I know that if $f \in L^{1}(R)$, and $g$ is the Fourier transform of $f$, and $|g|\in L^{1}(R)$, then for almost every $x\in R$, $f(x)=\int g(u)e^{2 \pi iux}du$. But if $f(x)=\int g(u)e^{2 \pi iux}du$ for almost every $x\in R$, can I conclude $g$ is the Fourier Transform of $f$? If so, why?
Thanks in advance!
Let $\mathcal{F}$ denote the Fourier transform. The assumption is that $f = \mathcal{F}^{-1}g$ a.e. This means that $f$ works the same as $\mathcal{F}^{-1}g$ in integrals. Therefore, $$ \mathcal{F}f(u) = \int f(x) e^{-i2\pi ux} dx = \int (\mathcal{F}^{-1}g)(x) e^{-i2\pi ux} dx = \mathcal{F}(\mathcal{F}^{-1}g)(u) \stackrel{\text{a.e.}}{=} g(u) . $$