Exercise 3, § 2

Let $f$ be a real function $\geqslant 0$, defined on an interval ]0, a] and regulated on this interval, such that the integral $g(x) = \int_0^x f(t)\,dt$ is convergent, but that $g$ does not have a right derivative at the point 0. Show that there exists a function $f_1$, regulated on ]0, a], such that $(f_1(x))^2 = f(x)$ for all $x$, that the integral $g_1(x) = \int_0^x f_1(t)\,dt$ converges, and that $g_1$ has a right derivative at 0. (First consider the case where $f$ is identical to a step function on every interval $[\varepsilon, a]$ (for $\varepsilon > 0$). In this case, divide each interval on which $f$ is constant into a large enough number of equal parts, and take $f_1$ to be constant on each of these intervals, the sign of $f_1$ differing on two consecutive such intervals. Proceed in the same way in the general case.)