Let $g(x,t)$ be a continous at $t_0$ uniformaly in x. Let $F$ be a distribution function for which $E_F|g(X,t_0)|<\infty$. Let $\{X_i\}$ be i.i.d with d.f F and suppose that $T_n\rightarrow_p t_0$. Then,
$\frac{1}{n}\sum_{i=1}^n g(X_i, T_n)\rightarrow_pE_Fg(X,t_0)$
Futher, the convergence in probability can be replaced by a.s convergence throughout.
My question: It seems that
According to WLLn, we have $\frac{1}{n}\sum_{i=1}^n g(X_i, t_0)\rightarrow_pE_Fg(X,t_0)$. Since $T_i \rightarrow t_0$, and $g(x,t)$ be continous at $t_0$ uniformaly in x, we have $ g(X_i, T_n)\rightarrow_p g(X,t_0)$ according to continous mapping, thus $\frac{1}{n}\sum_{i=1}^n g(X_i, T_n)\rightarrow_pE_Fg(X,t_0)$
In the statement about , we can replace convergence in probability with almost sure convergence.
That is $T_n\rightarrow_{a.s} t_0$, then $\frac{1}{n}\sum_{i=1}^n g(X_i, T_n)\rightarrow_{a.s} E_Fg(X,t_0)$ (continous mapping and SLLN)
In both cases the result follows from the (weak or strong) uniform law of large numbers (continuous mapping theorem doesn't apply to this situation). Specifically, let $Q_n(t):=n^{-1}\sum_{i=1}^n g(X_i,t)$ and $Q(t):=\mathsf{E}g(X,t)$. Then \begin{align} |Q_n(T_n)-Q(t_0)|&\le |Q_n(T_n)-Q(T_n)|+|Q(T_n)-Q(t_0)| \\ &\le \sup_{t\in \mathcal{T}}|Q_n(t)-Q(t)|+|Q(T_n)-Q(t_0)|, \end{align} where $\mathcal{T}$ is the parameter space. When $t\mapsto Q(t)$ is continuous the second term converges to $0$ (a.s. or in probability). The first term converges to $0$ by the referenced uniform LLN (assuming that $\mathcal{T}$ is compact and $\mathsf{E}\sup_{t\in\mathcal{T}}|g(X,t)|<\infty$).