[proofplan] We introduce a Gaussian convergence factor $e^{-t|u|^2/2}$ that makes all integrals absolutely convergent, rewrite the resulting expression as a spatial convolution with the Gaussian density $g_t$ (using [Fubini](/theorems/513) and the explicit Fourier transform of a Gaussian), recognise this as an approximation to the identity, and pass to the limit $t \to 0$ using the Dominated Convergence Theorem. The $L^1$ approximation property gives a.e. convergence along a subsequence, and DCT on the Fourier side gives full convergence. [/proofplan] [step:Define the Gaussian-regularised inversion integral $I_t(x)$] For $t > 0$, define \begin{align*} I_t(x) = \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} \hat{f}(u)\,e^{-i\langle u, x\rangle}\,e^{-t|u|^2/2}\,d\mathcal{L}^d(u). \end{align*} Since $|\hat{f}(u)| \leq \|f\|_1$ (by the [Riemann-Lebesgue Lemma](/theorems/526)) and $e^{-t|u|^2/2}$ is integrable, the product is in $L^1$, so $I_t(x)$ is well-defined. [/step] [step:Rewrite $I_t(x)$ as the Gaussian convolution $(f * g_t)(x)$] [claim:Regularised Integral Equals Gaussian Convolution] $I_t(x) = (f * g_t)(x)$ where $g_t(z) = (2\pi t)^{-d/2} e^{-|z|^2/(2t)}$. [/claim] [proof] Substitute $\hat{f}(u) = \int f(y)\,e^{i\langle u, y\rangle}\,d\mathcal{L}^d(y)$ into $I_t(x)$. The double integral is absolutely convergent: $\|f\|_1 \cdot (2\pi/t)^{d/2} < \infty$. By [Fubini's theorem](/theorems/513), exchange the order: \begin{align*} I_t(x) = \int_{\mathbb{R}^d} f(y)\left(\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} e^{i\langle u, y - x\rangle}\,e^{-t|u|^2/2}\,d\mathcal{L}^d(u)\right) d\mathcal{L}^d(y). \end{align*} Completing the square gives $\int e^{i\langle u, z\rangle} e^{-t|u|^2/2}\,d\mathcal{L}^d(u) = (2\pi/t)^{d/2} e^{-|z|^2/(2t)}$. Setting $z = y - x$: \begin{align*} \frac{1}{(2\pi)^d}\int e^{i\langle u, y-x\rangle}\,e^{-t|u|^2/2}\,d\mathcal{L}^d(u) = g_t(y - x). \end{align*} Therefore $I_t(x) = (f * g_t)(x)$. [/proof] [/step] [step:Show Gaussian convolutions approximate $f$ in $L^1$] [claim:Gaussian Approximation To Identity] $\|f * g_t - f\|_1 \to 0$ as $t \to 0$. [/claim] [proof] Since $g_t \geq 0$ and $\int g_t\,d\mathcal{L}^d = 1$, \begin{align*} \|f * g_t - f\|_1 \leq \int_{\mathbb{R}^d} \omega(y)\,g_t(y)\,d\mathcal{L}^d(y), \end{align*} where $\omega(y) = \|f(\cdot - y) - f(\cdot)\|_1$ is the $L^1$ translation modulus. Since $\omega(y) \to 0$ as $y \to 0$, $\omega(y) \leq 2\|f\|_1$, and $g_t$ concentrates at the origin as $t \to 0$, the standard approximation-to-the-identity argument gives $\int \omega(y)\,g_t(y)\,d\mathcal{L}^d(y) \to 0$. [/proof] [/step] [step:Pass to the limit $t \to 0$ using DCT and $L^1$ convergence] Since $\hat{f} \in L^1$ by hypothesis, the integrand in $I_t(x)$ is dominated by $|\hat{f}(u)| \in L^1$. As $t \to 0$, the integrand converges pointwise to $\hat{f}(u)\,e^{-i\langle u, x\rangle}$. By DCT, \begin{align*} I_t(x) \to \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} \hat{f}(u)\,e^{-i\langle u, x\rangle}\,d\mathcal{L}^d(u) \quad \text{for every } x. \end{align*} By the previous steps, $I_t = f * g_t \to f$ in $L^1$, so there exists a subsequence $t_k \to 0$ with $(f * g_{t_k})(x) \to f(x)$ a.e. Since the full limit exists from DCT, convergence holds along the full net: \begin{align*} f(x) = \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} \hat{f}(u)\,e^{-i\langle u, x\rangle}\,d\mathcal{L}^d(u) \quad \text{for } \mathcal{L}^d\text{-a.e. } x. \end{align*} [/step]