Let $(\Omega,\mathcal{F},\mathbb{P})$ be a standard (or Lebesgue-Rokhlin) probability space. That is, $(\Omega,\mathcal{F},\mathbb{P})$ is isomorphic to $[0,1]^{\mathbb{N}_{0}}$ endowed with the Borel $\sigma$-algebra $\mathcal{B}([0,1]^{\mathbb{N}_{0}})$ and the product of the Borel measures $\lambda^{\mathbb{N}_{0}}$. Let $\mathcal{L}^{\infty}$ be the set of all $\mathbb{R}$-valued random variables that are essentially bounded. Take any $X \in \mathcal{L}^{\infty}$.
Let $f: \mathbb{R} \to \mathbb{R}$ be a strictly increasing function (and thus, the inverse $f^{-1}$ exists). Consider a sequence $(f_{n})_{n \in \mathbb{N}}$ such that:
$f_{n}: \mathbb{R} \to \mathbb{R}$ is a strictly increasing function, for each $n \in \mathbb{N}$.
$f_{n} \to f$ uniformly on any compact set (as $n \to \infty$).
In this setting, I encountered the following statement:
\begin{align*} \lim_{n \to \infty} (f_{n})^{-1} \circ \mathbb{E}[f_{n}(X)] = (f)^{-1} \circ \mathbb{E}[f(X)]. \end{align*}
How is this equation justified? In particular, I cannot see how the fact that "$f_{n} \to f$ uniformly on any compact set" matters. Is the compactness related to the assumption that $(\Omega,\mathcal{F},\mathbb{P})$ is isomorphic to $[0,1]^{\mathbb{N}_{0}}$, which is compact (due to Tychonoff's theorem)?
Thank you.