A theorem in the paper "Noncommuting Random Products" by Furstenberg I have a question concerning the proof of theorem 2.5 at page 395 of the paper Noncommuting Random Products, by H. Furstenberg, Trans. Amer. Math. Soc., 1963.

The statement is as follows: 

Let $\mu$ be a probability measure on a group $G$. Let $M$ be a locally compact $G$-space, which does not admit a stationary measure for $\mu$, that is, there is no $\nu$ on $M$ such that $\mu*\nu=\nu$. Let $Z_n$ be a $\mu$-process on $M$, that is $Z_i$ are $M$ valued random variables such that the conditional distribution of $Z_{n+1}$ given $Z_{n},\dots,Z_0$ is $\mu*\delta_{Z_n}$. Let $\Delta\subset M$ be compact. Define $n(k)$ as the index $n$ for which $Z_n\in\Delta$ for the $k$-th time. Then $n(k)/k\to\infty$ with probability $1$.

The proof starts with stating that it is enough to show that for each compactly supported $\psi$ we have that with probability $1$
\begin{equation}
\frac{1}{n}\sum_0^{n-1}\psi(Z_k)\to 0
\end{equation}
This I do understand. However, Furstenberg proceeds to say that if this formula does not hold for some $\psi'$, then there exists a subsequence $n_i$ such that
\begin{equation}
\frac{1}{n_i}\sum_0^{n_i-1}\psi'(Z_k)\to \alpha\neq 0,
\end{equation}
and he seems to assume in what follows that neither $n_i$ nor $\alpha$ are random.

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Question:
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How is it that the second equation is the negation of the first? Is Furstenberg implicitly using here some "strong law of large numbers"?
During the proof Furstenberg picked (and fixed) a sample sequence $Z_n$. More precisely, 

1). It has been proved that (2.8) $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} F(Z_k) \to 0$ holds with probability $1$ for any $F=\tau f- f$. 

2). Now suppose (2.9) $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} \psi'(Z_k) \to 0$ fails for some $\psi'$. It means there is a positive probability such that the limit is $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} \psi'(Z_k) \not\to 0$. 

3). Taking the intersection of these two probabilities, one concludes that there is a positive probability (and take a sample sequence $Z_n$ from this) such that (2.8) holds along this sample sequence for all $F=\tau f- f$, while (2.9) gives a nonzero limit.

For this fixed sample sequence $Z_n$, passing to subsequences of $n_k$ if necessary, one gets a positive linear functional that is nonzero. At this moment we are done with the sample sequence and consider the measure induced by the linear functional.

This is my understanding. Hope it makes sense.
During the proof Furstenberg picked (and fixed) a sample sequence $Z_n$. More precisely, 

1). It has been proved that (2.8) $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} F(Z_k) \to 0$ holds with probability $1$ for any $F=\tau f- f$. 

2). Now suppose (2.9) $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} \psi'(Z_k) \to 0$ fails for some $\psi'$. It means there is a positive probability such that the limit is $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} \psi'(Z_k) \not\to 0$. 

3). Taking the intersection of these two probabilities, one concludes that there is a positive probability (and take a sample sequence $Z_n$ from this) such that (2.8) holds along this sample sequence for all $F=\tau f- f$, while (2.9) gives a nonzero limit.

For this fixed sample sequence $Z_n$, passing to subsequences of $n_k$ if necessary, one gets a positive linear functional that is nonzero. At this moment we are done with the sample sequence and consider the measure induced by the linear functional.

This is my understanding. Hope it makes sense.