Inverse Fourier transform and decay at infinity

Suppose $1<p<\infty$, $g:\mathbb{R} \to \mathbb{R}$. How can I see $\mathcal{F}^{-1}g \in L^p(\mathbb{R})$ as a condition on the decay of $g$ at infinity?

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I assume that $g$ is a locally integrable function. Then $\mathcal F^{-1}\in L^p$ implies the following decay property: for every $h>0$, $$\frac{1}{2h}\int_{x-h}^{x+h} g(t)\,dt \to 0 \quad \text{as } \ x\to\infty$$ Proof: $\frac{1}{2h}\int_{x-h}^{x+h} g(t)\,dt$ is the convolution of $g$ with $\phi=(2h)^{-1}\chi_{[-h,h]}$. The function $\mathcal F^{-1}\phi$ is not integrable (it's a sinc function), but is in $L^q$ for every $q>1$. By Hölder's inequality $$(\mathcal F^{-1}g)(\mathcal F^{-1}\phi)\in L^1$$ Applying $\mathcal F$, we find that $g*\phi$ is continuous and vanishes at infinity.
There could be more to say if we had an upper bound $p\le p_0<\infty$, which would allow to consider something more singular, like Riesz potentials of $g$.