I am trying to understand the proof of the second Paley-Wiener theorem, which states sufficient and necessary conditions for a function in $L^2(\mathbb{R})$ to have Fourier Transform with compact support. The idea of the proof is the following one:
Let $f\in L^2(\mathbb{R})$. For each $\varepsilon>0$, define $f_{\varepsilon}(x) = f(x)e^{-2\pi\varepsilon|x|}$. Note that $$\|f-f_\varepsilon\|_{L^2(\mathbb{R})} = \int_{\mathbb{R}} |f(x) - f_{\varepsilon}(x)|^2dx = \int_{\mathbb{R}} |f(x)|^2 (1 - e^{-2\pi\varepsilon|x|})^2dx.$$ Since the function inside the integral is dominated by $|f|^2\in L^1(\mathbb{R})$, we can apply the Dominated Convergence Theorem to obtain $$\lim_{\varepsilon\to 0}\|f-f_\varepsilon\|_{L^2(\mathbb{R})} = 0.$$ It means that $f_\varepsilon\to f$ in $L^2(\mathbb{R})$.
In order to show that the support of $\widehat{f}$ is contained in $[-\delta,\delta]$, it is enough to prove that $$\lim_{\varepsilon\to 0} \int_{\mathbb{R}}f_\varepsilon(x)e^{-2\pi ix\xi dx} = 0$$ for all $|\xi|>\delta$.
Question: Why can we interchange the limit and the integral in the last equality in order to get the desired result? I guess that it follows directly from the convergence that we proved in $L^2(\mathbb{R})$.
I have this idea: Since $f_\varepsilon\to f$ in $L^2(\mathbb{R})$, we have that $\widehat{f_\varepsilon}\to \widehat{f}$ in $L^2(\mathbb{R})$ by using the Plancherel's Theorem. Now because $\widehat{f_\varepsilon} \to 0$ in $L^2(\mathbb{R}\setminus[-\delta,\delta])$, it follows that $\widehat{f}=0$ in $L^2(\mathbb{R}\setminus[-\delta,\delta])$ (i.e., the equality holds almost everywhere in that set). Therefore, $\widehat{f}\in L^2([-\delta,\delta])$.
Is this argument right, or what is the idea behind that statement?