Let $\mathscr{G}=(G,\mathcal{B},\mu,T)$ be a measure-preserving dynamical system, and let $G$ be a locally compact topological group.

Let $(G,*)=\mathbb{R}/\mathbb{Z}$, and let $T:G\rightarrow G$, $T(r)=2r=r+r$$($mod $1)$.

Let $m$ be the Haar measure on $G$.

Is $m$ $T-$invariant?

I tried to show it myself, but ran into a problem because the Haar measure is invariant under multiplication, which means $+$.

And moreover, if the answer is yes, then Haar is unique?

  • $\begingroup$ The Haar measure is the Lebesgue measure of $\Bbb{R}$ then $d T(r) = 2 dr$ $\endgroup$ – reuns Dec 1 '20 at 21:10
  • $\begingroup$ @reuns What do you mean by $d$? $\endgroup$ – Or Shahar Dec 1 '20 at 21:12

Yes, $m$ is $T$-invariant even though Lebesgue's measure on $\mathbb R$ is not invariant under the transformation $x\mapsto 2x$. The reason is that, by definition, $m$ is $T$-invariant iff $$ m(T^{-1}(E)) = m(E), $$ for every measurable set $E$, and it so happens that, at least for all small enough arcs $A\subseteq G$, one has that $T^{-1}(A)$ consist of two arcs, each of which has half the length of $A$, and hence, together, they have the right measure.

Regarding uniqueness, it is a well known result that Haar measure is always unique (regardless of invariance by whatever transformations) up to scalar multiplication.

  • $\begingroup$ Thanks for answering, but on $[0,1]$ Haar measure is Lebesgue's measure up to scalar multiplication, right? $\endgroup$ – Or Shahar Dec 1 '20 at 23:45
  • $\begingroup$ $[0,1]$ is not a group. If you are asking whether Haar measure on $G$ is the same as Lebesgue's measure, then the answer is yes. $\endgroup$ – Ruy Dec 1 '20 at 23:46
  • $\begingroup$ Yes, I ment for $G$, thank you! $\endgroup$ – Or Shahar Dec 1 '20 at 23:47

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