[proofplan]
We prove the forward direction by saturating one positive-measure set under backward iterates and showing this saturation is almost invariant. Ergodicity then forces the saturation to have full measure, so any other positive-measure set must meet one of its backward iterates in positive measure; a small shift ensures the iterate is positive. For the converse, we test the criterion against an almost invariant set and its complement, then use induction to show all backward iterates of the almost invariant set remain inside it modulo null sets, contradicting the 2-set intersection condition unless the complement is null.
[/proofplan]
[step:Show that backward orbit saturations of positive sets have full measure]
Let $B \in \mathcal{B}$ satisfy $\mu(B) > 0$. Define the backward orbit saturation of $B$ to be the measurable set
\begin{align*}
\widetilde{B} := \bigcup_{m=0}^{\infty} T^{-m}(B).
\end{align*}
The set $\widetilde{B}$ belongs to $\mathcal{B}$ because $T$ is measurable and $\mathcal{B}$ is closed under countable unions.
Since
\begin{align*}
T^{-1}(\widetilde{B})
= T^{-1}\left(\bigcup_{m=0}^{\infty} T^{-m}(B)\right)
= \bigcup_{m=0}^{\infty} T^{-(m+1)}(B)
\subseteq \widetilde{B},
\end{align*}
and since $T$ is measure-preserving, we have
\begin{align*}
\mu(T^{-1}(\widetilde{B})) = \mu(\widetilde{B}).
\end{align*}
Therefore
\begin{align*}
\mu(\widetilde{B} \setminus T^{-1}(\widetilde{B}))
= \mu(\widetilde{B}) - \mu(T^{-1}(\widetilde{B}))
= 0,
\end{align*}
and hence
\begin{align*}
\mu(T^{-1}(\widetilde{B}) \triangle \widetilde{B}) = 0.
\end{align*}
Thus $\widetilde{B}$ is $T$-invariant modulo null sets. Since $B \subseteq \widetilde{B}$, we have $\mu(\widetilde{B}) \geq \mu(B) > 0$. By ergodicity, every almost invariant measurable set has measure $0$ or $1$, so
\begin{align*}
\mu(\widetilde{B}) = 1.
\end{align*}
[guided]
We start from a positive-measure set $B$ and collect all points whose forward orbit eventually lands in $B$. Formally, define
\begin{align*}
\widetilde{B} := \bigcup_{m=0}^{\infty} T^{-m}(B).
\end{align*}
Each set $T^{-m}(B)$ is measurable because $T$ is measurable, and the countable union is measurable because $\mathcal{B}$ is a $\sigma$-algebra.
The important property of $\widetilde{B}$ is that it is almost invariant. Indeed,
\begin{align*}
T^{-1}(\widetilde{B})
= T^{-1}\left(\bigcup_{m=0}^{\infty} T^{-m}(B)\right)
= \bigcup_{m=0}^{\infty} T^{-(m+1)}(B)
\subseteq \bigcup_{m=0}^{\infty} T^{-m}(B)
= \widetilde{B}.
\end{align*}
Thus the only possible difference between $\widetilde{B}$ and $T^{-1}(\widetilde{B})$ is the part of $\widetilde{B}$ that is not pulled back again. Because $T$ is measure-preserving,
\begin{align*}
\mu(T^{-1}(\widetilde{B})) = \mu(\widetilde{B}).
\end{align*}
Since $T^{-1}(\widetilde{B}) \subseteq \widetilde{B}$, subtracting measures gives
\begin{align*}
\mu(\widetilde{B} \setminus T^{-1}(\widetilde{B}))
= \mu(\widetilde{B}) - \mu(T^{-1}(\widetilde{B}))
= 0.
\end{align*}
The reverse difference $T^{-1}(\widetilde{B}) \setminus \widetilde{B}$ is empty, so
\begin{align*}
\mu(T^{-1}(\widetilde{B}) \triangle \widetilde{B}) = 0.
\end{align*}
Therefore $\widetilde{B}$ is invariant modulo null sets.
Now ergodicity applies: an almost invariant measurable set must have measure $0$ or full measure. Since $B \subseteq \widetilde{B}$ and $\mu(B) > 0$, the zero-measure alternative is impossible. Hence
\begin{align*}
\mu(\widetilde{B}) = 1.
\end{align*}
[/guided]
[/step]
[step:Use full-measure saturation to force a positive intersection at a positive time]
Let $A, B \in \mathcal{B}$ satisfy $\mu(A) > 0$ and $\mu(B) > 0$. Apply the previous step to $A$ and define
\begin{align*}
\widetilde{A} := \bigcup_{m=0}^{\infty} T^{-m}(A).
\end{align*}
Then $\mu(\widetilde{A}) = 1$. Since $\mu(B) > 0$, we have
\begin{align*}
\mu(B \cap \widetilde{A}) = \mu(B) > 0.
\end{align*}
Because
\begin{align*}
B \cap \widetilde{A}
= B \cap \left(\bigcup_{m=0}^{\infty} T^{-m}(A)\right)
= \bigcup_{m=0}^{\infty} \left(B \cap T^{-m}(A)\right),
\end{align*}
countable subadditivity implies that there exists $m \geq 0$ such that
\begin{align*}
\mu(B \cap T^{-m}(A)) > 0.
\end{align*}
To obtain a positive iterate, apply the same argument with $T^{-1}(A)$ in place of $A$. Since $T$ is measure-preserving,
\begin{align*}
\mu(T^{-1}(A)) = \mu(A) > 0.
\end{align*}
Hence there exists $m \geq 0$ such that
\begin{align*}
\mu(B \cap T^{-m}(T^{-1}(A))) > 0.
\end{align*}
Since $T^{-m}(T^{-1}(A)) = T^{-(m+1)}(A)$, setting $n := m+1$ gives $n \geq 1$ and
\begin{align*}
\mu(T^{-n}(A) \cap B) > 0.
\end{align*}
This proves the 2-set criterion under ergodicity.
[guided]
Let $A$ and $B$ be measurable sets with positive measure. We need to produce a positive integer $n$ for which $T^{-n}(A)$ meets $B$ in positive measure.
Apply the previous step to the set $A$. Define
\begin{align*}
\widetilde{A} := \bigcup_{m=0}^{\infty} T^{-m}(A).
\end{align*}
The previous step gives
\begin{align*}
\mu(\widetilde{A}) = 1.
\end{align*}
Since $B$ has positive measure and $\widetilde{A}$ has full measure,
\begin{align*}
\mu(B \cap \widetilde{A}) = \mu(B) > 0.
\end{align*}
But
\begin{align*}
B \cap \widetilde{A}
= B \cap \left(\bigcup_{m=0}^{\infty} T^{-m}(A)\right)
= \bigcup_{m=0}^{\infty} \left(B \cap T^{-m}(A)\right).
\end{align*}
If every set $B \cap T^{-m}(A)$ had measure $0$, then countable subadditivity would give $\mu(B \cap \widetilde{A}) = 0$, contradicting the previous positive-measure identity. Therefore there is some $m \geq 0$ such that
\begin{align*}
\mu(B \cap T^{-m}(A)) > 0.
\end{align*}
This gives a nonnegative time. The criterion requires a positive time, so we shift the target set once before applying the same argument. Because $T$ is measure-preserving,
\begin{align*}
\mu(T^{-1}(A)) = \mu(A) > 0.
\end{align*}
Applying the preceding argument to $T^{-1}(A)$ and $B$ gives some $m \geq 0$ with
\begin{align*}
\mu(B \cap T^{-m}(T^{-1}(A))) > 0.
\end{align*}
The preimage identity for iterates gives
\begin{align*}
T^{-m}(T^{-1}(A)) = T^{-(m+1)}(A).
\end{align*}
Now define $n := m+1$. Then $n \geq 1$ and
\begin{align*}
\mu(T^{-n}(A) \cap B) > 0.
\end{align*}
Thus ergodicity implies the stated 2-set intersection condition.
[/guided]
[/step]
[step:Propagate almost inclusion through all backward iterates]
Assume now that the 2-set criterion holds. Let $A \in \mathcal{B}$ satisfy
\begin{align*}
\mu(T^{-1}(A) \triangle A) = 0
\end{align*}
and $\mu(A) > 0$. Define the null set
\begin{align*}
N := T^{-1}(A) \setminus A.
\end{align*}
Then $\mu(N) = 0$, so $T^{-1}(A) \subseteq A$ modulo a null set.
We claim that for every integer $n \geq 1$,
\begin{align*}
\mu(T^{-n}(A) \setminus A) = 0.
\end{align*}
For $n=1$, this is the definition of $N$. Suppose the assertion holds for some $n \geq 1$. Since $T$ is measure-preserving,
\begin{align*}
\mu(T^{-1}(T^{-n}(A) \setminus A)) = \mu(T^{-n}(A) \setminus A) = 0.
\end{align*}
Moreover,
\begin{align*}
T^{-(n+1)}(A) \setminus A
\subseteq
\left(T^{-1}(T^{-n}(A) \setminus A)\right) \cup \left(T^{-1}(A) \setminus A\right).
\end{align*}
Both sets on the right have measure $0$, so finite subadditivity gives
\begin{align*}
\mu(T^{-(n+1)}(A) \setminus A) = 0.
\end{align*}
By induction, the claim holds for every $n \geq 1$.
[guided]
Now assume the 2-set criterion and let $A$ be almost invariant with positive measure. Almost invariance means
\begin{align*}
\mu(T^{-1}(A) \triangle A) = 0.
\end{align*}
In particular, the one-sided difference
\begin{align*}
N := T^{-1}(A) \setminus A
\end{align*}
has measure $0$. Thus $T^{-1}(A)$ is contained in $A$ up to a null set.
We need this containment not only for one iterate, but for every positive iterate. We prove by induction that for each integer $n \geq 1$,
\begin{align*}
\mu(T^{-n}(A) \setminus A) = 0.
\end{align*}
For $n=1$, this is exactly the statement $\mu(N)=0$.
Assume the statement holds for some fixed $n \geq 1$, so
\begin{align*}
\mu(T^{-n}(A) \setminus A) = 0.
\end{align*}
Because $T$ is measure-preserving, the preimage of this null set is again null:
\begin{align*}
\mu(T^{-1}(T^{-n}(A) \setminus A)) = \mu(T^{-n}(A) \setminus A) = 0.
\end{align*}
We now compare $T^{-(n+1)}(A)$ with $A$. If a point lies in $T^{-(n+1)}(A) \setminus A$, then it lies in $T^{-1}(T^{-n}(A))$ but not in $A$. There are two possibilities: either it already fails to lie in $T^{-1}(A)$, or it lies in $T^{-1}(A)$ but outside $A$. This gives the set inclusion
\begin{align*}
T^{-(n+1)}(A) \setminus A
\subseteq
\left(T^{-1}(T^{-n}(A) \setminus A)\right) \cup \left(T^{-1}(A) \setminus A\right).
\end{align*}
The first set on the right is null by the induction hypothesis and measure preservation; the second set is null by almost invariance. Finite subadditivity therefore gives
\begin{align*}
\mu(T^{-(n+1)}(A) \setminus A) = 0.
\end{align*}
This completes the induction and proves that every backward iterate $T^{-n}(A)$ is contained in $A$ modulo a null set.
[/guided]
[/step]
[step:Apply the 2-set criterion to rule out a positive-measure complement]
Assume for contradiction that $\mu(A^c) > 0$, where $A^c := X \setminus A$. Since $\mu(A) > 0$ and $\mu(A^c) > 0$, the 2-set criterion applied to the pair $A$ and $A^c$ gives an integer $n \geq 1$ such that
\begin{align*}
\mu(T^{-n}(A) \cap A^c) > 0.
\end{align*}
But the previous step gives $\mu(T^{-n}(A) \setminus A) = 0$, and
\begin{align*}
T^{-n}(A) \cap A^c = T^{-n}(A) \setminus A.
\end{align*}
Therefore
\begin{align*}
\mu(T^{-n}(A) \cap A^c) = 0,
\end{align*}
contradicting the criterion. Hence $\mu(A^c) = 0$, and since the ambient measure is normalized, $\mu(A)=1$.
Thus every almost invariant measurable set of positive measure has full measure. Equivalently, every almost invariant measurable set has measure $0$ or $1$, so $T$ is ergodic. This proves the converse direction and completes the proof.
[guided]
Assume, for contradiction, that the almost invariant set $A$ has complement of positive measure:
\begin{align*}
\mu(A^c)>0,\qquad A^c:=X\setminus A.
\end{align*}
Since $\mu(A)>0$ by the current case and the 2-set criterion is assumed, applying the criterion to the pair $A$ and $A^c$ gives an integer $n\geq 1$ such that
\begin{align*}
\mu(T^{-n}(A)\cap A^c)>0.
\end{align*}
But the previous step proved the stronger containment statement
\begin{align*}
\mu(T^{-n}(A)\setminus A)=0
\end{align*}
for every $n\geq 1$. Because $A^c=X\setminus A$, the two sets are identical:
\begin{align*}
T^{-n}(A)\cap A^c=T^{-n}(A)\setminus A.
\end{align*}
Hence
\begin{align*}
\mu(T^{-n}(A)\cap A^c)=0,
\end{align*}
contradicting the positive-measure conclusion from the criterion. Therefore $\mu(A^c)=0$. Since $\mu(X)=1$, this gives $\mu(A)=1$.
We have shown that every almost invariant set $A$ with $\mu(A)>0$ has $\mu(A)=1$; the remaining alternative is $\mu(A)=0$. Thus every almost invariant measurable set has measure $0$ or $1$, which is precisely ergodicity. This proves the converse implication.
[/guided]
[/step]
