I have troubles understanding the proof of the following theorem from my notes about Laplace transform $\mathcal{L}$.
Theorem: Let $f, g : [0, \infty) \to \mathbb{C}$ be functions of exponential increase; $|f(t)| \leq M e^{kt}, \; |g(t)| \leq Me^{kt}$. Then, if $\text{Re}(z) > k$, the following equality holds: $$ \mathcal{L} (f \ast g) = \mathcal{L}(f)\mathcal{L}(g) $$
Proof: We use Fubini's theorem to obtain: $$ \mathcal{L}(f \ast g)(z) = \int_0^{\infty} e^{-zt} (f \ast g)(t) dt = \int_0^{\infty} e^{-zt} \left( \int_0^t f(t-s)g(s)ds \right) dt = \int_0^{\infty} \left( \int_s^{\infty} f(t-s)g(s)e^{-z(t-s)} e^{-zs} dt \right) ds = \int_0^{\infty} g(s) e^{-zs} \left( \int_s^{\infty} f(t-s) e^{-z(t-s)} dt \right) ds = \mathcal{L}(f)(z) \cdot \mathcal{L}(g)(z) $$
If I recall correctly, professor claimed that we use Fubini's theorem in the third equality, to interchange the integrals. However, I do not understand, how exactly did we apply it and what happened. How do we get the new borders?