# Is the Fourier transform of a measurable function as a tempered distribution necessarily a complex Borel measure?

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?

• No! For example, the Fourier transform of $\mathrm{sign} x$ is $\operatorname{p.v.} (\pi x)^{-1}$. Apr 2, 2019 at 6:14
• @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. Apr 2, 2019 at 7:14