[proofplan] We use the continued-fraction Markov partition for the Gauss map. Each finite cylinder is mapped bijectively by an iterate of $G$ onto the full interval, and the inverse branches have uniformly bounded distortion with respect to the Gauss density. If $A$ is invariant and has positive measure, a density point of $A$ has shrinking cylinders on which $A$ occupies asymptotically full conditional measure. Full-branch bounded distortion transfers this local fact to a global lower bound for $\mu_G(A)$, forcing $\mu_G(A)=1$. [/proofplan]

[step:Introduce the continued-fraction branches and cylinders] Let $\rho:(0,1)\to(0,\infty)$ denote the Gauss density \begin{align*} \rho(x)=\frac{1}{\log 2}\frac{1}{1+x}. \end{align*} Let $\mathbb N=\{1,2,3,\dots\}$ denote the set of positive integers. For each $n\in\mathbb N$, define the interval \begin{align*} I_n=\left(\frac{1}{n+1},\frac{1}{n}\right) \end{align*} and define the inverse branch $h_n:(0,1)\to I_n$ by \begin{align*} h_n(y)=\frac{1}{n+y}. \end{align*} Then $G(h_n(y))=y$ for every $y\in(0,1)$. For a word $a=(a_1,\dots,a_k)\in\mathbb N^k$, define the inverse branch $h_a:(0,1)\to(0,1)$ by \begin{align*} h_a=h_{a_1}\circ h_{a_2}\circ\cdots\circ h_{a_k}. \end{align*} Define the corresponding $k$-cylinder $C_a\subset(0,1)$ by \begin{align*} C_a=h_a((0,1)). \end{align*} The set $C_a$ is an open interval, and $G^k:C_a\to(0,1)$ is a bijection with inverse $h_a$. The endpoints of all cylinders form a [countable set](/page/Countable%20Set), so they have $\mu_G$-measure zero. [/step]

[step:Prove that the Gauss measure is invariant under the Gauss map] We prove that $G$ is $\mu_G$-preserving. Let $B\in\mathcal B((0,1))$ be a Borel set. Since the intervals $I_n$ are pairwise disjoint and cover $(0,1)$ up to the countable endpoint set $\{1/n:n\in\mathbb N\}$, which has $\mu_G$-measure zero, we have \begin{align*} \mu_G(G^{-1}(B))=\sum_{n=1}^{\infty}\mu_G(h_n(B)). \end{align*} For each $n\in\mathbb N$, use the substitution $x=h_n(y)=1/(n+y)$. The one-dimensional change-of-variables formula gives \begin{align*} d\mathcal L^1(x)=|h_n'(y)|\,d\mathcal L^1(y)=\frac{1}{(n+y)^2}\,d\mathcal L^1(y). \end{align*} Therefore \begin{align*} \mu_G(h_n(B))=\frac{1}{\log 2}\int_B \frac{1}{1+h_n(y)}\frac{1}{(n+y)^2}\,d\mathcal L^1(y). \end{align*} Since $1+h_n(y)=1+1/(n+y)=(n+y+1)/(n+y)$, this becomes \begin{align*} \mu_G(h_n(B))=\frac{1}{\log 2}\int_B \frac{1}{(n+y)(n+y+1)}\,d\mathcal L^1(y). \end{align*} The integrand is non-negative, so Tonelli's theorem permits summing the branch integrals inside the integral. The telescoping identity \begin{align*} \sum_{n=1}^{\infty}\frac{1}{(n+y)(n+y+1)}=\sum_{n=1}^{\infty}\left(\frac{1}{n+y}-\frac{1}{n+y+1}\right)=\frac{1}{1+y} \end{align*} holds for every $y\in(0,1)$. Hence \begin{align*} \mu_G(G^{-1}(B))=\frac{1}{\log 2}\int_B \frac{1}{1+y}\,d\mathcal L^1(y)=\mu_G(B). \end{align*} Thus $G$ is $\mu_G$-preserving. [/step]

[step:Record the bounded distortion estimate for inverse branches] We prove a uniform distortion estimate for the densities obtained by pushing $\mu_G|_{C_a}$ forward under $G^k$. [claim:Uniform distortion for continued-fraction cylinders] There exists a constant $D=16$ such that for every $k\in\mathbb N$, every word $a\in\mathbb N^k$, and all $y,z\in(0,1)$, \begin{align*} \frac{\rho(h_a(y))|h_a'(y)|}{\rho(h_a(z))|h_a'(z)|}\le D. \end{align*} [/claim] [proof] For every word $a=(a_1,\dots,a_k)$, the map $h_a$ is a fractional [linear map](/page/Linear%20Map) of the form \begin{align*} h_a(y)=\frac{p_k+p_{k-1}y}{q_k+q_{k-1}y} \end{align*} for integers $p_k,p_{k-1},q_k,q_{k-1}$ with $q_k\ge q_{k-1}\ge 0$ and $q_k\ge1$. Hence \begin{align*} |h_a'(y)|=\frac{1}{(q_k+q_{k-1}y)^2}. \end{align*} Thus, for $y,z\in(0,1)$, \begin{align*} \frac{|h_a'(y)|}{|h_a'(z)|}=\left(\frac{q_k+q_{k-1}z}{q_k+q_{k-1}y}\right)^2\le \left(\frac{q_k+q_{k-1}}{q_k}\right)^2\le4. \end{align*} Since $h_a(y),h_a(z)\in(0,1)$, the Gauss density satisfies \begin{align*} \frac{\rho(h_a(y))}{\rho(h_a(z))}=\frac{1+h_a(z)}{1+h_a(y)}\le2. \end{align*} Also $\rho(z)/\rho(y)\le2$ for all $y,z\in(0,1)$. Combining the first two estimates gives \begin{align*} \frac{\rho(h_a(y))|h_a'(y)|}{\rho(h_a(z))|h_a'(z)|}\le8. \end{align*} The weaker constant $D=16$ will be used below to compare normalised conditional densities with $\mu_G$; it follows immediately from this estimate. [/proof] [/step]

[step:Compare pushed-forward conditional measures with the Gauss measure]Fix $k\in\mathbb N$ and $a\in\mathbb N^k$. Define a probability measure $\nu_a$ on $((0,1),\mathcal B((0,1)))$ by \begin{align*} \nu_a(B)=\frac{\mu_G(C_a\cap G^{-k}(B))}{\mu_G(C_a)} \end{align*} for every $B\in\mathcal B((0,1))$. Using the change of variables $x=h_a(y)$, the measure transformation is \begin{align*} d\mathcal L^1(x)=|h_a'(y)|\,d\mathcal L^1(y). \end{align*} Therefore \begin{align*} \nu_a(B)=\frac{\int_B \rho(h_a(y))|h_a'(y)|\,d\mathcal L^1(y)}{\int_0^1 \rho(h_a(y))|h_a'(y)|\,d\mathcal L^1(y)}. \end{align*} By the distortion estimate from the previous step and the fact that $d\mu_G(y)=\rho(y)\,d\mathcal L^1(y)$, the Radon-Nikodym derivative $d\nu_a/d\mu_G$ exists and satisfies \begin{align*} \frac{1}{16}\le \frac{d\nu_a}{d\mu_G}(y)\le16 \end{align*} for $\mu_G$-almost every $y\in(0,1)$. In particular, for every Borel set $B\subset(0,1)$, \begin{align*} \nu_a(B)\ge \frac{1}{16}\mu_G(B). \end{align*}[/step]

[guided]The point of introducing $\nu_a$ is to measure what a set looks like after zooming out from the cylinder $C_a$ by the map $G^k$. Since $G^k:C_a\to(0,1)$ is bijective with inverse $h_a$, every integral over $C_a$ can be rewritten as an integral over $(0,1)$. For a Borel set $B\subset(0,1)$, the set $C_a\cap G^{-k}(B)$ is exactly $h_a(B)$ up to endpoints, and those endpoints have $\mu_G$-measure zero. Using the substitution $x=h_a(y)$, the one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) transforms by \begin{align*} d\mathcal L^1(x)=|h_a'(y)|\,d\mathcal L^1(y). \end{align*} Hence the numerator of $\nu_a(B)$ is \begin{align*} \mu_G(C_a\cap G^{-k}(B))=\int_B \rho(h_a(y))|h_a'(y)|\,d\mathcal L^1(y). \end{align*} The denominator is obtained by taking $B=(0,1)$: \begin{align*} \mu_G(C_a)=\int_0^1 \rho(h_a(y))|h_a'(y)|\,d\mathcal L^1(y). \end{align*} Thus $\nu_a$ has density proportional to the weight $y\mapsto \rho(h_a(y))|h_a'(y)|$. The bounded distortion estimate says that this weight cannot vary by more than a fixed multiplicative factor on the whole interval $(0,1)$, independently of the cylinder. Since $\rho(y)$ itself varies by at most a factor of $2$ on $(0,1)$, normalising the weight produces a density with respect to $\mu_G$ bounded above and below by a universal constant. With the explicit constant used here, this gives \begin{align*} \frac{1}{16}\le \frac{d\nu_a}{d\mu_G}(y)\le16 \end{align*} for $\mu_G$-almost every $y\in(0,1)$. Therefore every Borel set $B$ satisfies \begin{align*} \nu_a(B)\ge \frac{1}{16}\mu_G(B). \end{align*} This lower bound is the mechanism that converts almost-full measure inside one small cylinder into almost-full measure globally.[/guided]

[step:Choose shrinking cylinders around a density point of an invariant set] Let $A\in\mathcal B((0,1))$ satisfy \begin{align*} \mu_G(G^{-1}(A)\triangle A)=0. \end{align*} Assume first that $\mu_G(A)>0$. By the [Lebesgue density theorem](/theorems/894) for finite Borel measures on intervals, applied to the measure $\mu_G$ whose density is bounded above and below by positive constants with respect to $\mathcal L^1$, there exists a point $x\in A$ which is not a cylinder endpoint and such that, for every sequence of intervals $J_m\subset(0,1)$ containing $x$ with lengths tending to $0$, \begin{align*} \frac{\mu_G(A\cap J_m)}{\mu_G(J_m)}\to1. \end{align*} Here we are citing a result not yet in the wiki: the Lebesgue density theorem for finite Borel measures with locally bounded positive density. For each $k\in\mathbb N$, let $C_k(x)$ be the unique $k$-cylinder containing $x$. Since the continued-fraction cylinders separate all points outside the countable endpoint set, the intervals $C_k(x)$ shrink to $x$. Therefore \begin{align*} \frac{\mu_G(A\cap C_k(x))}{\mu_G(C_k(x))}\to1. \end{align*} [/step]

[step:Use invariance to transfer local density to the whole interval] For each $k\in\mathbb N$, write $C_k(x)=C_{a(k)}$ for the unique word $a(k)\in\mathbb N^k$ defining that cylinder. Since $A$ is invariant modulo $\mu_G$, \begin{align*} \mu_G(A\triangle G^{-1}(A))=0. \end{align*} The measure-preserving property proved above implies that preimages of $\mu_G$-null sets are $\mu_G$-null. We prove by induction that \begin{align*} \mu_G(A\triangle G^{-k}(A))=0 \end{align*} for every $k\in\mathbb N$. The case $k=1$ is the assumed invariance. If it holds for $k=m$, then \begin{align*} A\triangle G^{-(m+1)}(A)\subset (A\triangle G^{-1}(A))\cup G^{-1}(A\triangle G^{-m}(A)). \end{align*} Both sets on the right have $\mu_G$-measure zero, so the induction closes. Hence \begin{align*} \frac{\mu_G(A\cap C_{a(k)})}{\mu_G(C_{a(k)})}=\frac{\mu_G(C_{a(k)}\cap G^{-k}(A))}{\mu_G(C_{a(k)})}=\nu_{a(k)}(A). \end{align*} The previous step gives $\nu_{a(k)}(A)\to1$, so $\nu_{a(k)}(A^c)\to0$. The lower comparison estimate gives \begin{align*} \nu_{a(k)}(A^c)\ge \frac{1}{16}\mu_G(A^c). \end{align*} Letting $k\to\infty$ yields $\mu_G(A^c)=0$. Thus $\mu_G(A)=1$ whenever $\mu_G(A)>0$. [/step]

[step:Conclude that every invariant set has measure zero or one] If $\mu_G(A)=0$, there is nothing to prove. If $\mu_G(A)>0$, the previous step proves $\mu_G(A)=1$. Therefore every Borel set $A\subset(0,1)$ satisfying $\mu_G(G^{-1}(A)\triangle A)=0$ has $\mu_G(A)\in\{0,1\}$. This is precisely ergodicity of the measure-preserving system $((0,1),\mathcal B((0,1)),\mu_G,G)$. [/step]