consider $f\in L^1(R_+)$ and define Laplace transform $$\mathcal{L}f(z):=\int_0^{\infty} f(s)e^{-zs}\mathbb{d}s. $$
How can I prove $$\lim_{\mathbf{Re}z\rightarrow\infty}\mathcal{L}f(z) = 0?$$
Intuitively it is obvious, but without exponential property I can't figure out how to tackle it.