A map is said to be topological mixing if given two sets $A$ and $B$ then there exists $N$ such that for all $n>N$
$$f^n(A) \cap B \neq \emptyset.$$
On the other hand, a measure $\mu$ is said to be mixing for a map $f$ if
$$\lim_{n \to \infty} \mu (f^{-n}(A) \cap B) = \mu (A) \mu(B)$$
for any $A,B$ measurable sets.
Let $\mathcal{A}$ be our collection of open sets. If $f$ is continuous, we say $f$ is topologically mixing if for all non-empty $A, B \in \mathcal{A}$, there exists an $N$ such that for all $n>N$,
$$f^n \cap B \ne \emptyset.$$
Let $\mathcal{B}$ be our collection of measurable sets. If $f$ is measurable, we say measure $\mu$ is mixing for $f$ if for all $A, B \in \mathcal{B}$,
$$\lim_{n \to \infty} \mu(f^{-n}(A) \cap B ) = \mu(A)\mu(B).$$
Claim: If $f$ is $\mathcal{A}$-measurable and there is some measure $\mu$ on $\mathcal{A}$ for which $\mu(A)>0$ for all non-empty $A \in \mathcal{A}$ and $\mu$ is mixing for $f$, then $f$ is continuous and topologically mixing for topology $\mathcal{A}$.
Proof: Consider non-empty sets $A, B \in \mathcal{A}$. By the stated properties of $\mu$, we know that $\mu(A)>0$ and $\mu(B)>0$.
Thus $\mu(A)\mu(B)>0$. Since $\mu$ is mixing for $f$, we know by the definition of limits that there exists an $N$ large enough so that for all $n>N$,
$$\mu(f^{-n}(A)\cap B) > 0.$$
This means that for all $n>N$, the set $f^{-n}(A)\cap B$ is non-empty.
I hope this gives you some intuition as to how they are similar!
