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I am stuck on this problem involving characteristic functions. Say you have two characteristic functions, $\phi_1$ and $\phi_2$, and you are looking at the set $A =\{t : \phi_1(t) = \phi_2(t)\}$. How can you prove (or at least tell) that this set is closed?

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1 Answer 1

Since $\phi_1$ and $\phi_2$ are continuous, $\phi_1 - \phi_2$ is continuous, and the set in question is closed because it is the preimage of $\{0\} \subset \mathbb{C}$ under $\phi_1 - \phi_2$. (Recall that a function is continuous if and only if the preimage of every closed set is closed.)

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A characteristic function (i.e. Fourier transform of a probability measure on $\mathbb R$) is always continuous. –  Robert Israel Oct 11 '11 at 17:41
    
@RobertIsrael: Ah, I was not sure of that. I'll edit my answer. –  Zhen Lin Oct 11 '11 at 17:43
    
oh right, I forgot about that property of characteristic functions. thanks a lot for your help! –  bbq Oct 11 '11 at 17:46

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