[Math] Proof of convolution theorem for Laplace transform

Question

The convolution theorem for Laplace transform states that
$$\mathcal{L}\{f*g\}=\mathcal{L}\{f\}\cdot\mathcal{L}\{g\}.$$
The standard proof uses Fubini-like argument of switching the order of integration:
$$\int_0^\infty d\tau \int_{\tau}^\infty e^{-st}f(t-\tau)g(\tau)\,dt=\int_0^\infty dt\int_0^te^{-st}f(t-\tau)g(\tau)\,d\tau$$
Fubini's theorem says that one can switch the order of integration. But what we have in the iterated integrals are not integrals, but limits of integrals (i.e., improper integrals). Are we justified to treat them like "proper" integrals and switch their order?

Answer 1

\begin{align*}\mathcal{L}\{f\}({s})\cdot\mathcal{L}\{g\}({s})&=\left(\int_0^\infty e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^\infty e^{-{s}u}{g}(u)\,du\right)\\ &=\lim_{{{L}}\to\infty}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right) \end{align*} By Fubini's theorem, \begin{align*}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right)&=\int_0^{{L}}\int_0^{{L}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du\\ &=\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du, \end{align*} where $R_{{L}}$ is the square region $$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}}.$$ Let $T_{{L}}$ be the triangular region $$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad t+u\leq {{L}}.$$ Provided that ${f}$ and ${g}$ are bounded by exponential functions, $$\lim_{{{L}}\to\infty}\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\lim_{{{L}}\to\infty}\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du.$$ Now, the function $$\varphi(v,u)=(v-u,u)$$ maps $D_{{L}}$ bijectively onto $T_{{L}}$, where $D_{{L}}$ is the triangular region $$0\leq v\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad v\geq u.$$ The component functions of $\varphi$ are $$t(v,u)=v-u\qquad\mbox{and}\qquad u(v,u)=u,$$ so the Jacobian of $\varphi$ is $$J\varphi=\det\begin{bmatrix}t_v&t_u\\u_v&u_u\end{bmatrix}=\det\begin{bmatrix}1&-1\\0&1\end{bmatrix}=1.$$ Hence, $$\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du.$$ By Fubini's theorem, \begin{align*}&\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du=\int_0^{{L}}\int_0^ve^{-{s}v}{f}(v-u){g}(u)\,du\,dv\\ \\ &=\int_0^{{L}}e^{-{s}v}\int_0^v{f}(v-u){g}(u)\,du\,dv=\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv. \end{align*} Hence, \begin{align*}\lim_{{{L}}\to\infty}\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du&=\lim_{{{L}}\to\infty}\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv\\ \\ &=\int_0^\infty e^{-{s}v}({f}\ast{g})(v)\,dv\\ \\ &=\mathcal{L}\{{f}\ast{g}\}({s}). \end{align*}