I'm trying to solve exercise 2.2.3 b) from Einsiedler and Ward's "Ergodic theory with a view towards number theory". The exercise considers an arbitrary metric space $X$ with a probability measure $\mu$ defined on its $\sigma$-algebra of Borel sets, and a Borel transformation $T$ of $X$ preserving $\mu$. In this setting we want to show, that the set of points $x\in X$ which are in the closure of their orbit $\{T^n(x): n\geqslant 1\}$ has measure 1.
So far I've reduced the problem to showing, that there are no metric probability preserving systems $(X, \mu, T)$ for which $\forall_{x\in X}d(x, T^n(x))\geqslant 1$ for all $n\geqslant 1$. This can be done by considering a possible counterexample $(X, T)$, noticing that for some $\varepsilon>0$ a set of positive measure $A$ must satisfy $$\forall_{a\in A}\forall_{n\geqslant 1}d(a, T^n(a))\geqslant \varepsilon,$$ and then considering the metric space $A$ with the scaled metric $d'=\frac{d}{\varepsilon}$ and scaled measure $\mu'=\frac{\mu}{\mu(A)}$, and the transformation $T'$ defined as $$T'(a)=T^n(a)\quad\text{where $n$ is the smallest positive integer such that} \quad T^n(a)\in A.$$ This transformation (defined almost everywhere by Poincaré recurrence theorem) can be checked to preserve $\mu$.
We can also assume that $T$ is invertible by considering the natural extension of the supposed counterexample $(X, T)$ before applying this reasoning.
Using Poincaré recurrence theorem, we can also see that in this case we must have $\mu'(B(a, 1))=0$ for all points $a\in A$. This clearly gives a contradiction if the space $X$ (or $A$) is separable, but in the general case I don't see a way to proceed.
I am starting to suspect that there might be some missing assumption in the statement, but the question in this generality still seems very interesting. Any ideas or directing to literature (or at least assuring me that the statement of the exercise is actually correct) would be very appreciated.
Update: I had a new idea - we can use A. H. Stone's theorem on the metric space $A$ with the open covering $\{B(a, \frac{1}{3}): a\in A\}$. The theorem tells us, that there exists a family $V_{n, a}$ indexed by $\mathbb{N}\times A$ of sets in $A$ that is a locally finite open covering, and satisfy $V_{n, a}\subset B(a, \frac{1}{3})$, and each subfamily $\{V_{n, a}: a\in A\}$ with fixed $n$ is discrete. This gives us countably many families, whose unions are then a countable covering. Thus one of them has positive measure. Repeating our procedure from before we can "restrict" our system to this family, and then we can make the metric between the members on the family always be 1 (this doesn't change the topology, since the family is discrete), which reduces the exercise to the following:
Show that for any indexing set $S$ and metric spaces $\{(X_s, d_s): s\in S\}$ of diameter at most $\frac{1}{3}$, any probability measure $\mu$ defined on the Borel sets of the space $(X, d)$, where $X=\sum\limits_{s\in S}X_s$ and $$ d(x, y)=d_s(x, y)\text{ if }x, y\in X_s\text{, and }1\text{ if $x$ and $y$ are in different }X_s,$$ and any injective Borel map $T:X\to X$ preserving $\mu$, for some $x\in X_s$ we have $T(x)\in X_s$.
This seems very close to being doable: this space seems to me almost discrete, and on discrete spaces $X$ you cannot define any probability preserving systems - otherwise picking in any way a single point from every orbit would give us a measurable set, whose countably many "translates" by $T$ partition the space, and so the measure of the whole space is either 0 or infinity. (This is a very similar argument to the one showing that the Vitali set is not measurable.)