Let \(f\in L^1(\RR^n)\), and suppose its Fourier transform \(\widehat f(\xi)=\int_{\RR^n}f(x)e^{-2\pi i x\cdot\xi}\,dx\) also lies in \(L^1(\RR^n)\). Then \[ F(x)\coloneqq\int_{\RR^n}\widehat f(\xi)e^{2\pi i x\cdot\xi}\,d\xi \] defines a bounded continuous function \(F\colon\RR^n\to\CC\), and \(F(x)=f(x)\) for almost every \(x\in\RR^n\). More precisely, \(F(x)=f(x)\) at every Lebesgue point \(x\) of \(f\), that is, every \(x\) with \[ \lim_{r\to0^+}\frac{1}{m(B_r(x))}\int_{B_r(x)}\abs{f(y)-f(x)}\,dy=0 ; \] in particular, \(F(x)=f(x)\) at every point \(x\) at which \(f\) is continuous.
Since \(\abs{\widehat f(\xi)e^{2\pi ix\cdot\xi}}=\abs{\widehat f(\xi)}\) and \(\widehat f\in L^1\), the bound \(\abs{F(x)}\leq\norm{\widehat f}_1\) holds and dominated convergence shows that \(F\) is continuous.
For \(t>0\), set \[ G_t(x)\coloneqq t^{-n}e^{-\pi \abs{x}^2/t^2}. \] Its Fourier transform is \[ \widehat{G_t}(\xi)=e^{-\pi t^2\abs{\xi}^2}. \] Indeed, in one dimension the function \[ H(\xi)\coloneqq\int_{\RR}e^{-\pi x^2}e^{-2\pi i x\xi}\,dx \] satisfies, by differentiation under the integral sign and integration by parts, \[ H'(\xi)=-2\pi\xi H(\xi), \qquad H(0)=1, \] so \(H(\xi)=e^{-\pi\xi^2}\). The \(n\)-dimensional identity follows by Fubini’s theorem, since \(e^{-\pi\abs{x}^2}=\prod_{j=1}^ne^{-\pi x_j^2}\), and the formula for \(G_t\) follows by the substitution \(x=ty\). Since \(G_t\) is even, the same computation gives \(\int_{\RR^n}e^{2\pi i w\cdot\xi}e^{-\pi t^2\abs{\xi}^2}\,d\xi=G_t(w)\) for \(w\in\RR^n\).
Fix \(x\in\RR^n\). By Fubini’s theorem T-4GPEF, \[ \begin{aligned} \int_{\RR^n}\widehat f(\xi) e^{2\pi i x\cdot\xi}e^{-\pi t^2\abs{\xi}^2}\,d\xi &=\int_{\RR^n}f(y) \qty{\int_{\RR^n}e^{2\pi i(x-y)\cdot\xi} e^{-\pi t^2\abs{\xi}^2}\,d\xi}dy\\ &=\int_{\RR^n}f(y)G_t(x-y)\,dy\\ &=(f*G_t)(x). \end{aligned} \] Fubini’s theorem applies because \[ \int_{\RR^n}\int_{\RR^n} \abs{f(y)}e^{-\pi t^2\abs{\xi}^2}\,d\xi\,dy=\norm{f}_1\,t^{-n}<\infty. \]
As \(t\to0^+\), dominated convergence on the left, with dominating function \(\abs{\widehat f}\), gives \[ \int_{\RR^n}\widehat f(\xi)e^{2\pi i x\cdot\xi}e^{-\pi t^2\abs{\xi}^2}\,d\xi \longrightarrow F(x). \] On the right, \(G_t(y)=t^{-n}G_1(y/t)\), where \(G_1\geq0\) has \(\int G_1=1\) and is radial and decreasing in \(\abs{y}\), so \((G_t)_{t>0}\) is an approximate identity with a radially decreasing integrable majorant. For such a family, \((f*G_t)(x)\to f(x)\) as \(t\to0^+\) at every Lebesgue point \(x\) of \(f\). Therefore \(F(x)=f(x)\) at every Lebesgue point of \(f\). By the Lebesgue differentiation theorem, almost every point of \(\RR^n\) is a Lebesgue point of \(f\), and every point of continuity of \(f\) is a Lebesgue point.