Equivalent condition for mixing Let $(X,\mathscr{A},\mu)$ be a measurable space and $T$ measure preserving map and define:
$T$ is mixing iff $\lim_{n\to\infty}\mu(A\cap T^{-n}B)=\mu(A)\mu(B)$ for all $A,B\in\mathscr{A}$.
How to prove that mixing property is equivalent with:
$\lim_{n\to\infty}\langle U_T^n f,g\rangle=\langle f,1\rangle\langle 1,g\rangle$ for all $f,g$ in dense subset of $L^2(\mu)$?
Here, $(U_Tf)(x)=f(Tx)$. 
One implication is direct (take characteristic functions for $f$ and $g$), but I'm having trouble to prove the second one. 
Any help or hint is welcome. Thanks in advance. 
 A: We denote the two statements we are interested in as follows, 

Mixing: A measure space $(X,\mathscr{A},\mu)$ with measure preserving map $T:X\rightarrow X$ is said to be mixing if for every $A,B\in \mathscr{A}$, $$\lim_{n\to\infty}\mu(A\cap T^{-n}B)=\mu(A)\mu(B).$$
Alternate Mixing: A measure space $(X,\mathscr{A},\mu)$ with measure preserving map $T:X\rightarrow X$ is said to be alternate mixing if for every $f,g\in D\subseteq L^2(\mu)$,
  $$\lim_{n\to\infty}\langle U_T^n f,g\rangle=\lim_{n\to\infty}\langle f(T^n),g\rangle=\langle f,1\rangle\langle 1,g\rangle=\mathbb{E}(f)\mathbb{E}(g).$$ Where $D$ is a dense subset of $L^2(\mu)$.

As stated by the OP, the implication (Alternate Mixing)$\implies$(Mixing) follows easily by considering indicator functions.
For the other implication, we will need the following lemma that follows from Lemma $3.13$ of Rudin's Real and Complex 
Analysis, $3$rd Ed, p. $69$. 

Lemma Let $(X,\mathscr{A},\mu)$ be any measure space. Then the set $S$ of all simple functions with finite support are dense in $L^2(\mu)$.

(Mixing)$\implies$(Alternate Mixing)
Assume that the statement of the mixing definition holds true. Take any $r,t\in S$. We define,
 $$ r(x)=\sum^n_{i=1}a_i\mathbf{1}_{A_i}(x)\qquad t(x)=\sum^m_{j=1}b_j\mathbf{1}_{B_j}(x)$$
Where $\bigcup_{i=1}^n{A_i}\subseteq X$ and $\bigcup_{j=1}^m{B_j}\subseteq X$ and also $\{a_i\}_{i=1}^n\cup\{b_j\}_{j=1}^m\subseteq \mathbb{R}$.
Consider then, 
$$\langle r(T^n),t\rangle=\int_X \left(\sum_{i}a_i\mathbf{1}_{T^{-n}A_i}\right) \left(\sum_{j}b_j\mathbf{1}_{B_j}\right)d\mu$$
$$=   \int_X \sum_{i,j}a_ib_j\ \mathbf{1}_{T^{-n}A_i\cap B_j}\ d\mu.$$
By the linearity of the integral, 
$$\langle r(T^n),t\rangle=\sum_{i,j}a_ib_j \int_X\mathbf{1}_{T^{-n}A_i\cap B_j}\ d\mu=\sum_{i,j}a_ib_j\ \mu(T^{-n}A_i\cap B_j).$$
Therefore, 
$$\lim_{n\rightarrow\infty}\langle r(T^n),t\rangle=\lim_{n\rightarrow\infty}\sum_{i,j}a_ib_j\ \mu(T^{-n}A_i\cap B_j)=\sum_{i,j}a_ib_j\ \lim_{n\rightarrow\infty}\mu(T^{-n}A_i\cap B_j).$$
By our assumption, we have that,
$$\lim_{n\to\infty}\langle U_T^n f,g\rangle=\lim_{n\rightarrow\infty}\langle r(T^n),t\rangle=\sum_{i,j}a_ib_j\mu(A_i)\mu(B_j)=\left(\sum_{i}a_i\mu(A_i)\right)\left(\sum_{j}b_j\mu(B_j)\right)$$ 
$$=\mathbb{E}(r)\mathbb{E}(t)=\langle r,1\rangle\langle 1,t\rangle.$$
And the required result follows. 
We can use the result we have just proven to prove that the alternate mixing definition holds true on all of $L^2(\mu)$. 
This will follow from the fact that any $f,g\in L^2(\mu)$ can be approximated arbitrarily well by two sequences of simple measurable functions in $S$. Using these sequences, the above argument can be modified to prove the general result. 
