 Solution

Uniform convergence on an interval $I$ means that for every $ϵ>0$ there is $N$ such that $|f_n(x)-f(x)|<ϵ$ for every $x\in I$ and every $n\ge N$. Equivalently, the supremum of this difference tends to zero. A uniform limit of continuous functions is continuous: at a point $x_0$, choose one $f_N$ uniformly within $ϵ/3$ of $f$, then use continuity of $f_N$ to make $|f_N(x)-f_N(x_0)|<ϵ/3$. The triangle inequality gives $|f(x)-f(x_0)|<ϵ$. Thus $f$ is integrable on the compact interval, andFor the half-line example, differentiating gives $f_n^{\prime }(x)=n^{-2}{e}^{-x/n}(1-x/n)$. The maximum is attained at $x=n$, where $f_n(n)=1/(en)$. Therefore $f_n\to 0$ uniformly on $[0,\infty )$. Nevertheless, the substitution $x=ny$ givesThe integral does not tend to zero. This is the phenomenon that uniformly vanishing functions can retain a nonzero integral: the finite-length estimate above has no finite analogue for the entire half-line.