suppose we have a sequence of continuous functions $(f_n)_{n\geq 1}$ with $f_n:[0,\infty)\rightarrow \mathbb{R}$ such that $\int_{0}^{\infty}\vert f_n(x) \vert<\infty$ for all $n\in\mathbb{R}$. Further, let $\lim\limits_{n\rightarrow \infty}f_n(x) = 0$ for all $x\geq 0$.
Is it true that $$\lim\limits_{n\rightarrow \infty}\int_{0}^{\infty}f_n(x)\cdot g(x) dx = 0,$$
if $g:[0,\infty)\rightarrow \mathbb{R}$ is a continuous and absolutely integrable function (i.e. $\int_{0}^{\infty}\vert g(x) \vert<\infty$)? Can I interchange the order of integration and the limit? (If needed, we may also assume that $\int_{0}^{\infty}\vert f_n(x)\cdot g(x) \vert<\infty$ for all $n$).
Best regards
If $f_n\to 0$ uniformly then the answer is affirmative (see the comments below). In general the answer is no, however. Here is a counter example. Consider the sequence $$f_n(x)=e^{-(x-n)^2+x}\;,$$ and let $g(x)=e^{-x}$. Then $f_n\to 0$ pointwise but not uniformly on $[0,\infty)$. By translation invariance of $dx$, we have $$\lim_{n\to \infty}\int_0^{\infty}e^{-(x-n)^2+x}e^{-x}dx=\lim_{n\to \infty}\int_0^{\infty}e^{-(x-n)^2}dx =\lim_{n\to\infty}\int_{-n}^{\infty}e^{-y^2}dy=\sqrt{\pi}$$