Suppose the process $X = \left\{X_{t}:t\in Z\right\}$ is strong mixing with the coefficient $\alpha(j) \rightarrow 0$ defined as \begin{equation} \alpha(j)=\sup_T\sup_{1\leq k\leq T-j}\sup\left\{\lvert P(A\cap B)-P(A)P(B)\rvert: B\in\mathcal{F}_{-\infty}^{k},A\in\mathcal{F}_{k+j}^{+\infty}\right\}, \end{equation} where $\mathcal{F}_{i}^{k}=\sigma\left(X_{l}:i\leq l\leq k\right)$.
Meanwhile, let $\beta = \{\beta_{t}:t\in Z\}$ be a process that is convergent to the constant $\beta_0$ in probability in the sense that for all $\epsilon>0\,,$ \begin{equation} \lim_{t\to\infty}P(|\beta_t - \beta_0|<\epsilon) = 0 \end{equation}
My aim is to show that the process $Y = \{Y_t\}$ where $Y_t = f(X_t,\beta_t)$ is also a strong mixing for a Borel function $f$.
I know that if $X$ and $\beta$ are two independent strong mixing then $Y$ is also a strong mixing. But, here, $\beta$ and $X$ are not independent.
In general no: let $Y_t=X_t+Z/(t+2)$, where $(X_t)_{t\geqslant 1}$ is i.i.d., $Z$ is independent of $(X_t)_{t\geqslant 1}$ and all these random variables take the values $1$ and $-1$ with probability $1/2$. Here $f(x,y)=x+y$ and $\beta_t=Z/(t+2)$ converges in probability to $0$. Let $A_t=\{Y_t\geqslant 1+1/(t+2)\}$; then $A_t=\{X_t=1\}\cap \{Z=1\}$ hence $\mathbb P\left(A_t\right)=1/4$ and $\mathbb P\left(A_t\cap A_2\right)=\mathbb P\left(\{X_t=1\}\cap\{X_2=1\}\cap\{Z=1\}\right)$, which gives $$ \alpha_{(Y_t)_t}(t-2)\geqslant \left\lvert \mathbb P\left(A_t\cap A_2\right)-\mathbb P\left(A_t\right)\mathbb P\left(A_2\right)\right\rvert=1/16. $$
There are also simpler counter-examples: if $Y_t=\beta_t=Y/t$, then you do not have a strong mixing sequence unless $Y$ is constant.