Continuous mapping theorem for a sequence of densities?

Question

Let ${f_n(x)}$ be a sequence of densities that uniformly converges to $f(x)$ almost surely, that is, $$ f_n(x) \xrightarrow[]{\text{a.s.}} f(x), \quad \text{uniformly},$$ or equivalently $$ \Pr\left( \lim_{n\rightarrow\infty} \sup_{x\in\mathbb{R}} | f_n(x) - f(x) | = 0 \right) = 1 . $$

Using the continuous mapping theorem, I would like to claim that $$ \psi\left(f_n(x)\right) \xrightarrow[]{\text{a.s.}} \psi\left(f(x)\right), \quad \text{uniformly},$$ or equivalently $$ \Pr\left( \lim_{n\rightarrow \infty} \sup_{x\in\mathbb{R}} | \psi\left(f_n(x)\right) - \psi\left(f(x)\right) | = 0 \right) = 1,$$ where $\psi$ is continuous on $\mathbb{R}_{>0}$.

Is this true? If not, can I at least claim that the $\psi\left(f_n(x)\right)$ converges pointwise to $\psi\left(f(x)\right)$ almost surely?

Answer 1

I believe this is not true and that you do not need to appeal to the continuous mapping theorem (which implies results stronger than what you are looking for here).

You're essentially asking if a continuous function preserves a uniform limit of functions. I'm being a little cavalier here but behaviors "almost surely" can essentially be treated as absolute behaviors in the measure theoretic sense.

For a counterexample:

Counterexamples to a continuous function preserving almost uniform convergence and convergence in measure.

noting that $z^2$ is not uniformly continuous, and above does not preserve the uniform convergence of $f_n \rightarrow f$.

The definition of continuity gives you pointwise convergence a.e. though.