Why $x(t)$ should be 1 in $(\frac{1}{2}, a_m]$ in proving the non convergence of the sequence in $C[0,1]$? Here is the part of the book I am speaking about:






But I am wandering in the equalities before the last three lines, the one on the right, why  $x(t)$ should be 1 in $(\frac{1}{2}, a_m]$ in proving the non convergence of the sequence in $C[0,1]$? I know that $x_m$ is the line $m(t - \frac{1}{2})$ in this part of the interval.
Could anyone help me answer this question, please?

For every $s \in \left(\frac12, 1\right]$ there exists some natural $k$ such that $a_m \le s$ for $m \ge k$, that is $s$ definitely ends up with being an element of $[a_m, 1]$ as $m$ progresses. Now $$\underbrace{0 = \lim_{m \to +\infty} \int_{a_m}^1 \lvert 1-x(t) \rvert \mathrm d t}_{\text{as your book has proved}} \ge \int_s^1 \lvert 1-x(t) \rvert \mathrm d t \ge 0 .$$ Thus, if $x$ has to be a continuous function, then it has to be equal to $1$ on the interval $[s, 1]$, in particular $x(s) = 1$.
