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Sep 13, 2014 at 23:35 comment added Joni Teräväinen You are right. I am fairly certain though that there should be an unbounded set of finite measure whose Fourier transform is not analytic.
Sep 12, 2014 at 1:49 comment added Nick Strehlke This is by now a rather old question, but I thought it might be worth pointing out that, in fact, $\hat{\mathbb 1}_A$ does extend to an entire function when $A$ is bounded. This is part of the Paley-Wiener family of results. The problem with the uniform approximation argument is that the convergence $\hat{s}_k\to\hat{f}$ takes place in $L^\infty(\mathbb R^n)$ and not in $L^\infty(\mathbb C^n)$.
Apr 25, 2014 at 11:09 history edited Joni Teräväinen CC BY-SA 3.0
Added references
Apr 24, 2014 at 9:40 history edited Joni Teräväinen CC BY-SA 3.0
fixed typo
Apr 23, 2014 at 18:30 history answered Joni Teräväinen CC BY-SA 3.0