= Solution
\b[No, under the absolute-integral convention.] Direct integration gives
$$
\widehat {\mathbf1_{[0,1]}}(\xi)=\frac{1-e^{-i\xi}}{i\xi},\qquad
\left|\widehat {\mathbf1_{[0,1]}}(\xi)\right|=\frac{2|\sin(\xi/2)|}{|\xi|}.
$$
On 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.
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