= Solution
\b[No.] Enumerate the rationals in $[0,1]$ as
$$
A=\{q_1,q_2,\ldots\},
$$
and define
$$
f(q_n)=n,\qquad f(x)=0\quad(x\notin A).
$$
The <function> is nonnegative and unbounded, while it is identically zero on $[0,1]\setminus A$. For any $r>0$, however, $f_r$ is positive on the dense set of rationals and zero on the dense set of irrationals. Every interval therefore has positive oscillation, and the upper and lower Darboux <integrals> differ. Thus even the truncations need not be Riemann integrable.
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