Given $f_{n} : [0, \infty) \mapsto \mathbb{R}$ a sequence of continuous functions that converge uniformaly to $f$
We know that $f_{n}$ is Riemann integrable any closed bounded interval $[a,b]$ and $$\lim_{n\to\infty} \int_{a}^{b} f_{n} \ dx = \int_{a}^{b} f(x) \ dx $$
However when the integral is improper $$\lim_{n\to\infty} \int_{a}^{\infty} f_{n} \ dx = \lim_{n\to\infty} \ \lim_{b\to\infty} \int_{a}^{b} f_{n} \neq \int_{a}^{\infty} f(x) \ dx $$ even though uniform convergence preserves limits?
I've seen a few counterexamples, but I don't understand what property of the improper integral makes this not possible.
If $(f_n)_{n\in\mathbb n}$ is a sequence of Riemann-integrable functions from $[0,a]$ into $\mathbb R$ wich converges uniformly to $f\colon[0,a]\longrightarrow\mathbb R$, then, for each $\varepsilon>0$, you have, if $n$ is large enough,$$(\forall x\in[0,a]):\bigl\lvert f(x)-f_n(x)\bigr\rvert<\varepsilon,$$and therefore,$$\left\lvert\int_0^af(x)\,\mathrm dx-\int_0^af_n(x)\,\mathrm dx\right\rvert<\varepsilon a.\tag1$$So, if $n$ is large enought, the numbers $\int_0^af(x)\,\mathrm dx$ and $\int_0^af_n(x)\,\mathrm dx$ are close to each other. But if are integrating over, say $[0,\infty)$, then you don't have $(1)$ and so the integrals $\int_0^\infty f_n(x)\,\mathrm dx$ give you no information about $\int_0^\infty f(x)\,\mathrm dx$.