As I read in this [post](https://mathoverflow.net/questions/300024/about-the-fourier-transform-of-the-logarithm-function) the Fourier transform of $\psi(\lambda) = \log{|\lambda|}$ must be interpreted in distributional sense and it is given by: 
$$\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$

where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$ (Schwartz space), for any $\nu >0$:

$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$

It follows that the convolution of $\mathscr{F}\{\psi\} * f$ is well defined as soon as $f$ decays fast enough around zero and at infinity. 

**Now my question is**: if you consider a measure $\mu$ that is compactly supported on $[a,b]$ with $a,b >0$ (therefore it is zero before $a$ and after $b$) is the convolution $\mathscr{F}\{\psi\} * \mu$ still well defined?

Ps: I posted before this question on MSE but I realised it was not the suitable website.