The strategy is to 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 $g_t$, recognise this as an approximation to the identity, and pass to the limit $t \to 0$ using the [Dominated Convergence Theorem](/theorems/511). The main tools are [Fubini's theorem](/theorems/513) (to exchange integration orders), the explicit Fourier transform of a Gaussian, and the $L^1$ approximation property of Gaussian kernels.

**Step 1 (Gaussian-regularised inversion integral).** 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 $\hat{f}(u)\,e^{-t|u|^2/2}$ is in $L^1(\mathbb{R}^d, \mathcal{L}^d)$, so $I_t(x)$ is well-defined for every $x$.

**Step 2 (Rewrite as a spatial convolution).**

[claim:Regularised Integral Equals Gaussian Convolution] $I_t(x) = (f * g_t)(x)$ for every $x \in \mathbb{R}^d$, where $g_t(z) = (2\pi t)^{-d/2} e^{-|z|^2/(2t)}$ is the Gaussian density. [/claim]

[proof] Substitute the definition $\hat{f}(u) = \int_{\mathbb{R}^d} f(y)\,e^{i\langle u, y\rangle}\,d\mathcal{L}^d(y)$ into $I_t(x)$: \begin{align*} I_t(x) = \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d} f(y)\,e^{i\langle u, y\rangle}\,d\mathcal{L}^d(y)\right) e^{-i\langle u, x\rangle}\,e^{-t|u|^2/2}\,d\mathcal{L}^d(u). \end{align*} The double integral is absolutely convergent: $|f(y)\,e^{i\langle u, y-x\rangle}\,e^{-t|u|^2/2}| = |f(y)|\,e^{-t|u|^2/2}$, and $\int |f(y)|\,d\mathcal{L}^d(y) \cdot \int e^{-t|u|^2/2}\,d\mathcal{L}^d(u) = \|f\|_1 \cdot (2\pi/t)^{d/2} < \infty$. 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*} The inner integral is $(2\pi)^{-d}$ times the Fourier transform of $u \mapsto e^{-t|u|^2/2}$ evaluated at $-(y - x)$. Completing the square (as in the Gaussian Fourier transform computation): $\int e^{i\langle u, z\rangle} e^{-t|u|^2/2}\,d\mathcal{L}^d(u) = (2\pi/t)^{d/2} e^{-|z|^2/(2t)}$. Setting $z = y - x$: \begin{align*} \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} e^{i\langle u, y-x\rangle}\,e^{-t|u|^2/2}\,d\mathcal{L}^d(u) = \frac{1}{(2\pi t)^{d/2}}\,e^{-|y-x|^2/(2t)} = g_t(y - x). \end{align*} Substituting back: $I_t(x) = \int_{\mathbb{R}^d} f(y)\,g_t(y - x)\,d\mathcal{L}^d(y) = (f * g_t)(x)$. [/proof]

**Step 3 (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)(x) - f(x) = \int_{\mathbb{R}^d} (f(x - y) - f(x))\,g_t(y)\,d\mathcal{L}^d(y). \end{align*} Taking absolute values and integrating in $x$ (using [Fubini](/theorems/513)): \begin{align*} \|f * g_t - f\|_1 \leq \int_{\mathbb{R}^d} \left(\int_{\mathbb{R}^d} |f(x-y) - f(x)|\,d\mathcal{L}^d(x)\right) g_t(y)\,d\mathcal{L}^d(y) = \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. It is standard that $\omega(y) \to 0$ as $y \to 0$ (continuity of translation in $L^1$) and $\omega(y) \leq 2\|f\|_1$ for all $y$. Since $g_t$ concentrates at the origin as $t \to 0$ (for any $\delta > 0$, $\int_{|y| > \delta} g_t(y)\,d\mathcal{L}^d(y) \to 0$), a standard approximation-to-the-identity argument gives $\int \omega(y)\,g_t(y)\,d\mathcal{L}^d(y) \to 0$. [/proof]

**Step 4 (Pass to the limit $t \to 0$).** Since $\hat{f} \in L^1(\mathbb{R}^d, \mathcal{L}^d)$ by hypothesis, the integrand in $I_t(x)$ satisfies $|\hat{f}(u)\,e^{-i\langle u,x\rangle}\,e^{-t|u|^2/2}| \leq |\hat{f}(u)| \in L^1$ for all $t > 0$. As $t \to 0$, the integrand converges pointwise to $\hat{f}(u)\,e^{-i\langle u, x\rangle}$. By the [Dominated Convergence Theorem](/theorems/511), \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 Step 2, $I_t(x) = (f * g_t)(x)$. By Step 3, $f * g_t \to f$ in $L^1$, so there exists a subsequence $t_k \to 0$ with $(f * g_{t_k})(x) \to f(x)$ for $\mathcal{L}^d$-a.e. $x$. Since the full limit exists (from DCT), the 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*}