TITLE: Fourier transform of a bounded function
QUESTION [10 upvotes]: This should really be well-known, but I was not able to find a definite answer to this question:
Is the Fourier transform of a bounded function always a borel measure (i.e. an order 0 distribution)?
In some sense, the distributional order corresponds to the order of a bounding polynomial and a bounded function can be bounded by an order zero polynomial. But I could not find any reference.
If the above statement is false, what is a counterexample?
If the statement is true: How does bounded variation correspond to all this? I.e. are there criterions for the Fourier transform to be of bounded variation?

REPLY [15 votes]: The answer is no : Fourier transform of the signum function is a constant times the distribution  $\lim_{\epsilon\to 0} 1_{|x|>\epsilon}/x$, of order $1$, also known as the principal value of $1/x$. Another keyword is Hilbert transform.