Let $g:\mathbf{X}\to\mathbb{R}_+$ be a positive continuous function where $\mathbf{X}\subseteq\mathbb{R}^d$. Let $f_n:\mathbf{X}\to\mathbb{R}$ where $n\in\mathbb{N}$ be a sequence of functions. Define the weighted supremum norm $\|\cdot\|_g$ by $$ \|f_1\|_g := \sup_{x\in\mathbf{X}}\frac{|f_1(x)|}{g(x)}. $$
Question: Let $Y$ be an integrable random variable taking values in $\mathbf{X}$. If $\|f_n\|_g < \infty$ for all $n$, is the family of random variables $(f_n(Y))_{n\in\mathbb{N}}$ uniformly integrable?
My thoughts: We know that if
$$
\sup_{n\in\mathbb{N}}\sup_{x\in\mathbf{X}}\frac{|f_n(x)|}{g(x)} < \infty.
$$
Then, with probability one:
$$
\sup_{n\in\mathbb{N}}\frac{|f_n(Y)|}{g(Y)} < \infty \implies \sup_{n\in\mathbb{N}} |f_n(Y)| < C g(Y)
$$
for some constant positive C. I know that if $g(Y)$ is integrable, then this implies $(f_n(Y))_{n\in\mathbb{N}}$ is uniformly integrable. So I am stuck here since I do not think the first inequality necessarily holds. Is there another way?
I don't think this is true but maybe I'm missing something.
Let's think about a very simple example: $g(x) = 1$, $f_n(x) = n$. Then, $\|f_n\|_g = n$ and $\|f_n\|_g < \infty$ for all $n$. Now, $(f_n(Y))_{n \in \mathbb{N}}$ does not depend on $Y$ and is clearly not uniformly integrable.