Swap $\limsup$ and $\mathbb{E}$ of sequence of functions evaluated at a random variable?

Question

I am a postgrad with more of a background in the functional analysis point of view of things but am recently needing to get the expectation involved and it has been a few years since I've done much probability.

I have a continous random variable $X$ and a function $g(X,a),$ continuous with respect to $a$ where $a\geq 0$ is some number/not a random variable, and $g$ evaluates to a real number (possibly negative). Consider a sequence $a_n\to 0$ where each $a_n>0.$ I want to know whether it is true that $$\limsup_{n\to\infty} \mathbb{E}[g(X,a_n)]\leq \mathbb{E}\left[\limsup_{n\to\infty }g(X,a_n)\right]. $$

I have been looking at things such as (reverse) Fatou's lemma, and dominated convergence theorem however that is applicable for when the $\limsup$ is related to a random variable, whereas here it is related to an ordinary sequence. My two lines of thought are:

• Consider the random variable (?) $Y_n:= g(X,a_n)$ and try to use Fatou's lemma/show it is uniformly integrable. Issue is I haven't seen this construction of a random variable before so am hesitant as to what I can/cannot do.
• Try and show that $F(a):=\mathbb{E}[g(X,a)]$ is continuous with respect to $a$. I have tried looking for results concerning this but so far have been unsuccessful

Any thoughts/pointers/guidance would be greatly appreciated!

Answer 1

If $g$ is measurable, then $Y_n=g(X,a_n)$ defines a random variable. There is no issue with that. As you stated, the problem then reduces to showing $$\limsup \mathbb EY_n\leq \mathbb E\limsup Y_n $$ I would guess that it will be hard to find an integrable upper bound for $Y_n$ without knowing anything further about $g$.

As to the continuity of $F$; there is a well known theorem, see for example here, that states if $|g(\cdot,a_n)|$ has an integrable upper bound $h(\cdot)$ and $g(x,\cdot)$ is continuous at $a$ for every $x$, then $F(\cdot)=\mathbb E[g(X,\cdot)]$ will also be continuous at $a$.

Again, we will need an upper bound of $g(\cdot,a_n)$ for applying this theorem. If $g(\cdot,a_n)$ would be bounded around a neighbourhood of $0$ for every $x$ one could make this work.