Sorry in advance if this is not sufficiently research-level, it is really more of a reference request since the proof is not difficult. Let $\mathcal{Y}$ be a compact set, let $\{X\_n\}$ denote a sequence of random variables, and let $f(x,y)$ and $g(y)$ be "nice" functions. Suppose that for each fixed $y\in\mathcal{Y}$, we have $$\liminf\_{n\to\infty} f(X\_n,y)\geq g(y)$$ almost surely. What is the appropriate theorem to cite that says that (for sufficiently nice functions), we have $$\liminf\_{n\to\infty} \min\_{y\in\mathcal{Y}} f(X\_n,y)\geq \min\_{y\in\mathcal{Y}} g(y)$$ almost surely?



Pointwise almost sure convergence implies global convergence
First here, without loss of generality (wlog) $g=0$ (otherwise, replace $f(x,y)$ and $g(y)$ by $f(x,y)-g(y)$ and $0$, respectively). Second, in view of the almost-sure (a.s.) condition and the a.s. desired conclusion, wlog each random variable $X\_n$ takes only one value, say $x\_n$ -- which let us assume for simplicity to be a real number.


So, the desired result takes this simplified form:



> 
> If  
> 
> $$\liminf\_{n\to\infty} f(x\_n,y)\ge0 \tag{1}$$
> for some "nice" function $f$ and all $y$ in a compact set $Y:=\mathcal Y$, then
> $$\liminf\_{n\to\infty}\min\_{y\in Y} f(x\_n,y)\ge0. \tag{2}$$
> 
> 
> 


Consider now the following example: Let $Y:=[-1,1]$ and let $f(x,y):=-xy^2\,e^{-xy^2}$ for real $x$ and $y\in Y$. Probably everyone will agree that this function $f$ is nice. Let $x\_n\to\infty$ (as $n\to\infty$). Then $f(x\_n,y)\to0$ for each $y\in Y$, so that (1) holds. However, $\min\_{y\in Y} f(x\_n,y)=-1/e$ for each large enough $n$ (namely, for each $n$ such that $x\_n\ge1$), so that (2) fails to hold.


The obvious reason for the just described phenomenon is that the sequence $(x\_n)$ is unbounded.


So, it is natural to assume that $|x\_n|\le M$ for some real $M>0$ and all $n$. It is also natural to assume that the "nice" function $f$ is continuous. So, $f$ is uniformly continuous on the compact set $[-M,M]\times Y$. Also, passing to a subsequence, wlog assume that $x\_n\to x\_\infty$ for some real $x\_\infty$.


So, $f(x\_n,y)\to f(x\_\infty,y)$ uniformly in $y\in Y$ and hence
$$\lim\_{n\to\infty}\min\_{y\in Y} f(x\_n,y)
=\min\_{y\in Y} f(x\_\infty,y)
=\min\_{y\in Y}\lim\_{n\to\infty}f(x\_n,y)\ge0,
$$
by (1). So, we have deduced, very simply, (2) from (1).


Essentially, all we used here is that any function continuous on a compact set $K$ is uniformly continuous on $K$. So, one can hardly expect a reference involving something more substantial than this.


