Exercise 10, § 6

In the interval $I = ]0,1[$ of $\mathbf{R}$, define the function $g$ by $g(t) = -\sqrt{n}$ for $\frac{1}{2}(\frac{1}{n} + \frac{1}{n+1}) < t \leq \frac{1}{n}$ and $g(t) = \sqrt{n}$ for $\frac{1}{n+1} < t \leq \frac{1}{2}(\frac{1}{n} + \frac{1}{n+1})$ ($n = 1,2,\ldots$), and extend $g$ by 0 in $\mathbf{R} - I$; $g$ is integrable for Lebesgue measure. Set

show that the function $f$, equal to $1/\sqrt{x}$ on $[ -1,+1 ]$ and to 0 elsewhere, which is integrable for Lebesgue measure, is such that $t \mapsto f(G(t))g(t)$ is not integrable on $I$.