= Solution
For a probability <measure-preserving system>, <ergodicity> means that every measurable set $A$ satisfying $T^{-1}A=A$ modulo null sets has measure zero or one. Equivalently, the only invariant functions are almost everywhere constant. The <pointwise ergodic theorem> states that, for $f\in L^1(\mu)$,
$$
\frac1N\sum_{j=0}^{N-1}f(T^jx)\longrightarrow\mathbb E(f\mid\mathcal I)(x)\quad\hbox{almost everywhere},
$$
where $\mathcal I$ is the invariant sigma-algebra; on a probability space the convergence also holds in $L^1$. For an <ergodic transformation>, the limit is the constant $\int f\,d\mu$.
For the <Gauss continued-fraction map>, the <Gauss measure> is the probability measure
$$
\boxed{d\nu(x)=\frac{dx}{(\log2)(1+x)},\qquad 0\leq x\leq1.}
$$
Its density integrates to one and is bounded above and below by positive constants, so it has the same null sets as <Lebesgue measure>. Apart from countably many endpoints, the inverse branches are $h_a(y)=1/(a+y)$, $a=1,2,\ldots$. For a bounded measurable test function $F$, substitution on each branch gives
$$
\begin{aligned}
\int F(Tx)\,d\nu(x)
&=\frac1{\log2}\sum_{a=1}^\infty\int_0^1\frac{F(y)}{(a+y)(a+y+1)}\,dy\\
&=\frac1{\log2}\int_0^1F(y)\sum_{a=1}^\infty\left(\frac1{a+y}-\frac1{a+y+1}\right)\,dy\\
&=\frac1{\log2}\int_0^1\frac{F(y)}{1+y}\,dy=\int F\,d\nu.
\end{aligned}
$$
Thus $T$ is a <measure-preserving transformation>. Its definition at zero and the rational branch endpoints does not affect these identities.
For <ergodicity>, we give the full shrinking-cylinder argument. A digit cylinder with first $n$ partial quotients $a_1,\ldots,a_n$ is an interval $I=h((0,1))$, where $h=h_{a_1}\circ\cdots\circ h_{a_n}$ and $T^n\circ h$ is the identity. The <continued fraction convergents> give
$$
h(y)=\frac{p_n+p_{n-1}y}{q_n+q_{n-1}y},\qquad
|h'(y)|=\frac1{(q_n+q_{n-1}y)^2},\qquad
|I|=\frac1{q_n(q_n+q_{n-1})}.
$$
Here $q_n\geq q_{n-1}>0$ after the initial step, and the recurrence $q_n=a_nq_{n-1}+q_{n-2}$ makes $q_n$ tend to infinity at least at the Fibonacci rate. Hence the nested cylinders of every irrational point shrink to that point. The <bounded distortion of continued-fraction cylinders> is uniform: the ratio of the largest to the smallest value of $|h'|$ is at most four. Since $1/(1+h(y))$ ranges between $1/2$ and $1$, the density
$$
w(y)=\frac{|h'(y)|}{(\log2)(1+h(y))}
$$
has maximum-to-minimum ratio at most eight. Consequently every measurable $E\subseteq(0,1)$ satisfies
$$
\frac{\nu(h(E))}{\nu(I)}\geq\frac18\operatorname{Leb}(E)\geq\frac{\log2}{8}\nu(E).
$$
The last inequality uses $d\nu/dx\leq1/\log2$.
Now suppose $T^{-1}A=A$ modulo null sets and $\nu(A)>0$. Choose an irrational <density point> $x\in A$, and let $I_n$ be its successive digit cylinders. The <Lebesgue density theorem> applies to these intervals even though they need not be centered: an interval of length $\ell$ containing $x$ lies in the centered interval of radius $\ell$, so its relative complement tends to zero. Comparability of the two densities then gives
$$
\frac{\nu(I_n\setminus A)}{\nu(I_n)}\longrightarrow0.
$$
But invariance gives $\mathbf1_A(h(y))=\mathbf1_A(y)$ almost everywhere for each inverse branch of $T^n$. Pulling the exceptional null sets through $h$ is legitimate since $h$ has a positive smooth Jacobian on this fixed branch. Thus the distortion estimate with $E=A^c$ gives
$$
\frac{\nu(I_n\setminus A)}{\nu(I_n)}=\frac{\nu(h(A^c))}{\nu(I_n)}\geq\frac{\log2}{8}\nu(A^c).
$$
Letting $n\to\infty$ forces $\nu(A^c)=0$. Therefore \b[the Gauss continued-fraction map is ergodic for the <Gauss measure>].
For an irrational $x$, the successive <continued fraction> digits satisfy $a_j(x)=\lfloor1/T^{j-1}x\rfloor$. The event that a digit equals two is the interval $(1/3,1/2]$, up to null endpoints. Apply the <pointwise ergodic theorem> to its indicator:
$$
\begin{aligned}
\frac1N|\{1\leq j\leq N:a_j(x)=2\}|
&\longrightarrow\nu((1/3,1/2])\\
&=\frac{\log(3/2)-\log(4/3)}{\log2}
=\frac{\log(9/8)}{\log2}
=\boxed{\log_2 9-3}.
\end{aligned}
$$
This holds for almost every $x$ with respect to both the <Gauss measure> and <Lebesgue measure>, since they have the same null sets.
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