A General Continous Mapping or "Slutsky" Theorem For Stochastic Processes While standard Slutsky or Continuous Mapping Theorems apply to non-random functions, are there versions that hold for random functions under suitable conditions?
For example, consider a stochastic process $X_n: \Omega \times D_1 \to D_2$, where $D_1,D_2$ are metric spaces. Suppose that for all $u$ in $D_1$, $X_n(u) \stackrel{a.s.}{\to} X(u)$. If $Y_n \in D_1$, and $Y_n \stackrel{a.s.}{\to } Y$ (we might even imagine $Y$ as being deterministic/constant), then does it follow (under suitable conditions) that $X_n(Y_n) \stackrel{a.s.}{\to} X(Y)$?
As a simple example of what I'm talking about, imagine the following extremely over complicated proof that for iid RV's $X_1,...,X_n$ with $E(X_i)=0$, $Var(X_i)=\sigma^2$,
$$S_n^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 \stackrel{a.s.}{\to} \sigma^2:$$
Define $$S_n^2(u )  =  \frac{1}{n} \sum_{i=1}^n (X_i - u)^2,  $$
so that $S_n^2 = S_n^2(\bar{X})$. By the SLLN, $S_n^2(u) \stackrel{a.s.}{\to} E(X_0-u)^2$ for each fixed $u \in \mathbb{R}$. Moreover, $\bar{X} \stackrel{a.s.}{\to}0$, hence we conclude $S_n^2(\bar{X}) \stackrel{a.s.}{\to} EX_0^2 = \sigma^2$. What extra details need to be given to make this argument rigorous?
 A: There is no necessity to speak about random functions. Namely, a random function $F(\cdot)$ is just some random element, and you can always consider the value $F(Y)$ as
$$
F(Y) = f(F,Y),
$$
where $f(x,y) = x(y)$ is a bivariate deterministic function.
Now, as usually, in order to conclude $f(F_n,Y_n) \to F_n(Y_n)$ from $F_n\to F, Y_n\to Y$, one needs $f$ to be continuous in a neighborhood of the support of $(F,Y)$ in a suitable sense.  Actually, at this point we can forget about randomness, as this becomes a purely analytical question.  Pointwise convergence of $F_n$ is rarely suitable, as $f$ is discontinuous w.r.t. it.
What is suitable is, for example, locally uniform convergence of $F_n$. In your case, the latter follows from the uniform law of large numbers: for any $[a,b]\subset\mathbb R$,
$$
\sup_{u\in [a,b]} \big|F_n(u) - F(u)\big| = \sup_{u\in [a,b]} \bigg|\frac1n \sum_{i=1}^n \big(X_n - u\big)^2 - \mathrm E (X-u)^2\bigg| \to 0,\quad n\to\infty,
$$
almost surely. Therefore,
$$
F_n(\overline X) \to F(0) = \sigma^2,\quad n\to\infty,
$$
almost surely, as desired.
