If $f_n(x)\xrightarrow{a.s}f(x)$ and $t_n\xrightarrow{a.s}t$ as ${n \to \infty}$, is this always true that $f_n(t_n)\xrightarrow{a.s}f(t)$ as ${n \to \infty}$?
Note that $t_n \xrightarrow{a.s} t$ means $t_n$ converges to $t$ almost surely, in other words, $P(\lim\limits_{n \to \infty}t_n=t)=1$.
To solve this problem, I have to conclude $P(\lim\limits_{n \to \infty}f_n(t_n)=f(t))=1$ from $P(\lim\limits_{n \to \infty}f_n(x)=f(x))=1$ and $P(\lim\limits_{n \to \infty}t_n=t)=1$.
Thus, If I prove $\lim\limits_{n \to \infty}f_n(t_n)=f(t)$ from $\lim\limits_{n \to \infty}f_n(x)=f(x)$ and $\lim\limits_{n \to \infty}t_n=t$, the proof is complete. How can I prove this?
Note that:
$$ f_n(x) \rightarrow f(x) \quad iff \quad (f_n(x)-f(x))\rightarrow0$$
So that one can write
$$ f_n(t_n)-f(t)=f_n(t_n)-f(t_n)+(f(t_n)-f(t)) $$
Of course $f_n(t_n)-f(t_n) \rightarrow 0 $, to prove your assumption, we need to prove: $f(t_n)-f(t) \rightarrow 0$, that hold if $f$ is continuos in $t$
On the other hand, if $f$ is not continuos in t$ this no longer holds, consider as a counter example:
Suppose $f=1$ for $x\geq 0$ and 0 otherwhise, $t_n=-1/n \quad tn \rightarrow0$ as but $f(t_n)=0 \ \ \forall n $ so it cannot converge to $f(0)=1$