[proofplan] We write out the Laplace transform of the convolution as a double integral, interchange the order of integration via Fubini's theorem (justified by absolute convergence for $\operatorname{Re} p$ sufficiently large), and perform the substitution $\tau = t - t'$ in the inner integral to factor the result into $\hat{f}(p) \cdot \hat{g}(p)$. [/proofplan] [step:Write the Laplace transform of the convolution as a double integral] By definition of the Laplace transform and the convolution $(f * g)(t) = \int_0^t f(t - t')g(t')\,d\mathcal{L}^1(t')$: \begin{align*} \mathcal{L}\{f * g\}(p) = \int_0^\infty \left(\int_0^t f(t - t')g(t')\,d\mathcal{L}^1(t')\right) e^{-pt}\,d\mathcal{L}^1(t). \end{align*} The region of integration is $\{(t, t') \in \mathbb{R}^2 : 0 \leq t' \leq t < \infty\}$. [/step] [step:Interchange the order of integration via Fubini's theorem] We rewrite the region as $\{(t, t') : 0 \leq t' < \infty,\; t \geq t'\}$. Since $f$ and $g$ are of exponential order (with $|f(t)| \leq M_1 e^{\sigma_1 t}$ and $|g(t)| \leq M_2 e^{\sigma_2 t}$), the integrand $|f(t-t')g(t')e^{-pt}|$ is bounded by $M_1 M_2 e^{(\sigma_1 + \sigma_2 - \operatorname{Re} p)t}$, which is integrable over the region when $\operatorname{Re} p > \sigma_1 + \sigma_2$. By Fubini's theorem (the absolute convergence condition is satisfied), we interchange the order: \begin{align*} \mathcal{L}\{f * g\}(p) = \int_0^\infty g(t') \left(\int_{t'}^\infty f(t - t')e^{-pt}\,d\mathcal{L}^1(t)\right) d\mathcal{L}^1(t'). \end{align*} [guided] Why can we interchange the order of integration? Fubini's theorem requires the integrand to be absolutely integrable over the product domain. We verify this. Since $f$ and $g$ are of exponential order, there exist constants $M_1, M_2, \sigma_1, \sigma_2$ with $|f(s)| \leq M_1 e^{\sigma_1 s}$ and $|g(s)| \leq M_2 e^{\sigma_2 s}$ for $s \geq 0$. For $(t, t')$ in the region $0 \leq t' \leq t$: \begin{align*} |f(t - t')g(t')e^{-pt}| \leq M_1 e^{\sigma_1(t - t')} \cdot M_2 e^{\sigma_2 t'} \cdot e^{-(\operatorname{Re} p)t} = M_1 M_2\, e^{(\sigma_1 - \operatorname{Re} p)t + (\sigma_2 - \sigma_1)t'}. \end{align*} Integrating over $0 \leq t' \leq t < \infty$ yields a finite result when $\operatorname{Re} p > \sigma_1 + \sigma_2$ (the integral over $t$ converges due to the exponential decay, and the $t'$-integral is bounded by the $t$-range). With absolute integrability established, Fubini's theorem permits interchanging the order, giving the iterated integral with $t'$ as the outer variable and $t \geq t'$ as the inner range. [/guided] [/step] [step:Substitute $\tau = t - t'$ in the inner integral to factor the transform] In the inner integral, substitute $\tau = t - t'$, so $t = \tau + t'$ and $d\mathcal{L}^1(t) = d\mathcal{L}^1(\tau)$. When $t = t'$, $\tau = 0$; when $t \to \infty$, $\tau \to \infty$. The inner integral becomes: \begin{align*} \int_{t'}^\infty f(t - t')e^{-pt}\,d\mathcal{L}^1(t) = \int_0^\infty f(\tau)e^{-p(\tau + t')}\,d\mathcal{L}^1(\tau) = e^{-pt'}\int_0^\infty f(\tau)e^{-p\tau}\,d\mathcal{L}^1(\tau) = e^{-pt'}\hat{f}(p). \end{align*} Substituting back into the outer integral: \begin{align*} \mathcal{L}\{f * g\}(p) = \int_0^\infty g(t')e^{-pt'}\,d\mathcal{L}^1(t') \cdot \hat{f}(p) = \hat{f}(p) \cdot \hat{g}(p). \end{align*} This completes the proof that the Laplace transform converts convolution to pointwise multiplication. [/step]