The strategy is to verify that the convolution 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 the translation invariance of $\mathcal{L}^d$.

**Step 1 ($f * \nu$ is integrable).**

[claim:Convolution Is Integrable] If $f \in L^1(\mathbb{R}^d, \mathcal{L}^d)$ and $\nu$ is a finite Borel measure on $\mathbb{R}^d$, then $f * \nu \in L^1(\mathbb{R}^d, \mathcal{L}^d)$ with $\|f * \nu\|_1 \leq \|f\|_1 \cdot \nu(\mathbb{R}^d)$. [/claim]

[proof] The function $(x, y) \mapsto |f(x - y)|$ is $\mathcal{B}(\mathbb{R}^d) \otimes \mathcal{B}(\mathbb{R}^d)$-measurable. By [Fubini's theorem](/theorems/513) (applied to the $\sigma$-finite 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 $\int |f(x-y)|\,d\mathcal{L}^d(x) = \int |f(z)|\,d\mathcal{L}^d(z) = \|f\|_1$. Therefore \begin{align*} \int_{\mathbb{R}^d} |(f * \nu)(x)|\,d\mathcal{L}^d(x) \leq \int_{\mathbb{R}^d} \|f\|_1\,\nu(dy) = \|f\|_1 \cdot \nu(\mathbb{R}^d) < \infty. \end{align*} In particular, $(f * \nu)(x)$ is finite for $\mathcal{L}^d$-a.e. $x$, and $f * \nu \in L^1$. [/proof]

**Step 2 (Compute the Fourier transform).** For any $u \in \mathbb{R}^d$, \begin{align*} \widehat{f * \nu}(u) &= \int_{\mathbb{R}^d} (f * \nu)(x)\,e^{i\langle u, x\rangle}\,d\mathcal{L}^d(x) = \int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d} f(x - y)\,\nu(dy)\right) e^{i\langle u, x\rangle}\,d\mathcal{L}^d(x). \end{align*}

[claim:Fubini Exchange And Translation] The double integral is absolutely convergent and equals $\hat{f}(u)\,\hat{\nu}(u)$. [/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 Step 1. By [Fubini's theorem](/theorems/513), we may exchange the order of integration: \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 $z = x - y$ (so $x = z + y$, $d\mathcal{L}^d(x) = d\mathcal{L}^d(z)$): \begin{align*} \int_{\mathbb{R}^d} f(x-y)\,e^{i\langle u, x\rangle}\,d\mathcal{L}^d(x) = \int_{\mathbb{R}^d} f(z)\,e^{i\langle u, z+y\rangle}\,d\mathcal{L}^d(z) = 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) = \int_{\mathbb{R}^d} e^{i\langle u, y\rangle}\,\hat{f}(u)\,\nu(dy) = \hat{f}(u)\int_{\mathbb{R}^d} e^{i\langle u, y\rangle}\,\nu(dy) = \hat{f}(u)\,\hat{\nu}(u). \end{align*} [/proof]