[proofplan] We prove the Fourier inversion formula by introducing a Gaussian convergence factor $e^{-\varepsilon|\xi|^2/2}$, exchanging order of integration via Fubini, computing the resulting Gaussian Fourier transform explicitly, recognising the outcome as a convolution with a normalised Gaussian (an approximation to the identity), and passing to the limit $\varepsilon \to 0$ using the dominated convergence theorem and the $L^1$ convergence of approximate identities. [/proofplan] [step:Regularise via a Gaussian convergence factor] Define for $\varepsilon > 0$: \begin{align*} I_\varepsilon(x) &= \int_{\mathbb{R}^n} \hat{f}(\xi) \, e^{ix \cdot \xi} \, e^{-\varepsilon|\xi|^2/2} \, d\mathcal{L}^n(\xi). \end{align*} Since $\hat{f} \in L^\infty(\mathbb{R}^n)$ (by $\|\hat{f}\|_{L^\infty} \leq \|f\|_{L^1}$) and $e^{-\varepsilon|\xi|^2/2}$ is in $\mathcal{S}(\mathbb{R}^n)$, the product $\hat{f}(\xi) e^{-\varepsilon|\xi|^2/2}$ belongs to $L^1(\mathbb{R}^n)$, so $I_\varepsilon(x)$ is well-defined for every $x$. [/step] [step:Rewrite $I_\varepsilon$ as a spatial convolution via Fubini] Substituting $\hat{f}(\xi) = \int_{\mathbb{R}^n} f(y) e^{-iy \cdot \xi} \, d\mathcal{L}^n(y)$ and exchanging the order of integration (justified by Fubini, since $f \in L^1$ and the Gaussian provides absolute convergence): \begin{align*} I_\varepsilon(x) &= \int_{\mathbb{R}^n} f(y) \left(\int_{\mathbb{R}^n} e^{i(x-y)\cdot\xi} \, e^{-\varepsilon|\xi|^2/2} \, d\mathcal{L}^n(\xi)\right) d\mathcal{L}^n(y). \end{align*} The inner integral is the [Fourier transform](/page/Fourier%20Transform) of the Gaussian $\xi \mapsto e^{-\varepsilon|\xi|^2/2}$ evaluated at $-(x - y)$. The standard Gaussian computation gives \begin{align*} \int_{\mathbb{R}^n} e^{i(x-y)\cdot\xi} \, e^{-\varepsilon|\xi|^2/2} \, d\mathcal{L}^n(\xi) &= \left(\frac{2\pi}{\varepsilon}\right)^{n/2} e^{-|x-y|^2/(2\varepsilon)}. \end{align*} [/step] [step:Identify the approximation to the identity and express $I_\varepsilon$ as $(2\pi)^n(f * \varphi_\varepsilon)$] Define $\varphi_\varepsilon(z) = (2\pi\varepsilon)^{-n/2} e^{-|z|^2/(2\varepsilon)}$, a normalised Gaussian with $\int_{\mathbb{R}^n} \varphi_\varepsilon \, d\mathcal{L}^n = 1$. Then \begin{align*} I_\varepsilon(x) &= (2\pi)^n \int_{\mathbb{R}^n} f(y) \, \varphi_\varepsilon(x - y) \, d\mathcal{L}^n(y) = (2\pi)^n \, (f * \varphi_\varepsilon)(x). \end{align*} The family $\{\varphi_\varepsilon\}_{\varepsilon > 0}$ is an approximation to the identity: each $\varphi_\varepsilon$ is non-negative, integrates to $1$, and concentrates at the origin as $\varepsilon \to 0$. The standard $L^1$ theory gives $f * \varphi_\varepsilon \to f$ in $L^1$ as $\varepsilon \to 0$. If $f$ is continuous at $x$, then $f * \varphi_\varepsilon(x) \to f(x)$ pointwise. [/step] [step:Pass to the limit $\varepsilon \to 0$ using the dominated convergence theorem] Since $\hat{f} \in L^1(\mathbb{R}^n)$ by hypothesis, the [Dominated Convergence Theorem](/theorems/4) (with dominating function $|\hat{f}|$) gives \begin{align*} \lim_{\varepsilon \to 0} I_\varepsilon(x) &= \int_{\mathbb{R}^n} \hat{f}(\xi) \, e^{ix \cdot \xi} \, d\mathcal{L}^n(\xi). \end{align*} Combining with the previous step: \begin{align*} \int_{\mathbb{R}^n} \hat{f}(\xi) \, e^{ix \cdot \xi} \, d\mathcal{L}^n(\xi) &= (2\pi)^n \lim_{\varepsilon \to 0} (f * \varphi_\varepsilon)(x) = (2\pi)^n f(x) \end{align*} for $\mathcal{L}^n$-almost every $x$ (extracting an a.e.-convergent subsequence from the $L^1$ convergence $f * \varphi_\varepsilon \to f$). If $f$ is continuous, the pointwise convergence of the approximation to the identity upgrades this to equality for every $x \in \mathbb{R}^n$. [/step]