Let $f$ be piecewise function $f: \mathbb{R} \rightarrow \mathbb{C}$ such that $ \int_{-\infty}^{\infty}|f(x)| d x<+\infty$
Statement: There is necessary exist $g$ such that $ \int_{-\infty}^{\infty}|g(\omega)| d \omega<+\infty$ and $f(x)= \int_{-\infty}^{\infty}g(\omega)e^{i\omega x} d \omega$
Answer : False
take $f(x)$=the indicator $\mathbb{\chi}_{[-1,1]}$
I dont understand this counter example, how does it contradict the statement?
$\int e^{-iwx}f(x)dx=\frac {\sin w} w$. This implies that $f$ is the Fourier Transform of $g(w)=\frac {\sin w }w$ but $g$ is not inetgrable.