Fourier transform of $L^1$ function is continuous function

Question

I would like to know if my proof is correct. Define Fourier transform of $f\in L^1(\mathbb{R}^n)$ as $$\widehat{f}(x)=\int_{\mathbb{R}^n}e^{-ix\cdot y}f(y)\text{d}y.$$ I want to show that $\widehat{f}$ is continuous. So pick any sequence $x_n\rightarrow x$. Now $g_n(y)=e^{-ix_n\cdot y}f(y)$ is measurable for all $n\in\mathbb{N}$, $g_n(y)\rightarrow e^{-ix\cdot y}f(y)=g(y)$ for all $y$ and $|e^{-ix_n\cdot y}f(y)|=|f(y)|$ where $|f(y)|$ is integrable. So by dominated convergence theorem we have $$\lim_{n\rightarrow \infty}\widehat{f}(x_n)=\lim_{n\rightarrow \infty}\int_{\mathbb{R}^n}e^{-ix_n\cdot y}f(y)\text{d}y=\lim_{n\rightarrow\infty}\int_{\mathbb{R}^n}g_n(y)\text{d}y=\int_{\mathbb{R}^n}g(y)\text{d}y=\int_{\mathbb{R}^n}e^{-ix\cdot y}f(y)\text{d}y=\widehat{f}(x).$$ Therefore $\hat{f}$ is continuous. Am I correct?