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$$($mod $1)$.
Let $m$ be the Haar measure on $G$.
Is $m$ $T-$invariant?
I tried to show it myself, but ran into a problem because the Haar measure is invariant under
multiplication, which means $+$.
And moreover, if the answer is yes, then Haar is unique?
Best Answer
Yes, $m$ is $T$-invariant even though Lebesgue's measure on $\mathbb R$ is not invariant under the transformation $x\mapsto 2x$. The reason is that, by definition, $m$ is $T$-invariant iff $$ 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.