[proofplan]
We partition $(0,1)$ into the inverse branches of the Gauss map, ignoring the countable endpoint set where the usual formula gives value $0$. On each branch, the inverse map is $u_n(y)=1/(n+y)$, and a one-dimensional change of variables transforms the Gauss density into the summand $\big((n+y)(n+y+1)\big)^{-1}$. Summing these branch contributions gives a telescoping series equal to the original density, which proves invariance for every bounded Borel measurable [test function](/page/Test%20Function).
[/proofplan]
[step:Discard the endpoint set where the branch formula is exceptional]
Let $N=\{1/n:n\in\mathbb{N}\}$. Since $N$ is countable and $\mu_G$ is absolutely continuous with respect to $\mathcal{L}^1$, we have $\mu_G(N)=0$. Therefore the value of $G$ on $N$ does not affect the integral of the bounded measurable function $f\circ G$.
For each $n\in\mathbb{N}$, define the open branch interval
\begin{align*}
I_n=\left(\frac{1}{n+1},\frac{1}{n}\right).
\end{align*}
The intervals $(I_n)_{n\in\mathbb{N}}$ are pairwise disjoint, and
\begin{align*}
(0,1)\setminus N=\bigcup_{n=1}^{\infty} I_n.
\end{align*}
Since $f$ is bounded and $\mu_G$ is a finite measure, the function $f\circ G:(0,1)\to\mathbb{R}$ is absolutely integrable with respect to $\mu_G$. Countable additivity of the integral over the disjoint measurable decomposition above therefore gives
\begin{align*}
\int_0^1 f(G(x))\,d\mu_G(x)=\sum_{n=1}^{\infty}\int_{I_n} f(G(x))\,d\mu_G(x).
\end{align*}
[/step]
[step:Write the Gauss map on each branch and introduce its inverse]
Fix $n\in\mathbb{N}$. On $I_n$, one has $n<1/x<n+1$, so $\lfloor 1/x\rfloor=n$ and
\begin{align*}
G(x)=\frac{1}{x}-n.
\end{align*}
Define the inverse branch as the map $u_n:(0,1)\to I_n$. For $y\in(0,1)$, set
\begin{align*}
u_n(y)=\frac{1}{n+y}.
\end{align*}
Then $u_n$ is a $C^1$ decreasing bijection from $(0,1)$ onto $I_n$, and for every $y\in(0,1)$,
\begin{align*}
G(u_n(y))=y.
\end{align*}
Also,
\begin{align*}
|u_n'(y)|=\frac{1}{(n+y)^2}.
\end{align*}
[/step]
[step:Transform each branch integral to the common variable $y$]
Using the definition of $\mu_G$, the $n$th branch contribution is
\begin{align*}
\int_{I_n} f(G(x))\,d\mu_G(x)
=
\frac{1}{\log 2}
\int_{I_n} \frac{f(G(x))}{1+x}\,d\mathcal{L}^1(x).
\end{align*}
Apply the substitution $x=u_n(y)=1/(n+y)$. Since $u_n$ is a $C^1$ bijection from $(0,1)$ onto $I_n$, the one-dimensional change-of-variables formula gives
\begin{align*}
d\mathcal{L}^1(x)=\frac{1}{(n+y)^2}\,d\mathcal{L}^1(y).
\end{align*}
Moreover,
\begin{align*}
1+u_n(y)=1+\frac{1}{n+y}=\frac{n+y+1}{n+y}.
\end{align*}
Therefore
\begin{align*}
\int_{I_n} f(G(x))\,d\mu_G(x)
=
\frac{1}{\log 2}
\int_0^1
f(y)\frac{1}{(n+y)(n+y+1)}
\,d\mathcal{L}^1(y).
\end{align*}
[guided]
Fix $n\in\mathbb{N}$ and define the branch interval
\begin{align*}
I_n=\left(\frac{1}{n+1},\frac{1}{n}\right).
\end{align*}
The point of passing to the inverse branch is that $G$ is not globally one-to-one, but it is one-to-one on this interval $I_n$. On $I_n$, the integer part of $1/x$ is exactly $n$, so $G(x)=1/x-n$. Solving $y=1/x-n$ for $x$ gives the inverse branch $u_n:(0,1)\to I_n$ defined by
\begin{align*}
u_n(y)=\frac{1}{n+y}.
\end{align*}
This map is continuously differentiable, bijective, and decreasing. Its derivative satisfies
\begin{align*}
u_n'(y)=-\frac{1}{(n+y)^2},
\end{align*}
so the absolute Jacobian in the one-dimensional change of variables is
\begin{align*}
|u_n'(y)|=\frac{1}{(n+y)^2}.
\end{align*}
Now compute the branch integral from the definition of the Gauss measure:
\begin{align*}
\int_{I_n} f(G(x))\,d\mu_G(x)
=
\frac{1}{\log 2}
\int_{I_n}\frac{f(G(x))}{1+x}\,d\mathcal{L}^1(x).
\end{align*}
Substituting $x=u_n(y)$ changes the domain from $I_n$ to $(0,1)$ and changes [Lebesgue measure](/page/Lebesgue%20Measure) by
\begin{align*}
d\mathcal{L}^1(x)=\frac{1}{(n+y)^2}\,d\mathcal{L}^1(y).
\end{align*}
The composition simplifies because $G(u_n(y))=y$. The density also transforms explicitly:
\begin{align*}
1+u_n(y)=1+\frac{1}{n+y}=\frac{n+y+1}{n+y}.
\end{align*}
Thus
\begin{align*}
\frac{1}{1+u_n(y)}\frac{1}{(n+y)^2}
=
\frac{n+y}{n+y+1}\frac{1}{(n+y)^2}
=
\frac{1}{(n+y)(n+y+1)}.
\end{align*}
Hence the transformed branch contribution is
\begin{align*}
\int_{I_n} f(G(x))\,d\mu_G(x)
=
\frac{1}{\log 2}
\int_0^1
f(y)\frac{1}{(n+y)(n+y+1)}
\,d\mathcal{L}^1(y).
\end{align*}
[/guided]
[/step]
[step:Sum the branch densities and telescope]
For each $y\in(0,1)$ and $n\in\mathbb{N}$,
\begin{align*}
\frac{1}{(n+y)(n+y+1)}
=
\frac{1}{n+y}-\frac{1}{n+y+1}.
\end{align*}
Thus, for every $m\in\mathbb{N}$,
\begin{align*}
\sum_{n=1}^{m}\frac{1}{(n+y)(n+y+1)}
=
\frac{1}{1+y}-\frac{1}{m+1+y}.
\end{align*}
Letting $m\to\infty$ gives the pointwise identity
\begin{align*}
\sum_{n=1}^{\infty}\frac{1}{(n+y)(n+y+1)}
=
\frac{1}{1+y}.
\end{align*}
Since $f$ is bounded, define the finite constant
\begin{align*}
M=\sup_{y\in(0,1)}|f(y)|.
\end{align*}
For each $m\in\mathbb{N}$, define the partial-sum integrand $S_m:(0,1)\to\mathbb{R}$ by
\begin{align*}
S_m(y)=f(y)\sum_{n=1}^{m}\frac{1}{(n+y)(n+y+1)}.
\end{align*}
The functions $S_m$ converge pointwise on $(0,1)$ to
\begin{align*}
y\mapsto f(y)\sum_{n=1}^{\infty}\frac{1}{(n+y)(n+y+1)}.
\end{align*}
Moreover, the telescoping estimate gives the domination
\begin{align*}
|S_m(y)|\leq \frac{M}{1+y}
\end{align*}
for every $m\in\mathbb{N}$ and $y\in(0,1)$, and $y\mapsto M/(1+y)$ is integrable on $(0,1)$ with respect to $\mathcal{L}^1$. Applying the [dominated convergence theorem](/theorems/4) to the sequence $(S_m)_{m\in\mathbb{N}}$ yields
\begin{align*}
\int_0^1 f(G(x))\,d\mu_G(x)=\frac{1}{\log 2}\int_0^1 f(y)\sum_{n=1}^{\infty}\frac{1}{(n+y)(n+y+1)}\,d\mathcal{L}^1(y).
\end{align*}
Using the telescoping identity,
\begin{align*}
\int_0^1 f(G(x))\,d\mu_G(x)
=
\frac{1}{\log 2}
\int_0^1 \frac{f(y)}{1+y}\,d\mathcal{L}^1(y).
\end{align*}
[/step]
[step:Identify the recovered integral with integration against $\mu_G$]
By the definition of $\mu_G$,
\begin{align*}
\frac{1}{\log 2}
\int_0^1 \frac{f(y)}{1+y}\,d\mathcal{L}^1(y)
=
\int_0^1 f(y)\,d\mu_G(y).
\end{align*}
Combining this identity with the previous step gives
\begin{align*}
\int_0^1 f(G(x))\,d\mu_G(x)=\int_0^1 f(y)\,d\mu_G(y).
\end{align*}
Renaming the dummy variable $y$ as $x$ gives the stated invariance formula.
[/step]
