[proofplan] We verify that the convolution $f * \nu$ is integrable using [Fubini's theorem](/theorems/513) and the finiteness of $\nu$, then compute its Fourier transform by exchanging the order of integration and applying translation invariance of $\mathcal{L}^d$. [/proofplan] [step:Verify $f * \nu \in L^1$ via Fubini and the finiteness of $\nu$] [claim:Convolution Is Integrable] $\|f * \nu\|_1 \leq \|f\|_1 \cdot \nu(\mathbb{R}^d)$. [/claim] [proof] By [Fubini's theorem](/theorems/513) applied to the product of $\mathcal{L}^d$ and $\nu$, \begin{align*} \int_{\mathbb{R}^d} \left(\int_{\mathbb{R}^d} |f(x-y)|\,\nu(dy)\right) d\mathcal{L}^d(x) = \int_{\mathbb{R}^d} \left(\int_{\mathbb{R}^d} |f(x-y)|\,d\mathcal{L}^d(x)\right) \nu(dy). \end{align*} For fixed $y$, the substitution $z = x - y$ (translation invariance of $\mathcal{L}^d$) gives $\int |f(x-y)|\,d\mathcal{L}^d(x) = \|f\|_1$. Therefore \begin{align*} \int_{\mathbb{R}^d} |(f * \nu)(x)|\,d\mathcal{L}^d(x) \leq \|f\|_1 \cdot \nu(\mathbb{R}^d) < \infty. \end{align*} [/proof] [/step] [step:Compute $\widehat{f * \nu}$ by exchanging integration order and using translation invariance] [claim:Fubini Exchange And Translation] $\widehat{f * \nu}(u) = \hat{f}(u)\,\hat{\nu}(u)$. [/claim] [proof] The integrand $(x, y) \mapsto f(x-y)\,e^{i\langle u, x\rangle}$ satisfies $|f(x-y)\,e^{i\langle u,x\rangle}| = |f(x-y)|$, whose double integral is $\|f\|_1 \cdot \nu(\mathbb{R}^d) < \infty$. By [Fubini's theorem](/theorems/513), we exchange the order: \begin{align*} \widehat{f * \nu}(u) = \int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d} f(x - y)\,e^{i\langle u, x\rangle}\,d\mathcal{L}^d(x)\right)\nu(dy). \end{align*} For the inner integral, substitute $z = x - y$: \begin{align*} \int_{\mathbb{R}^d} f(x-y)\,e^{i\langle u, x\rangle}\,d\mathcal{L}^d(x) = e^{i\langle u, y\rangle}\int_{\mathbb{R}^d} f(z)\,e^{i\langle u,z\rangle}\,d\mathcal{L}^d(z) = e^{i\langle u, y\rangle}\,\hat{f}(u). \end{align*} Substituting back: \begin{align*} \widehat{f * \nu}(u) = \hat{f}(u)\int_{\mathbb{R}^d} e^{i\langle u, y\rangle}\,\nu(dy) = \hat{f}(u)\,\hat{\nu}(u). \end{align*} [/proof] [/step]