A positive "Fourier transform" is integrable

Let $f\in L^1_\mathbb C(\mathbb R^n)$. I once read, in one of my old exam, that if $\hat{f}(\mathbb R^n)\subset\mathbb R_+$, then $\hat{f}\in L^1(\mathbb R^n)$. As far as I remember, the professor gave an identity using mollifiers but I'm not 100% sure. Any hints on how to prove that ?

Thanks!

If $f(x)$ is continuous at $x = 0$ it is true: Let $\phi(x)$ be a Schwartz function with positive Fourier transform, such as a Gaussian, and define $\phi_{\epsilon}(x) = {1 \over \epsilon^n} \phi({x \over \epsilon})$. Then $f \ast \phi_{\epsilon}(x)$ converges to $f(x)$ for almost all $x$ (this is a property of approximations to the identity), while ${\scr{F}}(f \ast \phi_{\epsilon}) = \hat{f}(\xi)\hat{\phi}(\epsilon x)$.

Notice that $\hat{f}(\xi)\hat{\phi}(\epsilon \xi)$ is in $L^1$ since $\hat{f}$ is bounded and $\hat{\phi}(x)$ is a Schwartz function. So the Fourier inversion formula applies to $\hat{f}(\xi)\hat{\phi}(\epsilon \xi)$ and you have that $${\scr F}^{-1}(\hat{f}(\xi)\hat{\phi}(\epsilon \xi)) = f \ast \phi_{\epsilon}(x)$$ This holds at all $x$ since both sides are continuous functions.
Plugging in $x = 0$ gives $$\int_{{\mathbb R}^n} \hat{f}(\xi)\hat{\phi}(\epsilon \xi)\,d\xi = (2\pi)^{n \over 2}f \ast \phi_{\epsilon}(0)$$ Now let $\epsilon$ go to zero and you get $$\int_{{\mathbb R}^n} \hat{f}(\xi)\,d\xi = (2\pi)^{n \over 2}f (0)$$ I don't believe it's true necessarily if $f(x)$ is not continuous at $x = 0$... you can take a function $g(x)$ with $g(x) = c|x|^{-\epsilon}$ near the origin and then let $f(x) = g(x) \ast \bar{g}(-x)$; its Fourier transform will be $|\hat{g}(\xi)^2|$ which will be nonnegative.