# A sufficient condition for a probability measure to have compact support

Consider a probability measure $\mu$ on, let's say, $\mathbb R$.

Is there a necessary and sufficient condition so that $\mu$ has compact support $Supp(\mu)$ ?

I agree this question is too vague, and it may tempting to answer it by quoting the definition of compactness for the support, so let me be more precise.

Imagine you know the Fourrier transform $F_\mu$ of $\mu$ (which contains all the information concerning $\mu$), that is $$F_\mu(t)=\int e^{itx}\mu(dx),\qquad t\in\mathbb R,$$ or its Cauchy-Stieltjes transform $C_\mu$ (similar story), i.e. $$C_\mu(z)=\int \frac{\mu(dx)}{z-x},\qquad z\in {\mathbb C }\setminus Supp(\mu),$$ is there any (necessary and) sufficient condition on $F_\mu$ or $C_\mu$ to force $Supp(\mu)$ to be a compact set ?

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Perhaps en.wikipedia.org/wiki/Hausdorff_moment_problem is relevant? –  Qiaochu Yuan May 5 '13 at 5:15

The Paley-Wiener theorem as it is presented in W.Rudin's "Functional analysis" (Theorem 7.23), will it be satisfactory for you? In your situation: $\mu$ has compact support if and only if $F_\mu$ can be extended as an entire function to $\mathbb C$, and for some $C>0$ and $r>0$ $$|F_\mu(z)|\le C\cdot e^{r\cdot |{\rm Im} z|},\qquad z\in {\mathbb C}.$$
Yes, it does ! By any chance do you know if there is a similar characterization involving $C_\mu$ ? –  Adrien Hardy May 5 '13 at 9:22