Let's consider $T: (0,\hspace{-0.05 in}1] \rightarrow [0,\hspace{-0.05 in}1)$ which is defined as $Tx = \{ \frac{1}{x} \}$.
It can be shown that Lebesgue measure is not invariant under this map.
However, there exists a measure which has that feature: $\mu([a,\hspace{-0.03 in}b]) = \log_{2}{\frac{1+a}{1+b}}$.
How can one show that it has that feature, i.e.
$\mu(T^{-1}\hspace{-0.03 in}(A))= \mu(A)$ for all measurable subsets $A$ of $[0,\hspace{-0.05 in}1]$?
Any help would be much appreciated
My favourite tool for this question is the transfer operator.
Let:
$$g(x) := \frac{1}{\ln (2)} \frac{1}{1+x}.$$
Then $\mu (a) = \int_A g(x) dx$, so $g$ is the density of $\mu$ with respect to the Lebesgue measure. Then $\mu$ is $T$-invariant if and only if, for any measurable and bounded function $f$,
$$\int_{[0,1]} f \circ T \ d \mu = \int_{[0,1]} f \ d \mu,$$
that is,
$$\int_{[0,1]} f \circ T (x) \cdot g(x) \ d x = \int_{[0,1]} f (x) \cdot g(x) \ d x.$$
But the transformation $T$ has countably many branches $\psi_n (x) := 1/x-n$, each being defined on $[1/(1+1),1/n]$ with image $[0,1]$, so that:
$$\int_{[1/(1+1),1/n]} f \circ T (x) \cdot g(x) \ d x = \int_{[0,1]} f(y) \cdot \frac{1}{(n+y)^2} g \left( \frac{1}{n+y} \right) \ dy,$$
and :
$$\int_{[0,1]} f \circ T (x) \cdot g(x) \ d x = \int_{[0,1]} f(y) \cdot \sum_{n \geq 1} \frac{1}{(n+y)^2} g \left( \frac{1}{n+y} \right) \ dy = \int_{[0,1]} f (y) \cdot g(y) \ d y.$$
The later inequality holds for any $f$ if and only if:
$$g(y) = \sum_{n \geq 1} \frac{1}{(n+y)^2} g \left( \frac{1}{n+y} \right),$$
which, knowing the explicit formula for $g$, you can check.
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The transfer operator here is the dual operator to the composition by $T$. By the change of variables formula, for $\mathcal{C}^2$ transformation of the interval with countably many strictly monotonous branches, it can be expressed as:
$$\mathcal{L} g (y) = \sum_{x : \ T(x) = y} \frac{1}{|T' (x)|} g(x).$$
A measure with density $g$ with respect to the Lebesgue measure is invariant if and only if $\mathcal{L}(g) = g$.