Find borel measure for on $X=[1,\infty)$ such that for every $s\geq 1$ the following holds $\mu(T_sA)=\mu(A)$ for every measureable $A$.

Question

For every $s\geq1$ we define the transformation $T_s(x)=s\cdot x$. Find a borel measure which is not the zero measure on $X=[1,\infty)$ such that for every $s\geq 1$ the following holds $\mu(T_sA)=\mu(A)$ for every measureable $A$.
Hint: Log(ab)=log(a)+log(b) and use lebesgue stieltjes measure.
I feel really lost here any hint or more would really appreciated.

Answer 1

Define $\mu$ by $$ \mu(A) = \int_A \frac1x \,dx. $$ It suffices to check the condition for intervals $[a,b)$. We can calculate $$ \mu[sa,sb) = \int_{sa}^{sb} \frac1x \,dx = \log(sb) - \log(sa) = \log(sb/sa) = \log(b/a) = \log(b) - \log(a) = \int_{a}^{b} \frac1x \,dx = \mu[a,b). $$