and since $\varphi$ is surjective you have that:
$$m(\varphi^{-1}(G))=m(G)$$
So the push-forward measure $\mu=m\circ\varphi^{-1}$ gives $G$ the same measure. Since Haar measures are unique up-to multiplication by a constant, we have that if $\mu$ is also Haar, than $\mu=cm$ for some $c\in\mathbb{R}$, but since we already know that $m(G)=\mu(G)$ this forces $c=1$.
It is left to check that $\mu$ is indeed a Haar measure, but this seems rather simple:
Let $g\in G$ be arbitrary, $g'\in G$ such that $\varphi(g')=g$ (existence given by surjectivity), and let $E\subseteq G$ be mesurable. Then 
$$\varphi^{-1}(gE)=\lbrace x\in G\mid \varphi(x)\in gE\rbrace=\lbrace x\mid g^{-1}\varphi(x)=\varphi(g'^{-1}x)\in E\rbrace=g'\varphi^{-1}(E)$$
and
$$\mu(gE)=m(\varphi^{-1}(gE))=m(g'\varphi^{-1}(E))=m\varphi^{-1}(E)=\mu(E)$$
There still remains to show that $\varphi^{-1}(E)$ is still measurable, and this is where the continuity assumption kicks in :)
