Let $X=[0,1)$, $\tau(x)=<\frac{1}{x}>$ (the fractional part of 1/x) for $x \neq 0$ and $\tau(0)=0$. Let $d\mu = \frac{dx}{\log2 (1+x)}$. Show that $\tau$ is a measure-preserving transformation.
$\textit{Work so far:}$
To show that $\tau$ is a measure-preserving transformation of $X$, we need to show that
(i) $\tau^{-1}(E)$ is measurable whenever $E$ is measurable
(ii) $\mu(\tau^{-1}(E)) = \mu(E)$ for any measurable $E \subseteq X$
For (i) we have a few different characterizations for measurable $E$. The one I tried working with was that $E$ is measurable iff $E$ is almost an $F_\sigma$ or $G_\delta$ set. If $E$ is, say, almost $G_\delta$, then we can write $$E = G \cup Z$$ where $G$ is a $G_\delta$ set and $Z$ is a set of measure zero. In which case, $$ \tau^{-1}(E) = \tau^{-1}(G \cup Z) = \tau^{-1}(G) \cup \tau^{-1}(Z)$$ ...but I'm not sure how to show that $\tau^{-1}(G)$ and $\tau^{-1}(Z)$ are measurable given how $\tau$ is defined.
For (ii), if we assume that $E$ is measurable then $$\mu(\tau^{-1}(E)) = \int_{\tau^{-1}(E)} 1 d\mu= \int_X \chi_{\tau^{-1}(E)}d\mu = \int_X \chi_E(\tau(x)) d\mu $$ $$=\int_E <\frac{1}{x}> \frac{1}{\log 2} \frac{dx}{1+x}$$ Since $<\frac{1}{x}> \leq 1$ then by monotinicity of the Lebesgue integral, $$\cdots \leq \int_E \frac{1}{\log 2} \frac{dx}{1+x} = \mu(E)$$ How might we show "$\geq$"?
There is also a hint given: $$\sum_{k=1}^\infty \frac{1}{(x+k)(x+k+1)} = \frac{1}{1+x}$$ ...but it has not yet proven useful.
This is problem 8 in chapter 6 of "Real Analysis" by Stein and Shakarchi. I'm trying to do this problem along with #10 so that I can better understand the connection between measure and continued fractions. Thanks!
Hints: It suffices to look at intervals $I$ as the generate the Borel sets (and you show easily that $\tau^{-1} I$ is a union of intervals, whence measurable). For the second you need to integrate over these inverse images of $I$. For this note that $$ \tau^{-1} (y) =\{\frac{1}{n+y} : n\geq 1\}$$ and use this to carry out the integrals. You should arrive at the above-mentioned telescopic sum.