In Folland, Real Analysis, Theorem 8.35, he defines the following: $$\Phi \in L^1(\mathbb{R}^n) \cap C_0, \Phi(0) =1, \phi = \Phi^\vee := \int f(\xi) e^{2\pi i \xi \cdot x} d\xi,$$ where $C_0$ is the space of continuous functions vanishing at $\infty$. Further, he assumes $\phi \in L^1(\mathbb{R}^n)$.
He then note's that if $\phi_t(x) := t^{-n} \phi(t^{-1}x)$, $\Phi(t \xi) = (\phi_t)^\wedge(\xi)$, where $f^\wedge(\xi) := \int e^{2\pi i \xi \cdot x} f(x) dx$ is the Fourier transform of $f$. I can prove this step.
Finally, he notes then that $\int \phi(x) dx = \Phi(0) = 1.$
Why?
Since $\phi^\wedge = (\Phi^\vee)^\wedge = \Phi$ a.e. by Fourier inversion theorem and that $\phi, \Phi \in L^1$ by assumption. and $\phi^\wedge$, $\Phi$ are both continuous, hence $\phi^\wedge = \Phi$ everywhere. Hence: $$\Phi(0) = \phi^\wedge(0) = \int e^{2\pi i 0 \cdot x} \phi(x) dx = \int \phi(x) dx.$$