boundary of limsup and liminf for sequence of averages with convergent subsequence I'm having difficulty following the proof of Lemma III.1.7 in "Probability and Stochastics" (Cinlar). The lemma is:

Let $(x_n)$ be a sequence of positive numbers and denote $\overline x_n = \frac{1}{n} \sum_{i=1}^n x_i$
Let $N = (n_k)_{k \in \mathbb{N}}$ be a subsequence of $\mathbb{N}$ such that $\lim_{k \rightarrow \infty} \frac{n_{k+1}}{n_k} = r > 0$
If the sequence $(\overline x_n)$ converges along $N$ to $x$ (so the subsequence $(x_{n_k})$ converges to $x$)
Then $\frac{x}{r} \leq \lim \inf \overline x_n \leq \lim \sup \overline x_n \leq rx$

To begin with, this chain of inequalities implies that $\frac{x}{r} \leq rx$ which implies $r^2 \geq 1$. Allowing this is a small oversight with the intention that we should have $r \geq 1$, I'm having trouble further in the proof.
We begin by observing that for $n_k \leq n < n_{k+1}$, noting positivity of $x_n$ and that each summand (in the averages) occurs in the sum on the right we get
$\frac{n_k}{n_{k+1}} \overline x_{n_k} \leq \overline{x}_n \leq \frac{n_{k+1}}{n_k} \overline x_{n_{k+1}}$
Now the argument states simply that taking limits ($n \rightarrow \infty$, which implies also $k \rightarrow \infty$ to satisfy $n_k \leq n < n_{k+1}$) yields $LHS \rightarrow \frac{x}{r}$ and $RHS \rightarrow rx$; and true enough (with $r$ and $x_n$ positive). But where do the $\lim \inf$ and $\lim \sup$ fit into this?
 A: Basically, since $\frac{n_k}{n_{k+1}} \overline x_{n_k} \leq \overline{x}_n \leq \frac{n_{k+1}}{n_k} \overline x_{n_{k+1}}$ holds for all $n \in \mathbb{N}$ (in the sense that for a given $n$ there always exists a $k_n$ such that it holds), it'll still be true for $n$s belonging to a sequence $(m_l)_l$ such that the subsequence $(\overline{x}_{m_l})_l$ tends to $\liminf \overline{x}_n$, and same for $\limsup \overline{x}_n$ (I specify "tends" and not "converges" since at this step they could be equal to $+\infty$, at least for the latter, but that's just some detail). The existence of those subsequences is a good exercise if you haven't shown this before.
Taking the limit when $l \to \infty$ in the resulting inequalities (which I won't write because it's index-bonanza, what with all the $x$s, $k$s, $n$s, $m$s, $l$s, I prefer writing subsequences in the form $(u_{\varphi(n)})_n$ for that reason... but that's just a preference!) will get you $\frac{x}{r} \leq \lim \inf \overline x_n$ for $\liminf$ and $\lim \sup \overline x_n \leq rx$ for $\limsup$, and the middle inequality $\lim \inf \overline x_n \leq \lim \sup \overline x_n$ is true in general, thus we do have the desired result.
EDIT: To try and clarify what I said in view of your valid apprehension (I will admit what I said above is probably not clear at all), let me write down my anwser more thoroughly. I'll switch to subsequence maps for the notations, if I may:
Let $\varphi$ be the subsequence map corresponding to $N$, meaning: $\overline{x}_{\varphi(n)} \xrightarrow[n \to \infty]{} x$ and $u_n := \frac{\varphi(n+1)}{\varphi(n)} \xrightarrow[n \to \infty]{} r$.
As you said, for each $n \in \mathbb{N}$ there exists a unique $k(n) \in \mathbb{N}$ such that: $\varphi(k(n)) \leq n < \varphi(k(n)+1)$, and for that $k(n)$ we get:
$$\frac{\varphi(k(n))}{\varphi(k(n)+1)} \overline x_{\varphi(k(n))} \leq \overline{x}_n \leq \frac{\varphi(k(n)+1)}{\varphi(k(n))} \overline x_{\varphi(k(n)+1)}$$
A bit more succintly, with the notation $u_n$ I introduced:
$$\frac{\overline x_{\varphi(k(n))}}{u_{k(n)}} \leq \overline{x}_n \leq u_{k(n)}\, \overline x_{\varphi(k(n)+1)}$$
Now, define $\psi$ another subsequence map such that $\overline{x}_{\psi(n)} \xrightarrow[n \to \infty]{} \liminf \overline{x}$. Along that subsequence, we have:
$$\frac{\overline x_{\varphi(k(\psi(n)))}}{u_{k(\psi(n))}} \leq \overline{x}_{\psi(n)} \leq u_{k(\psi(n))}\, \overline x_{\varphi(k(\psi(n))+1)}$$
Even if it means extracting a subsequence of $\psi(n)$ by taking out $n$s for which $k(\psi(n))$ are duplicates (except for one copy of course), we can assume WLOG that $k \circ \psi$ is also a subsequence map ($k$ is not a subsequence map though in general), this is possible due to the definition of the $k(n)$s.
We are on the final stretch: since $k \circ \psi$ is a subsequence map, we obtain that: $u_{k(\psi(n))} \xrightarrow[n \to \infty]{} r$, that: $\overline{x}_{\varphi(k(\psi(n)))} \xrightarrow[n \to \infty]{} x$ and that $\overline{x}_{\varphi(k(\psi(n))+1)} \xrightarrow[n \to \infty]{} x$, hence we can take the limit in the inequalities, and:
$$\frac{x}{r} \leq \liminf \overline{x}_n \leq rx$$
Since the same can be done for $\limsup$, and we always have $\liminf \leq \limsup$, we get the desired result.
