Q:

Sawyer's proof that a certain lim sup of sets has full measure

Given $(X, \mathcal{A}, \mu)$ a probability space, let $\mathcal{F}$ be a family of $\mu$-invariant measurable functions, closed under composition, with the following property:
If $A$ is a measurable set such that $A = T^{-1} (A)$ modulo null-sets for every $T$ in $\mathcal{F}$, then either $\mu(A) = 0$ or $\mu(A) = 1$.
The objective is to conclude the following Lemma: 

Given a collection $\left\{ E_{n} \right\}_{n \geq 1}$  of measurable
  sets satisfying $\sum_{n=1}^{\infty} \mu (E_{n}) = \infty$, there
  exists $T_{1}, \ldots, T_{n}, \ldots \in \mathcal{F}$ such that:
\begin{equation*} \mu \left( \bigcap_{N \geq 1} \bigcup_{n \geq N}
 T^{-1}_{n} (E_{n})\right) = \mu \left( \limsup_{n \geq 1} T^{-1}_{n}
 (E_{n}) \right) \end{equation*}
has full measure in $X$.

This strongly resembles of Borel-Cantelli's theorem, and the proof provided indeed follows the same line of thought as in the theorem's proof. It relies on the following result: 
Under the hypotheses in the definition, given any two measurable sets $A, B$ and any constant $\theta > 1$, one may find $T \in \mathcal{F}$ satisfying $\mu \left(A \cap T^{-1} (B) \right) \leq \theta \mu (A) \mu (B)$.
Thus, if $A_{n} = E_{n}^{\mathrm{c}}$ and we choose arbitrary $T_{1} \in \mathcal{F}$, $\theta_{1} > 1$, substituting A for $T_{1}^{-1} (A_{1})$ and $B$ for $A_{2}$, the aforementioned result yields $T_{2} \in \mathcal{F}$ such that $$\mu \left( T^{-1}_{1} (A_{1}) \cap T^{-1}_{2} (A_{2}) \right) \leq \theta_{1} \mu (A_{1}) \mu (A_{2}).$$
In general, we proceed inductively: at step $m$ choose arbitrary $\theta_{m} > 1$ and take $A$ as $T^{-1}_{1} (A_{1}) \cap \ldots \cap T^{-1}_{m} (A_{m})$, $B$ as $A_{m+1}$, obtaining $T_{m+1}$ such that:
\begin{equation}
\mu \left( \bigcap_{n=1}^{m+1} T^{-1}_{n} (A_{n}) \right) \leq \left( \prod_{n=1}^{m} \theta_{n} \right) \prod_{n=1}^{m+1} \left( 1 - \mu(E_{n}) \right) 
\end{equation}
Now, the above inequality is exactly of the kind used in Borel-Cantelli's proof. As long as we pick $(\theta_{n})_{n \geq 1}$ in such a way that its product converges, the above estimate allows us to conclude that $\mu \left( \bigcup_{n=1}^{\infty} T^{-1}_{n} (E_{n}) \right) = 1$.
The problem, however, is that we'd like this conclusion to actually hold for $\bigcup_{n = N}^{\infty} T^{-1}_{n} (E_{n})$. One cannot simple repeat the previous construction starting from $T^{-N} (A_{N})$, since this could possibly yield a different sequence $\left( T_{n} \right)_{n \geq N}$ at each time.
The author stops the proof right here, claiming that this conclusion is sufficient to:
(1) Choose integers $N_{k}$ with the property that $$\mu \left( \bigcup_{n = N_{k}}^{N_{k+1}} T^{-1}_{n} (E_{n}) \right) \geq 1 - 1/k^2$$ and,
(2) From (1) conclude that $$\bigcup_{n = N}^{\infty} T^{-1}_{n} (E_{n})$$ has full measure for every $N$.
Now, I can see why (1) implies (2). However, I haven't been able to conclude (1). Does it follow simply from the inequality obtained and some measure-theoretical argument or could it use some specific property of the family $\mathcal{F}$? Any help would be appreciated.
PS: This is Lemma 2 of the article Maximal Inequalities of Weak Type, by S. Sawyer, http://www.jstor.org/stable/1970516, p. 165, although I've changed the notation to a more contemporary one.

A:

By the monotone continuity of $\mu$,
\begin{align}
   \mu\left(\bigcup_{n=1}^\infty T_n^{-1}(E_n)\right) &=
      \lim_{m\to\infty} \mu\left(\bigcup_{n=1}^mT_n^{-1}(E_n)\right) \;.
\end{align}
Once you find $T_n$ such that the left-hand side is $1$, you can choose $N_1$ such that
\begin{align}
   \mu\left(\bigcup_{n=1}^{N_1} T_n^{-1}(E_n)\right) &\geq
      1- 2^{-1}
\end{align}
(I use $1-2^{-k}$ instead of $1-1/k^2$ for clarity at this step.)
Now, forget the choices of $T_n$ for $n>N_1$, which you will not need.  Find (using the same procedure) new $T_{N_1+1}, T_{N_1+2}, \ldots, T_{N_2}$ with $N_2>N_1$ such that
\begin{align}
   \mu\left(\bigcup_{n=N_1}^{N_2} T_n^{-1}(E_n)\right) &\geq
      1- 2^{-2} \;.
\end{align}
The same procedure is applicable because we still have $\sum_{n>N_1}\mu(E_n)=\infty$.
Repeating this, you get an entire sequence $T_1, T_2, \ldots$ satisfying
\begin{align}
   \mu\left(\bigcap_{N\geq 1}\bigcup_{n\geq N} T_n^{-1}(E_n)\right) &= 1 \;.
\end{align}

