# 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?

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Do you mean an automorphism $\varphi$? If so, simply check that the new measure is left (or right) invariant as well to see that it is proportional to Haar measure. (the factor is called the module of the automorphism). See e.g. Hewitt-Ross, section 15.28. – t.b. Sep 18 '11 at 19:40
No.. Endomorphism. Say, consider the pushforward of m by $\varphi$ defined as: $\mu(A) = m(\varphi^{-1}(A))$. Now, $\mu$ is supposed to be the same as $m$(perhaps by uniqueness of Haar measure, it is enough to prove Haar property for $\mu$). – Lit Sep 18 '11 at 19:47
Okay, then consider the endomorphism $\varphi: G \to G$ that sends everything to the neutral element... Perfectly continuous but you won't get Haar measure by that procedure unless $G = \{1\}$. – t.b. Sep 18 '11 at 19:59
Oops, I checked the statement again and it includes a surjectivity condition ... Sorry. – Lit Sep 18 '11 at 20:03

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 :)

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So it is not true in general for locally compact groups, and compactness is necessary if the claim is true. Btw, a measure preserving transformation is defined to be something such that measure of a measurable set is the same as measure of its inverse image, not the forward image. The idea still works, though. – Lit Sep 18 '11 at 20:35
This can be generalized to $x\mapsto\alpha\cdot x$, for $\alpha\in\mathbb R\setminus\{0,1,-1\}$. – Asaf Karagila Sep 18 '11 at 20:36
@Lit: Thank you, that makes the proof even correct :) – kneidell Sep 18 '11 at 20:43
Thanks, accepted the answer! But, again, sorry, $\mu(gE)$ should be $m(\varphi^{-1}(gE)$, etc.. – Lit Sep 18 '11 at 20:45
@Lit: Dear kneidell and Lit, It is not the surjectivity of $\varphi$ that ensures $m(G) = \mu(G)$; this would be true for any $\varphi$, since one always has $\varphi^{-1}(G) = G$ if $G$ is the codomain of $\varphi$. Where surjectivity is used is in the last displayed equation (which is incorrect as written), namely: you have to choose $g'$ such that $\varphi(g') = g$ (this is where surjectivity is used), and then one finds that $\varphi^{-1}(gE) = g'\varphi^{-1}(E)$ (using the fact that $\varphi$ is an endomorphism). The rest of the chain of equalities then holds, but with $g'$ ... – Matt E Dec 21 '11 at 20:32