Let:
- $X_n \rightarrow X$ in distribution.
- $f_n(x,t)$ be a sequence of differentiable functions from $\mathbb{R}_+$ to $\mathbb{R}_+$ that converge uniformly to $f(x,t)$.
- $f_n(x,t) \le g(t)$ with $\int |g(t)| d t < \infty$.
- $f_n(X_n,t) \rightarrow f(X,t)$ in distribution.
I want to show that $$ \int_0^\infty f_n(X_n,t) dt \rightarrow \int_0^\infty f(X,t) d t $$ in distribution.
If this is true, one could try to prove it combining the results in [1] and [2]. Essentially, it would amount to looking at $$ \int_0^\infty f_n(X_n,t) dt = \int_0^\infty [f_n(X_n,t) - f(X_n,t)] dt + \int_0^\infty f(X_n,t) dt $$ If I'm not mistaken, the second integral on the right-hand side converges in distribution to $\int_0^\infty f(X,t) dt$: $X_n$ converges to $X$ in distribution, $f$ is continuous, and the integral itself is continuous (because of [2]), so we can apply the continuous mapping theorem. Now, it would remain to show that the first integral goes to 0 in distribution. Uniform continuity suggests that we could take the limit inside the integral, but the fact that $X_n$ is a random variable is confusing me.
I think that an alternative way to prove it might be showing uniform convergence of $\int_0^\infty f_n(x,t) dt$ to $\int_0^\infty f(x,t) d t$, but I'm not sure this is easier.
Sources
[1] prove a continuous mapping theorem for $g_n(X_n)\stackrel{d}{\to}g(X)$?
[2] Continuity of parameter dependent integral (source needed)