Any $x\in(0,1)$ can be written as a (regular) continued fraction $$ x = \cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}} = [a_1,a_2,a_3,\dotsc] $$ An irrational number has a unique expansion, while a rational number has exactly two. Defining \begin{align} a_1(x)&=\left\lfloor\frac{1}{x}\right\rfloor & T(x) &= \begin{cases} \frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor & \text{if } x\neq0\\ 0 & \text{if } x=0 \end{cases} \end{align} we can compute all the partial quotients $a_i(x)=a_1(T^{i-1}(x))$ (that is, until $T^{i-1}(x)=0$).
It can be shown that the map $T$ (called the Gauss map) admits an ergodic invariant absolutely continuous measure with density $$ d\mu = \frac{1}{\ln(2)}\,\frac{dx}{1+x} $$
I have trouble understanding the proof of ergodicity found in the article A continued fraction titbit by M. Keane.
He associates to each $\alpha\in(0,1)$ a pair $(a,c)$ in $$ \Delta=\left\{ (a,c)\in\mathbb{R}^2 : a>0,c>0,a+c<1 \right\} $$ such that $\alpha$ is a root of $p(X)=aX^2+\sqrt{1-4ac}X-c$. The set $\Delta$ is chosen in such a way that the other root of $p(X)$ is less than $-1$. Then he notes that $T$ induces a map $S$ on $\Delta$ $$ S(a,c)=(c,a+n\sqrt{1-4ac}-cn^2) $$ where $n$ is maximal such that $a+n\sqrt{1-4ac}-cn^2>0$ and then he proves that there is an $S$ invariant absolutely continuous measure on $\Delta$ with density $$ d\nu=\frac{da\,dc}{\sqrt{1-4ac}} $$
Then he notes that the subset of $\Delta$ corresponding to a given $\alpha\in(0,1)$ $$ F(\alpha)=\left\{ (a,c)\in\Delta : a\alpha^2+\sqrt{1-4ac}\alpha-c=0 \right\} $$ is just the line segment between $(0,\alpha)$ and $(\frac{1}{1+\alpha},\frac{\alpha}{1+\alpha})$ and that the other root $\tilde{\alpha}$ of $p(X)$ corresponds to $$ P(\omega)=\left\{ (a,c)\in\Delta : a\left(-\frac{1}{\omega}\right)^2+\sqrt{1-4ac}\left(-\frac{1}{\omega}\right)-c=0 \right\} $$ which is the line segment between the points $(\omega,0)$ and $(\frac{\omega}{1+\omega},\frac{1}{1+\omega})$, where $\omega=-\frac{1}{\tilde{\alpha}}$.
Writing $d\nu$ in terms of $\alpha$ and $\omega$ and integrating over $\omega$ he then finds the Gauss measure (up to normalisation) as an invariant measure for $T$.
To prove ergodicity he says that the "future" $\sigma$-algebra (by which I assume he means the $\sigma$-algebra of the images of $S$) $\mathcal{F}$ is the union of sets $F(\alpha)$ while the "past" $\sigma$-algebra (by which I assume he means the $\sigma$-algebra of the preimages of $S$) $\mathcal{P}$ is the union of sets $P(\omega)$. He then says that an invariant subset should be a member of both and that this implies that it or its complement has zero measure (which gives ergodicity).
Here is my issue: I believe that the families of lines $F(\alpha)$ and $P(\omega)$ form a double ruling of $\Delta$, which then implies that if a set can be written both as a union of $F(\alpha)$ and as a union of $P(\omega)$, then it is either $\Delta$ or $\varnothing$, but for example $\mathbb{Q}\cap(0,1)$ is a $T$-invariant set and it doesn't correspond to either.