A measure is defined as:
$\mu ([a,b]) = \int_a^b \frac{1}{1+x} = ln \left( \frac{1+b}{1+a} \right)$
The function $\Phi_k$ is defined as:
- $\Phi_k = kx$ for: $x \in \left(0, \frac{1}{k} \right)$
- $\Phi_k = \{ \frac{1}{x} \} = \frac{1}{x} \mod 1 $ for: $x \in \left( \frac{1}{k}, 1 \right)$
Show that the measure $\mu(x)$ is invariant for $\Phi_k$.
By invariant I mean that: for every $[a,b]$ it is true that: $\mu (\Phi_k^{-1}([a,b])) = \mu([a,b])$
I found a simpler, similar example, explained right here: https://personalpages.manchester.ac.uk/staff/Charles.Walkden/ergodic-theory/lecture14.pdf
I know that function $\Phi_k$ looks like that:
In the pink points the graphs (blue, green, purple, black) get cut and the rest of $\Phi_k$ is exchanged for the red part.
I was unable to emulate the thinking form the link into my case. Any help would be much appreciated.
It suffices to show that
$$ \int_{0}^{1} \varphi(\Phi_k(x)) \, \mu(\mathrm{d}x) = \int_{0}^{1} \varphi(x) \, \mu(\mathrm{d}x) $$
for all test functions $\varphi \in C([0,1])$. However,
\begin{align*} \int_{0}^{1} \varphi(\Phi_k(x)) \, \mu(\mathrm{d}x) &= \int_{0}^{\frac{1}{k}} \varphi(kx) \, \frac{\mathrm{d}x}{1+x} + \int_{\frac{1}{k}}^{1} \varphi(\{1/x\}) \, \frac{\mathrm{d}x}{1+x} \\ &= \int_{0}^{1} \varphi(x) \, \frac{\mathrm{d}x}{k+x} + \int_{1}^{k} \varphi(\{x\}) \, \frac{\mathrm{d}x}{x(x+1)} \\ &= \int_{0}^{1} \varphi(x) \, \frac{\mathrm{d}x}{k+x} + \sum_{j=1}^{k-1} \int_{0}^{1} \varphi(x) \, \frac{\mathrm{d}x}{(x+j)(x+j+1)} \\ &= \int_{0}^{1} \varphi(x) \biggl( \frac{1}{k+x} + \sum_{j=1}^{k-1} \frac{1}{(x+j)(x+j+1)} \biggr) \, \mathrm{d}x \\ &= \int_{0}^{1} \varphi(x) \cdot \frac{1}{1+x} \, \mathrm{d}x \\ &= \int_{0}^{1} \varphi(x) \, \mu(\mathrm{d}x). \end{align*}
Therefore the desired conclusion follows.