By setting for every integer $k \geqslant 1$, $a_k = \frac{1}{k} - \frac{1}{2^{k+1}}$ and $b_k = \frac{1}{k} + \frac{1}{2^{k+1}}$, show that we can choose an integer $k_0 \geqslant 1$ such that: $$\forall k \geqslant k_0, \quad b_{k+1} < a_k.$$ Deduce that the function $f : ]0,1[ \longrightarrow \mathbf{R}$ defined by: $$f : t \longmapsto \begin{cases} k^2 \cdot 2^{k+1} \cdot (t - a_k), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k, a_k + \frac{1}{2^{k+1}}\right] \\ k^2 \cdot 2^{k+1} \cdot (b_k - t), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k + \frac{1}{2^{k+1}}, b_k\right] \\ 0, & \text{otherwise} \end{cases}$$ is a well-defined and continuous function on $]0,1[$, integrable on $]0,1[$ and that this function $f$ does not belong to the set $\mathscr{D}_{0,1}$.
By setting for every integer $k \geqslant 1$, $a_k = \frac{1}{k} - \frac{1}{2^{k+1}}$ and $b_k = \frac{1}{k} + \frac{1}{2^{k+1}}$, show that we can choose an integer $k_0 \geqslant 1$ such that:
$$\forall k \geqslant k_0, \quad b_{k+1} < a_k.$$
Deduce that the function $f : ]0,1[ \longrightarrow \mathbf{R}$ defined by:
$$f : t \longmapsto \begin{cases} k^2 \cdot 2^{k+1} \cdot (t - a_k), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k, a_k + \frac{1}{2^{k+1}}\right] \\ k^2 \cdot 2^{k+1} \cdot (b_k - t), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k + \frac{1}{2^{k+1}}, b_k\right] \\ 0, & \text{otherwise} \end{cases}$$
is a well-defined and continuous function on $]0,1[$, integrable on $]0,1[$ and that this function $f$ does not belong to the set $\mathscr{D}_{0,1}$.