I'm working my way through a Fourier analysis textbook and came across the classic result that a function cannot be both band- and time-limited, that is, if both $f$ and $\hat{f}$ are compactly supported then $f \equiv 0$. It also says that in fact if $f$ vanishes on any interval and $\hat{f}$ is compactly supported then $f \equiv 0$. My question is: does this result also hold for sets of positive measure, not just intervals?
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If the support of $\hat f$ is contained in $[-M,M]$, then $$ f(x) = \int_{-M}^M \hat f(\xi)e^{ix \xi }\,d\xi $$ (ignoring various constants involving $\pi$). Expand the exponential function into a power series. Since the series converges uniformly on $[-M,M]$, it can be integrated term by term, resulting in a uniformly convergent series $$ f(x) = \sum_{n=0}^\infty \frac{i^n x^n}{n!} \int_{-M}^M \hat f(\xi)\xi^n \,d\xi $$ Therefore, $f$ is a real-analytic function: it is locally represented by a power series. The Identity Theorem for real analytic functions says that the zero set of $f$ has no points of accumulation. In particular, it is countable, hence of measure zero.
Let me qualify the above statement with: when dealing with Fourier transform, functions that agree up to a null set are indistinguishable. So, the precise statement is that $f$ has a representative that is real-analytic. The conclusion about the zero set being of measure zero makes sense since this property is independent of the choice of a representative.
