If $x=[a_0,a_1,\dots]$ show that $\mu$-almost every $x \in (0,1/N]$ is infinitely recurrent 
Let $G$ be the Gauss map,
  $$G(x)= \begin{cases}
0 & \text{if} \ x=0 \\ 
 \{\frac{1}{x} \}=\frac{1}{x} \ \mathrm{mod} \ 1 & \text{if $0<x\leq 1$}\end{cases}$$
and $\mu$ be the Gauss measure given by integrating the density $\frac{1}{\ln 2}\left(\frac{1}{1+x}\right)$.
Let $N >0$ be a given positive integer. Show that for $\mu$-almost every $x \in (0,\frac{1}{N}]$ there are infinitely many entries $a_i$ in the continued fraction expansion of $x=[a_0,a_1,\dots,a_n,\dots]$ such that $a_i \geq N$.

I believe this can be solved by applying the strong form of the Poincaré recurrence theorem. As $a_n = \left[\frac{1}{G^n(x)}\right]$ if $a_i \geq N$ then $\frac{1}{N+1} \leq G^n(x)\leq \frac{1}{N}$.
We know that $\mu((0,\frac{1}{N}])>0$ (easily shown). Therefore $\mu$-almost every $x \in B=\left(0\,\frac{1}{N}\right]$ is infinitely recurrent to $B$.

Prove the conclusion of the previous part holds for $\mu$-almost every $x \in [0,1]$.

How does this work. I think it would be the same proof, but this does not feel right (and seems to easy).
These are the versions of the Poincaré recurrence that I know. 


and this may be useful

Also you may assume that $G$ is ergodic wrt $\mu$.
 A: While the question being asked can be settled using ergodicity of the Gauss map and the ergodic theorem, as I explained in comments to the question, it can also be explained using only the ergodicity alone (which is itself a nontrivial feature of the Gauss map, so it's not like we're getting something for free). I'll state the result in a general context, as an ergodic refinement of the Poincare recurrence theorem.
Theorem: If $(X,\mu,T)$ is an ergodic dynamical system with $0 < \mu(X) < \infty$, then for each measurable subset $A \subset X$ with $\mu(A) > 0$, $\mu$-almost all $x \in X$ have $T^n(x) \in A$ for infinitely many $n$.
The distinction from the Poincare recurrence theorem is that the conclusion is about $\mu$-almost all $x \in X$ rather than $\mu$-almost all $x \in A$.
Proof: For $m \geq 0$, let $X_m$ be the set of $x \in X$ such that $T^n(x) \in A$ for some $n \geq m$. In set-theoretic notation, $X_m = \bigcup_{n \geq m} T^{-n}(A)$.  This set is measurable and $\mu(X_m) > 0$ since 
$\mu(X_m) \geq \mu(T^{-m}(A)) = \mu(A) > 0$.
Check $T^{-1}(X_m) \subset X_m$ (that is, if $T(x) \in X_m$ then $x \in X_m$). Since $X_m$ and $T^{-1}(X_m)$ have equal $\mu$-measure, the complement $X_m  - T^{-1}(X_m)$ has measure $0$. Outside of $X_m  - T^{-1}(X_m)$, check that $\chi_{X_m}(x) = \chi_{X_m}(Tx)$ by looking separately at $x \in T^{-1}(X_m)$ (both sides are $1$) and at $x \in X - X_m$ (both sides are $0$). Therefore $\chi_{X_m}(x) = \chi_{X_m}(Tx)$ for $\mu$-almost all $x \in X$, so $\chi_{X_m} \colon X \rightarrow \{0,1\}$ is an almost invariant function on $(X,\mu,T)$. Therefore the set $\{x \in X : \chi_{X_m}(x) = 1\} = X_m$ is an almost invariant set, and when the dynamical system is ergodic its almost invariant sets have measure $0$ or full measure. As $\mu(X_m) > 0$, we must have $\mu(X_m) = \mu(X)$: $\mu$-almost all points of $X$ are in $X_m$, so the complement of $X_m$ has $\mu$-measure zero.
What is the complement of $X_m$? It's the $x \in X$ such that $T^n(x) \not\in A$ for all $n \geq m$. A countable union of measure $0$ sets has measure $0$, so $\mu$-almost all $x \in X$ are outside the complement of every $X_m$, meaning $\mu$-almost all $x \in X$ are in every $X_m$.  To say $x$ lies in every $X_m$ means $T^n(x) \in A$ for some $n \geq m$ no matter what $m$ is, and that's another way of saying $T^n(x) \in A$ for infinitely many $n$.  Thus $\mu$-almost all $x \in X$ have their $T$-orbit $\{x,T(x),T^2(x),\dots\}$ meet $A$ infinitely often provided $\mu(A) > 0$.
