Edit: @saz answered the original question so I modified it to use compact convergence (uniform convergence on all compacts sets) instead to see if it still holds.
Suppose:
$\cdot$ $X:\Omega\to\mathbb{R}$ is an integrable random variable
$\cdot$ Continuous functions $f:\mathbb{R}\to\mathbb{R}$ and $f_n:\mathbb{R}\to\mathbb{R}$ are such that $f(X)$ and $f_n(X)$ is integrable for all $n\in\mathbb{N}$
$\cdot$ $\sup_{n\in\mathbb{N}}f_n(x) < \infty$ for all $x\in\mathbb{R}$
$\cdot$ $f_n$ converges to $f$ compactly as $n\to\infty$.
Does it hold that $\sup_{i\in\mathbb{N}} f_n(X)$ is integrable?
What I know so far: $$ \sup_{n\in\mathbb{N}} |f_n(X)| \leq f(X) + \sup_{n\in\mathbb{N}} |f_n(X) - f(X)| $$ So if $\sup_{n\in\mathbb{N}} |f_n(X) - f(X)|$ is integrable, then $\sup_{n\in\mathbb{N}} |f_n(X)|$ is also integrable. But how do I show $\sup_{n\in\mathbb{N}} |f_n(X) - f(X)|$ is integrable?