$\mathcal{L}(f\ast g) = \mathcal{L}f \mathcal{L}g$
where $\mathcal{L}$ is the Laplace Transform.
This is my book's proof, however I do not understand how they have the infinitesimal $ds$ instead of $dt$ after their substitution.
$$\mathcal{L}(f\ast g)(z) = \int_{0}^\infty (f\ast g)(t)e^{-zt}\ dt \\ = \int_0^\infty\int_0^tf(t-s)g(s)\ ds e^{-zt} \ dt \\ = \int_0^\infty\int_0^t f(t-s)e^{-z(t-s)}g(s)e^{-zs}\ ds \ dt \\ = \int_0^\infty\int_0^\infty f(u)e^{-zu}g(s)e^{-zs}\ du \ ds \\ = \mathcal{L}f \mathcal{L}g$$ where we have changed variables $u=t-s$.
I'm not sure how they used this substitution because in step 3, if they let $u=t-s$, then if $t$ is constant, shouldn't it be $du = -ds$?
So they shouldn't have a $ds$ and instead, should have a $dt$ after the substitution right?
Perhaps these intermediate steps help to clarify:
$$\begin{align}\mathcal{L}(f\ast g)(z) &= \int_{0}^\infty (f\ast g)(t)e^{-zt}\ dt \\ &= \int_0^\infty\left(\int_0^tf(t-s)g(s)\ ds \right)e^{-zt} \ dt \\ &= \int_0^\infty\left(\int_0^t f(t-s)e^{-z(t-s)}g(s)e^{-zs}\ ds\right) \ dt \\ &\color{blue}{= \int_0^\infty\left(\int_s^\infty f(t-s)e^{-z(t-s)}\ dt\right)g(s)e^{-zs} \ ds}\\ &\color{blue}{= \int_0^\infty\left(\int_0^\infty f(u)e^{-zu}\ du\right)g(s)e^{-zs} \ ds} \\ &= \left(\int_0^\infty f(u)e^{-zu}\ du\right)\left(\int_0^\infty g(s)e^{-zs} \right)\ ds \\ &= \mathcal{L}f \mathcal{L}g\end{align}$$
The substitution had $s$ as fixed, and was a substitution with respect to $t$.