For $x \in [0,1)$ then the continued fraction representation of $$x=0 + \cfrac{1}{a_1(x)+\cfrac{1}{a_2(x)+\cfrac{1}{a_3(x)+\cfrac{1}{\dots}}}}$$ can be written as $[0; a_1(x), a_2(x), a_3(x), \dotsc]$ where $a_j(x)$ is a positive integer.
Define $F(n_1, n_2, \dots, n_k)$ to be a set of all $x \in [0,1)$ such that $$a_1(x) = n_1, a_2(x) = n_2, \dots, a_k(x) = n_k$$ (first $k$ digits in the continued fraction expansion is $n_1, \dots, n_k$).
Define $\mu$ be the (Gauss) measure, $$\mu(A) = \int_A \frac{1}{\log{2}}\frac{dt}{t+1}.$$
Then $$\mu(F(n_1, n_2, ..., n_k)) = \mu(F(n_k, n_{k-1}, ..., n_1)).$$
I want to verify the result. I study Ergodic Theory of Number using the online material : https://www-fourier.ujf-grenoble.fr/sites/default/files/files/fichiers/lecturenotesdajani2013.pdf
This fact is listed on the page $38$ of the pdf file. I check the book that mentioned as reference, but it is one of the exercise without proof.
I try the case when the length is 2: $$\mu(F(m, n)) = \frac{1}{\log(2)}\int_{F(m, n)} \frac{dt}{t+1} = \frac{\log(x+1)}{\log(2)}|_{F(m, n)}.$$
Since $F(m, n) = \Big[\frac{n}{nm+1}, \frac{n+1}{m(n+1)+1}\Big)$ and $F(n, m) = \Big[\frac{m}{nm+1}, \frac{m+1}{n(m+1)+1}\Big),$ by a simple simplification $$\log(x+1)|_{F(m,n)} = \log\Big(\frac{n + nm+1}{mn+1} \cdot \frac{m(n+1)+1}{n+1+m(n+1)+1}\Big)$$ and $$\log(x+1)|_{F(n, m)} = \log\Big(\frac{m+nm+1}{mn+1} \cdot \frac{n(m+1)+1}{m+1+n(m+1)+1}\Big).$$
So $$\mu(F(m,n)) = \mu(F(n,m)).$$
For general case of length $k$, it is quite hard to find the actual form of $F(n_1, n_2, \dots, n_k)$ and $F(n_k, \dots, n_1)$. So I wonder if there is a more straigth forward way to see why this identity holds.