if the inequality $$\int_{0}^{\infty}[f(t)-g(t)]dt\le 0$$ holds such that
$f(t)$ and $g(t)$ are (uniformly) continues functions, $0\le f(t)$, $0\le g(t)\lt k$ and $\lim \limits_{t\rightarrow+\infty}g(t)=0$, where $k$ is a positive constant.
then one can conclude that $$\lim \limits_{t\rightarrow+\infty}f(t)=0?$$
A proof or a counterexample .
No. Consider $k > 1$, and let $g(x) = \mathbf{1}_{(0, 1)}(x)$ smoothed out so that the total added area is $\epsilon$. Then let $$f(x) = \sum_{k=1}^{\infty} \mathbf{1}_{(k, k+2^{-k})}(x)$$ smoothed out arbitrarily so that the total added area is $\epsilon$ (say, $\epsilon/2^k$ for the $k$th bump). We will have $$\int_0^{\infty} \left(f(x)-g(x)\right)\,\mathrm{d}x = \left[\sum_{k=1}^{\infty} 2^{-k}\right]-1 = 0$$ but $\lim\limits_{x\to \infty} f(x)$ does not exist.