Exercise 1, § 2

Let $\alpha$ and $\beta$ be two finite real numbers such that $\alpha < \beta$. Show that if $\gamma$ and $\delta$ are two numbers such that $\alpha < \gamma \leq \delta < \beta$ then there exists a real function $f$ defined on an interval $[0, a]$, taking only the values $\alpha$ and $\beta$, such that, on all the interval $[\varepsilon, a]$ (for $\varepsilon > 0$), $f$ is a step function and that, if one puts $g(x) = \int_0^x f(t)\, dt$, one has

(*) (take for $g$ a function whose graph is a broken line whose consecutive sides have gradients $\alpha$ and $\beta$ and whose peaks occur alternately on the lines $y = \gamma x, \ y = \delta x$ for $\gamma \neq \delta$, or on the line $y = \gamma x$ and the parabola $y = \gamma x + x^2$ for $\gamma = \delta$).

By the same method show that, whether $\gamma$ and $\delta$ are finite or not ($\gamma < \delta$), there exists a real function $f$ defined on $[0, a]$, such that on every interval $[\varepsilon, a]$ (for $\varepsilon > 0$), $f$ is a step function, that the integral $g(x) = \int_0^x f(t)\, dt$ exists, and that one again has the relations (*).