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?
 A: 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 $x(s) = 1$.
A: We say that $x_m(t) \to x(t)$ (pointwise) if for every value for $t$ we can show that $\forall \epsilon > 0\;\exists M_{t,\epsilon}: |x_m(t)-x(t)|<\epsilon\;\;\forall m>M_{t,\epsilon}$. The specific $M_{t,\epsilon}$ may vary by your choice of $t$ (as shown by the subscripts -- added for clarity).
The fact that $x(t)=0$ on $\left[0,\frac12\right)$ and $x(t)=1$ on $\left(a_m,1\right]$ is pretty trivial.
For $t \in \left(\frac12,a_m\right]$ note that $a_m = \frac12 + \frac1m$. Now pick a $t' \in \left(\frac12,a_m\right)$. We know that $0<x_m(t')=m\left(t'-\frac12\right)<1$. What would be a good candidate for $x(t')$ (i.e., the pointwise limit of $x_m(t'))$?
If we pick $x(t')>1$ we know that won't work because our function is constrained to the range $[0,1]$. If we pick $x(t')=c\in (0,1)$, then $x_m(t')$ will exceed $x(t')$ when $t'>a_k$ for some $k>0$ since $x_k(t') \in \left(a_k,1\right] \implies x_k(t')=1>c$. Specifically, we can select $k>\left(t'-\frac12\right)^{-1}$. For all $m>k$ $|x_m-x(t)|=|x_m(t)-c| > 0$ and so $c$ cannot be the limiting value.
This leaves us with $x(t')=1$ as the pointwise limit for $x_m(t')$ for any $t' \in \left(\frac12,1\right]$
We can also see this via the following (more direct) line of reasoning:
$$a_m=\frac12 + \frac1m \xrightarrow{m\to \infty} \frac12 \implies (a_m,1] \xrightarrow{m\to \infty} \left(\frac12,1\right] \implies x_m(t') \xrightarrow{m\to \infty} 1 \;\;\forall t'>\frac12$$
The author used a geometric argument in Figures 9 and 10 to show that the sequence of functions $x_m$ are Cauchy for the metric $d(x_m,x_n)=\int_0^1|x_m(t)-x_n(t)|dt$, and these functions, being continuous and defined on $[0,1]$ are members of $\mathcal{C}([0,1])$.
However, as shown in the author's proof and above, $x_m(t) \xrightarrow{m\to \infty} \mathbf{1}_{>0}(t) \notin \mathcal{C}([0,1])$.
By definition of a complete metric space, $\mathcal{C}([0,1])$ is complete $iff$ the limit of all Cauchy sequences in $\mathcal{C}([0,1])$ are also members of $\mathcal{C}([0,1])$. The indicator function $\mathbf{1}_{>0}(t)$ is not continuous at $t=\frac12$ and so $\mathcal{C}([0,1])$ is not complete.
