For $n\geq 1$ and let $g_n(x) = \sin^2(x+ \frac{1}{n}) , x\in [0,\infty)$ and $f_n(x)=\int_{0}^{x}g_n(t)dt$ , then
- {$f_n(x)$} converges point wise to a function on $[0,\infty)$ , but does not converge uniformly on $[0,\infty)$
- {$f_n(x)$} does not converge point wise to any function on $[0,\infty)$
- {$f_n(x)$} converges uniformly to a function on $[0,1]$
- {$f_n(x)$} converges uniformly to a function on $[0,\infty)$
What I answered is {$f_n(x)$} converges uniformly to a function on $[0,\infty)$ because the limit converges to $\sin^2(x)$ which is bounded continuous function