 Solution

No, under the absolute-integral convention. Direct integration givesOn a fixed positive-length portion of every period, the numerator is bounded below. Summing the resulting lower bounds gives a divergent harmonic series, so the Fourier transform is not in $L^1$. Also there is no continuous almost-everywhere representative: continuity would force it to be $1$ on $(0,1)$ and $0$ on $(-\infty ,0)$, giving incompatible limits at $0$. Symmetrically truncated conditional inversion does hold at interior continuity points and gives half the jump value at the endpoints; this does not contradict the answer.