The convolution means
\begin{equation*}
f(x)*g(x) = \int\limits_{-\infty}^{\infty} f(\xi)g(x-\xi)\,d\xi = \int\limits_{-\infty}^{\infty} f(x-\xi)g(\xi)\,d\xi
\end{equation*}
derive from the Dirac Delta function
\begin{equation*}
f(\alpha) = \int\limits_{-\infty}^{\infty}\delta(\alpha-x)f(x)\,dx.
\end{equation*}
\begin{equation*}
\mathscr{F} \{f(x) * g(x)\} = \int\limits_{-\infty}^{\infty}e^{-isx}\{f(x) * g(x)\}\,dx
\end{equation*}
\begin{equation*}
=\int\limits_{x=-\infty}^{\infty}\frac{1}{2\pi}\left[ \int\limits_{\xi=-\infty}^{\infty}f(\xi)g(x-\xi)\,d\xi\right] e^{-isx}\,dx
\end{equation*}
(on changing the order of integration)
\begin{equation*}
=\int\limits_{\xi=-\infty}^{\infty}\left[f(\xi)\int\limits_{x=-\infty}^{\infty}g(x-\xi) e^{-isx}\,dx\right]\,d\xi
\end{equation*}
\begin{equation*}
=\int\limits_{\xi=-\infty}^{\infty}f(\xi)\left[\int\limits_{x=-\infty}^{\infty}g(x-\xi) e^{-is(x-\xi)}\,d(x-\xi)\right]e^{-is\xi}\,d\xi; \quad x-\xi=\lambda
\end{equation*}
\begin{equation*}
=\int\limits_{\xi=-\infty}^{\infty} f(\xi)e^{-is\xi}\,d\xi\left[ \int\limits_{\lambda=-\infty}^{\infty}g(\lambda)e^{-is\lambda}\,d\lambda\right]
\end{equation*}
\begin{equation*}
=\int\limits_{\xi=-\infty}^{\infty} f(\xi)e^{-is\xi}\,d\xi\cdot\mathscr{F}\{g(\lambda)\}
\end{equation*}
\begin{equation*}
=\mathscr{F}\{f(\xi\}\cdot \mathscr{F}\{g(\lambda)\} = F(s).G(s).
\end{equation*}
If \(x\) is time and \(s\) is frequency then convolution of two signal in time domain corresponds to the multiplication of their Fourier transform in frequency domain.
