[proofplan] We prove $\widehat{f * g}(\xi) = \hat{f}(\xi)\hat{g}(\xi)$ by writing out the left-hand side as a double integral, justifying the exchange of integration order via Fubini--Tonelli (using $\|f\|_{L^1}\|g\|_{L^1} < \infty$), performing the substitution $z = x - y$ to factor the exponential, and recognising the two resulting single integrals as $\hat{f}(\xi)$ and $\hat{g}(\xi)$. [/proofplan] [step:Write out the double integral] By definition of the [Fourier transform](/page/Fourier%20Transform) and the [convolution](/page/Convolution): \begin{align*} \widehat{f * g}(\xi) &= \int_{\mathbb{R}^n} (f * g)(x) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} \left(\int_{\mathbb{R}^n} f(x - y) \, g(y) \, d\mathcal{L}^n(y)\right) e^{-ix \cdot \xi} \, d\mathcal{L}^n(x). \end{align*} [/step] [step:Justify the exchange of integration order via Fubini--Tonelli] The double integral is absolutely convergent: by Tonelli's theorem and the translation-invariance of Lebesgue measure ($\int_{\mathbb{R}^n} |f(x-y)| \, d\mathcal{L}^n(x) = \|f\|_{L^1}$ for every $y$), \begin{align*} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(x-y)| \, |g(y)| \, d\mathcal{L}^n(y) \, d\mathcal{L}^n(x) &= \|f\|_{L^1} \, \|g\|_{L^1} < \infty. \end{align*} Fubini's theorem therefore permits exchanging the order of integration. [/step] [step:Change variables and factor the exponential] After applying Fubini and substituting $z = x - y$ (so $x = z + y$ and $d\mathcal{L}^n(x) = d\mathcal{L}^n(z)$): \begin{align*} \widehat{f * g}(\xi) &= \int_{\mathbb{R}^n} g(y) \left(\int_{\mathbb{R}^n} f(z) \, e^{-i(z+y) \cdot \xi} \, d\mathcal{L}^n(z)\right) d\mathcal{L}^n(y). \end{align*} Since $e^{-i(z+y)\cdot\xi} = e^{-iz\cdot\xi} \, e^{-iy\cdot\xi}$, the inner integral factors: \begin{align*} \int_{\mathbb{R}^n} f(z) \, e^{-i(z+y)\cdot\xi} \, d\mathcal{L}^n(z) &= e^{-iy\cdot\xi} \int_{\mathbb{R}^n} f(z) \, e^{-iz\cdot\xi} \, d\mathcal{L}^n(z) = e^{-iy\cdot\xi} \, \hat{f}(\xi). \end{align*} [/step] [step:Conclude by recognising the remaining integral as $\hat{g}(\xi)$] Substituting back: \begin{align*} \widehat{f * g}(\xi) &= \hat{f}(\xi) \int_{\mathbb{R}^n} g(y) \, e^{-iy \cdot \xi} \, d\mathcal{L}^n(y) = \hat{f}(\xi) \, \hat{g}(\xi). \end{align*} [/step]