[proofplan]
We prove the recurrence statement by applying the multiple recurrence positivity theorem to the indicator function of $B$. The key point is that the time average of the measures of the multiple intersections has a positive lower bound, so not every term in the average can vanish. Once one positive term is found, the corresponding time $r$ gives exactly the intersection required in the theorem.
[/proofplan]
[step:Convert the intersection measure into an integral of shifted indicator functions]
Let $\mathbb{1}_B: X \to \{0,1\}$ denote the indicator function of $B$. Since $B \in \mathcal B$, the function $\mathbb{1}_B$ is $\mathcal B$-measurable, belongs to $L^\infty(X,\mathcal B,\mu)$, and satisfies
\begin{align*}
\int_X \mathbb{1}_B(x)\,d\mu(x)=\mu(B)>0.
\end{align*}
For each $r\in\mathbb N$, define the measurable set
\begin{align*}
A_r:=B\cap T^{-r}B\cap T^{-2r}B\cap\cdots\cap T^{-(k-1)r}B.
\end{align*}
Here $T^{-jr}B$ denotes the preimage $(T^{jr})^{-1}(B)$ under the $jr$-fold iterate $T^{jr}:X\to X$, with $T^0=\operatorname{id}_X$; no invertibility of $T$ is being assumed. Because $T:X\to X$ is measurable and $B\in\mathcal B$, each $T^{-jr}B$ belongs to $\mathcal B$, hence $A_r\in\mathcal B$. For every $x\in X$,
\begin{align*}
\mathbb{1}_{A_r}(x)=\prod_{j=0}^{k-1}\mathbb{1}_B(T^{jr}x),
\end{align*}
where the product is finite and takes values in $\{0,1\}$. Therefore
\begin{align*}
\mu(A_r)=\int_X \prod_{j=0}^{k-1}\mathbb{1}_B(T^{jr}x)\,d\mu(x).
\end{align*}
[/step]
[step:Apply the mean multiple recurrence theorem to the positive set $B$]
We use the [Mean Furstenberg Multiple Recurrence Theorem](/theorems/MEAN-FURSTENBERG-MULTIPLE-RECURRENCE): if $(X,\mathcal B,\mu,T)$ is a probability measure-preserving system, $f:X\to[0,\infty)$ is a bounded measurable function with $\int_X f\,d\mu>0$, and $k\in\mathbb N$, then
\begin{align*}
\liminf_{N\to\infty}\frac{1}{N}\sum_{r=1}^{N}\int_X\prod_{j=0}^{k-1}f(T^{jr}x)\,d\mu(x)>0.
\end{align*}
This is the nontrivial ergodic recurrence input, established separately in its mean functional form, so the present argument is the deduction of the set recurrence statement from that prior theorem. The hypotheses are satisfied with $f=\mathbb{1}_B$: the system is probability measure-preserving by hypothesis, $\mathbb{1}_B$ is bounded and non-negative, and its integral is $\mu(B)>0$. Hence
\begin{align*}
\liminf_{N\to\infty}\frac{1}{N}\sum_{r=1}^{N}\mu(A_r)>0.
\end{align*}
[guided]
We need a prior result that turns the positive measure of $B$ into positivity of multiple return averages. The input is the [Mean Furstenberg Multiple Recurrence Theorem](/theorems/MEAN-FURSTENBERG-MULTIPLE-RECURRENCE). It applies to a probability measure-preserving system and to a bounded non-negative measurable function $f:X\to[0,\infty)$ whose integral is positive, and it gives
\begin{align*}
\liminf_{N\to\infty}\frac{1}{N}\sum_{r=1}^{N}\int_X\prod_{j=0}^{k-1}f(T^{jr}x)\,d\mu(x)>0.
\end{align*}
This theorem is the deep recurrence theorem being used as an external input; the present proof only converts its mean functional conclusion into the displayed set recurrence conclusion.
We verify the hypotheses for $f=\mathbb{1}_B$. The system is probability measure-preserving by the theorem statement. Since $B\in\mathcal B$, the indicator function $\mathbb{1}_B:X\to\{0,1\}$ is measurable and bounded. It is non-negative because its only values are $0$ and $1$. Finally,
\begin{align*}
\int_X \mathbb{1}_B(x)\,d\mu(x)=\mu(B)>0.
\end{align*}
Therefore the mean multiple recurrence theorem gives
\begin{align*}
\liminf_{N\to\infty}\frac{1}{N}\sum_{r=1}^{N}\int_X\prod_{j=0}^{k-1}\mathbb{1}_B(T^{jr}x)\,d\mu(x)>0.
\end{align*}
For each $r\in\mathbb N$, let
\begin{align*}
A_r:=B\cap T^{-r}B\cap T^{-2r}B\cap\cdots\cap T^{-(k-1)r}B,
\end{align*}
where $T^{-jr}B=(T^{jr})^{-1}(B)$ and $T^0=\operatorname{id}_X$. For every $x\in X$, membership in $A_r$ is equivalent to $T^{jr}x\in B$ for every $j\in\{0,\dots,k-1\}$. Hence
\begin{align*}
\mathbb{1}_{A_r}(x)=\prod_{j=0}^{k-1}\mathbb{1}_B(T^{jr}x).
\end{align*}
Integrating this identity with respect to $\mu$ gives
\begin{align*}
\mu(A_r)=\int_X\prod_{j=0}^{k-1}\mathbb{1}_B(T^{jr}x)\,d\mu(x).
\end{align*}
Substituting this into the mean recurrence inequality yields
\begin{align*}
\liminf_{N\to\infty}\frac{1}{N}\sum_{r=1}^{N}\mu(A_r)>0.
\end{align*}
This is stronger than the theorem asks for: it says the average size of the multiple return intersections is bounded away from zero along arbitrarily large initial intervals, so at least one individual intersection must have positive measure.
[/guided]
[/step]
[step:Extract one positive return time from the positive average]
Suppose, for contradiction, that $\mu(A_r)=0$ for every $r\in\mathbb N$. Then for every $N\in\mathbb N$,
\begin{align*}
\frac{1}{N}\sum_{r=1}^{N}\mu(A_r)=0,
\end{align*}
which contradicts
\begin{align*}
\liminf_{N\to\infty}\frac{1}{N}\sum_{r=1}^{N}\mu(A_r)>0.
\end{align*}
Therefore there exists $r\in\mathbb N$ such that $\mu(A_r)>0$. Since $\mathbb N$ starts at $1$, this $r$ satisfies $r\ge 1$, and by the definition of $A_r$ we have
\begin{align*}
\mu\left(B\cap T^{-r}B\cap T^{-2r}B\cap\cdots\cap T^{-(k-1)r}B\right)>0.
\end{align*}
This is the desired conclusion.
[/step]
