 Solution

Uniform convergence means that for every $\epsilon >0$ there is $N$ such that $|s_n(x)-s(x)|<\epsilon$ for every $x\in \mathbb{R}$ and every $n\ge N$. Equivalently, ${\sup }_x|s_n(x)-s(x)|\to 0$. The nonzero constant functions $s_n=1/n$ converge uniformly to zero. In contrast, the continuous moving bumps $s_n(x)=\max (0,1-|x-n|)$ converge pointwise to zero but have supremum one for every $n$.

On the compact interval,Thus uniform convergence permits the stated interchange with integration.

Choose $\boxed{s(x)=1/(1+|x|)}$. It is continuous, nonnegative and tends to zero at infinity, but its improper integral diverges logarithmically. This even choice also makes the prescribed cutoff continuous at both $n$ and $-n$: beyond either endpoint it remains at $s(n)$ for another unit of distance and then decreases as $s(n)/(|x|-n{)}^2$.

Inside $[-n,n]$, $s_n=s$. Outside, both functions lie between zero and $1/(n+1)$, so ${\sup }_x|s_n-s|\le 1/(n+1)\to 0$. Nevertheless each cutoff has finite integral:Here the value of the minimum at $t=0$ is its continuous limiting value one. This example shows that uniform convergence on an unbounded domain does not by itself ensure convergence of the integrals over that domain.