Let $T:X \to X$ be homeomorphism on a compact manifold $X$. We say that $\mu$ is $T-$invariant measure if $\mu(T^{-1}(A))=\mu(A)$ for all Borel measurable set $A$.
I read the following sentence on some paper:
If $\mu$ is some $T$-invariant measure and $x\in A$ then, with a frequency that is asymptotically bounded away from zero, the iterates $T^{n}$ will $\epsilon$-approach the support of $\mu$ for any $\epsilon>0.$
Does the above sentence imply $A \cap \text{support}(\mu)\neq \emptyset?$
It's not true in general. For instance, consider $X = \mathbb{T}^{2}$ the two-torus and let $T(x,y) = (x + \alpha, y)$ for some $\alpha \in \mathbb{R} \setminus \mathbb{Q}$. The probability measure $\mu$ obtained from integration on the line $[0,1] \times \{0\}$ is invariant for $T$, but the orbit $\{T^{n}(0,1/2)\}_{n \in \mathbb{N}}$ stays a distance $\frac{1}{2}$ away.
What would be true is ``$\mu$-a.e. $x \in X$ has the property that, with a frequency that is asymptotically bounded away from zero, the iterates $T^{n}$ will $\epsilon$-approach the support of $\mu$ for any $\epsilon > 0$." It follows from the ergodic theorem. (Also, by compactness of $X$, you can decompose $\mu$ into a mixture of ergodic, invariant probability measures --- this is convenient in the proof.)
If you knew $\mu$ was ergodic, then you could strengthen the above statement to ``$\mu$-a.e. $x \in X$ has the property that, for any $y$ in the support of $\mu$, with a frequency that is asymptotically bounded away from zero, the iterates $T^{n}(x)$ will $\epsilon$-approach $y$." The point is when things are ergodic, (most) orbits visit the entire support. (In the non-ergodic case, which parts of the support will be seen depends on the ergodic decomposition.)