# Haar measure- $T-$ invariant

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?

• The Haar measure is the Lebesgue measure of $\Bbb{R}$ then $d T(r) = 2 dr$ – reuns Dec 1 '20 at 21:10
• @reuns What do you mean by $d$? – 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.
• Thanks for answering, but on $[0,1]$ Haar measure is Lebesgue's measure up to scalar multiplication, right? – Or Shahar Dec 1 '20 at 23:45
• $[0,1]$ 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. – Ruy Dec 1 '20 at 23:46
• Yes, I ment for $G$, thank you! – Or Shahar Dec 1 '20 at 23:47