Solution
= Solution
\b[No.] Define
$$
f(0)=0,\qquad f(x)=\frac1x\quad(0<x\leq1).
$$
It is continuous at every point of $(0,1)$ and is unbounded. Its truncation is Riemann integrable, but
$$
\int_0^1\min\left(\frac1x,r\right)\,dx
=\int_0^{1/r}r\,dx+\int_{1/r}^1\frac{dx}{x}
=1+\log r\longrightarrow\infty.
$$
\b[Thus it is not improperly integrable with a finite <integral>.]