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Let $u\in\mathcal{S}'(\mathbb{R}^n)$. Suppose that $u$ is also a measurable function on $\mathbb{R}^n$. Is it true that the Fourier transform $\hat{u}$ as a tempered distribution is always a complex Borel measure?

I believe that this is the case with more assumptions on $u$. For instance, say $u$ is integrable. fixing a bounded open set $\Omega$ in $\mathbb{R}^n$, for any a test function $\phi\in \mathcal{D}(\Omega)$, $$|\langle \hat{u},\phi\rangle|=\big|\int_{\mathbb{R}^n}u(x)\int_{\Omega}\phi(p)e^{-ix\cdot p}dp\big|\le \|u\|_{1}\|\phi\|_{\infty}|\Omega|$$ and from there on the Riesz–Markov–Kakutani representation theorem could be used.

But is $\hat{u}$ a complex Borel measure under weaker assumptions on $u$, or none at all other than that $u$ is a measurable function?

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    $\begingroup$ No! For example, the Fourier transform of $\mathrm{sign} x$ is $\operatorname{p.v.} (\pi x)^{-1}$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 2, 2019 at 6:14
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    $\begingroup$ @Cabbage By Bochner's theorem, the functions whose Fourier transforms are measures are exactly the complex functions "of positive definite type", i.e. the linear span of the positive-definite functions. In particular, they are all continuous. So there is something wrong with your argument. For example, the indicator function of $[-1,1]$ is integrable, but its Fourier transform, the sinc function (up to normalization) does not define a measure because it is not absolutely integrable. $\endgroup$
    – Robert Furber
    Commented Apr 2, 2019 at 7:14

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