if $1 - \int\limits_0^\infty {g(s)ds > } 0$ have we $1 - \int\limits_0^\infty {{e^{ - s}}g(s)ds > } 0$?

Question

Let $g$ be non increasing positive differentiable function such that $g(0)>0$. We assume that $$1 - \int\limits_0^\infty g(s)ds > 0$$ Do we have $$1 - \int\limits_0^\infty {{e^{ - s}}g(s)ds > } 0$$ ? if this is not true, what conditions can we put on $g$ to ensure the above inequality? thanks.

Answer 1

Of course: $$1 - \int_0^{\infty}e^{-s}g(s)ds = 1-\int_0^{\infty}g(s)ds+\int_0^{\infty}(1-e^{-s})g(s)ds >0$$