I will begin this question by mentioning what the terms mean in the statement.
I have a probability space $(X,\mathcal{B},\mu)$ and a measurable map $T:X\to X$. I assume that $T$ is bijective, and $T^{-1}$ is also measurable. Suppose in addition that $X$ is a bounded metric space with metric $d$. By uniformly expanding, I mean there is a constant $C>1$ such that $d(Tx,Ty)\ge C d(x,y)$ for all $x,y\in X$.
Exactness has two formulations in ergodic theory; I will state them both. Firstly, we say that a nonsingular map $T$ is exact if $A\in \bigcap_{n\ge 0}T^{-n}(\mathcal{B})$ implies $\mu(A)\in \{0,1\}$. If $T$ is measure-preserving, then $T$ is exact if one has $T(A)\in \mathcal{B}$ for all $A\in \mathcal{B}$, and if $\mu(A)>0$, then $\lim_{n\to \infty}\mu(T^nA)=1$.
Now that the definitions have been stated, is it true that if $T:X\to X$ is as described in the first paragraph, then it is exact? To me, it seems as if it should be true intuitively if considering $X$ to be some interval and $\mu$ to be the Lebesgue measure (from the second definition of exactness), however I am unsure if it is true in the generality above ($X$ bounded metric space, $\mu$ a probability measure).
Thanks in advance!