[proofplan] We prove the Riemann--Lebesgue lemma in two parts. [Uniform continuity](/page/Uniform%20Continuity) of $\hat{f}$ follows from the dominated convergence theorem applied to $|f(x)||e^{-ix \cdot h} - 1|$, with the dominator $2|f|$ independent of $\xi$. Decay at infinity is proved first for indicator functions of rectangles (explicit computation), then extended to simple functions by linearity, and finally to all of $L^1$ by a density argument using the bound $\|\hat{f}\|_{L^\infty} \le \|f\|_{L^1}$ and the closedness of $C_0(\mathbb{R}^n)$ under uniform limits. [/proofplan] [step:Prove uniform continuity of $\hat{f}$ via the dominated convergence theorem] Fix $\xi \in \mathbb{R}^n$ and let $h \to 0$ in $\mathbb{R}^n$. Estimate: \begin{align*} |\hat{f}(\xi + h) - \hat{f}(\xi)| &= \left|\int_{\mathbb{R}^n} f(x) \, e^{-ix \cdot \xi}\left(e^{-ix \cdot h} - 1\right) d\mathcal{L}^n(x)\right| \leq \int_{\mathbb{R}^n} |f(x)| \, |e^{-ix \cdot h} - 1| \, d\mathcal{L}^n(x). \end{align*} The integrand satisfies $|f(x)| \, |e^{-ix \cdot h} - 1| \leq 2|f(x)|$ and $|e^{-ix \cdot h} - 1| \to 0$ pointwise as $h \to 0$. Since $2|f| \in L^1(\mathbb{R}^n)$, the [Dominated Convergence Theorem](/theorems/4) gives $|\hat{f}(\xi + h) - \hat{f}(\xi)| \to 0$. The bound $2|f|$ is independent of $\xi$, so for any $\varepsilon > 0$, the threshold $|h| < \delta$ ensuring the [integral](/page/Integral) is below $\varepsilon$ depends only on $f$ and $\varepsilon$, establishing uniform continuity. [/step] [step:Compute the Fourier transform of indicator functions of rectangles and verify decay] Let $R = \prod_{j=1}^n [a_j, b_j]$ and $f = \mathbb{1}_R$. By Fubini's theorem: \begin{align*} \hat{f}(\xi) &= \prod_{j=1}^n \int_{a_j}^{b_j} e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j) = \prod_{j=1}^n \frac{e^{-ia_j \xi_j} - e^{-ib_j \xi_j}}{i\xi_j} \end{align*} for $\xi$ with all components nonzero. Each factor is bounded by $2/|\xi_j|$, so as $|\xi| \to \infty$ at least one component $|\xi_j| \to \infty$, forcing the product to zero. Hence $\hat{f} \in C_0(\mathbb{R}^n)$ for indicator functions of rectangles. [/step] [step:Extend to simple functions by linearity] A [simple function](/page/Simple%20Function) $s = \sum_{k=1}^N c_k \, \mathbb{1}_{R_k}$, where each $R_k$ is a rectangle, has Fourier transform $\hat{s} = \sum_{k=1}^N c_k \, \widehat{\mathbb{1}_{R_k}}$. Since $C_0(\mathbb{R}^n)$ is a vector space, $\hat{s} \in C_0(\mathbb{R}^n)$. [/step] [step:Extend to all $L^1$ functions by density and uniform approximation] Let $f \in L^1(\mathbb{R}^n)$. Simple functions built from rectangles are dense in $L^1(\mathbb{R}^n)$, so there exists a [sequence](/page/Sequence) $\{s_m\}_{m \geq 1}$ of such simple functions with $\|f - s_m\|_{L^1} \to 0$. The bound \begin{align*} \|\hat{f} - \hat{s}_m\|_{L^\infty} &= \|\widehat{f - s_m}\|_{L^\infty} \leq \|f - s_m\|_{L^1} \to 0 \end{align*} shows $\hat{s}_m \to \hat{f}$ uniformly on $\mathbb{R}^n$. Since $C_0(\mathbb{R}^n)$ is closed under [uniform convergence](/page/Uniform%20Convergence) (it is a closed subspace of $L^\infty$), the [limit](/page/Limit) $\hat{f}$ belongs to $C_0(\mathbb{R}^n)$. [/step]