This exercise is from the book Foundations of Ergodic Theory by Viana, p. 9, exercise 1.2.5. part a.)
Let $(X,\mathcal{F},\mu)$ be a probability space and $f:X\to X$ be a $\mu$-preserving mapping in the sense that $\forall A\in\mathcal{F}:\mu(A) = \mu(f^{-1}(A))$. Assume that $A\in\mathcal{F}:\mu(A) > 0$ and that $\left(n_j\right)_{j=1}^\infty$ is the sequence of non-negative integers such that $\mu(A\cap f^{-n_j}(A)) > 0$ for all $j=1,2,\dots$.
The task is the show that for any increasing sequence $k_1<k_2<\cdots$ there exists $j>i\geq 1$ such that $\mu(A\cap f^{-(k_j - k_i)}(A)) > 0$. A given hint is to first show that for every $N\in\mathbb{N}$ such that $\frac{1}{N} < \mu(A)$ there exists $j\in\mathbb{N}$ such that $0\leq n_j\leq N$ and $\mu(A\cap f^{-n_j}(A)) > 0$.
A minor detail which has occurred to me is that instead of "any" sequence $k_1<k_2<\cdots$ we should probably only look at subsequences of $(n_j)_{j\in\mathbb{N}}$. Since it is not true that if $(a_n)_{n=1}^\infty, (b_n)_{n=1}^\infty$ are any two sequences of non-negative integers, then $\exists n\in\mathbb{N}:\exists j>i: a_n = b_j - b_i$. But I digress.
As of now I have not made any progress in the main task or the given hint. Namely,
(Progress on hint:) I have not managed to deduce any relationship with connects increasing power of $f$ to decreasing lower bound of $\mu(A\cap f^{-n_j}(A))$. Poincarés recurrence theorem gives us that $\mu$ almost every point of $A$ visits $A$ for infinitely many powers of $f$. Therefore $B := \limsup_{n\to\infty} A\cap f^{-n}(A)$ has a non-zero measure and by Borel-Cantelli lemma $\sum_{n=1}^\infty \mu(A\cap f^{-n}(A)) = +\infty$.
(Progress on the main task:) While we can add as many pre-images of $f$ to $\mu$'s argument, $\mu(A) = \mu(f^{-M}(A)),\forall M\geq 0$, I have not managed to connect this to the difference between $n_j,n_i$. Something tells me that we would like to use a proof by contradiction by bounding $A\cap f^{-n_j}(A)$ above by some $A\cap f^{-(n_{j_1} - n_{j_2})}(A)$, but so far this has not lead to anything. The biggest problem that I think I am having is that I have no way of knowing with which indices the points of $A\cap f^{-n_j}(A)$ may visit $A$ again, and hence it is difficult to say anything about $\mu(A\cap f^{-(n_j - n_i)}(A))$.
We should not just consider subsequences of $(n_j)_j$: the increasing sequence $k_\bullet$ should be arbitrary.
To be clear, in the exercise it is implied that $n_1<n_2<\cdots$ is the complete enumeration of all naturals $n$ with $\mu(A\cap f^{-n}(A))>0$, and such a strictly increasing sequence exists by recurrence theorems.
Let’s think about that hint. Motivation-wise, I just reflected that they suggested a bound on $N$ for a reason and I also want to show that there is a nontrivial intersection, this felt like a “find a finite cover” problem.
Take $N\in\Bbb N$ so large such that $1/N<\mu(A)$. Suppose to the contradiction that there is no $j$ with $1\le n_j<N$. Then it follows that the family of sets $(f^{-k}(A))_{1\le k\le N}$ have $\mu$-null intersections. Because otherwise we get, for some $1\le k<k’\le N$, $0<\mu(f^{-k}(A)\cap f^{-k’}(A))=\mu(A\cap f^{-(k’-k)}(A))$ thus there exists $j,n_j=k’-k$ but then $1\le n_j<N$ will hold.
Ok, well that then implies: $$\mu\left(\bigcup_{k=1}^N f^{-k}(A)\right)=\sum_{k=1}^N\mu(f^{-k}(A))=N\cdot\mu(A)>1$$Contradicting the fact we are working in a probability space. Therefore, such $j$ with $1\le n_j<N$ always exist. Done!
Now suppose to the contrary that there is a counterexample $k_1<k_2<\cdots$.
The hint’s purpose is not as a lemma, but rather to make us observe that proving the statement of the hint is very similar to proving the intended result. Note that the hint is essentially just helping us to prove the intended result applied to $k_1:=1,k_2:=2,\cdots$.
Since $\mu(f^{-k_i}(A)\cap f^{-k_j}(A))=\mu(A\cap f^{-(k_j-k_i)}(A))=0$ for all $1\le i<j$, if $N:=1+\lceil \mu(A)^{-1}\rceil$ then we get: $$\mu\left(\bigcup_{n=1}^N f^{-k_n}(A)\right)=\sum_{n=1}^N\mu(f^{-k_n}(A))=N\cdot\mu(A)>1$$
Which is the same contradiction!