# What is the absurd in this proof of the non completeness of ($C[0,1],\| \cdot\|_1)$

I have the following proof in my notes:

$$(C[0,1],\| \cdot\|_1)$$ is not complete

Consider the sequence of functions $$f_n(t)= \begin{cases} \frac{1}{\sqrt t} &, & 1/n \le t \le 1 \\ \sqrt n &,&0\le t \le \frac{1}{n} \end{cases}$$

It can easily be shown that it is Cauchy in $$\lVert \cdot \rVert_1$$ So we have that $$\forall \varepsilon \exists N_\varepsilon$$ such that $$\lVert f_n -f \rVert_\infty \le \varepsilon \forall m,n >N_\varepsilon$$

Suppose $$f_n$$ converges, Then $$\exists f \in C[0,1]$$such that $$\lVert f_n-f\rVert_1 \to 0.$$

Then $$f_{n_k} \to f \text{ a.e in} [0,1]$$

$$f(t)=\lim_{k \to \infty} f_{n_k}(t)= \frac{1}{\sqrt t} \text{ a.e in} [0,1]$$. Absurd.

I'm just failing to understand the last line. Can someone please explain

1. what is the absurd?

2. And how does pointwise convergence follows from a.e convergence (last line). Am I supposed to understand this $$f_{n_k} \to f \text{ a.e in} [0,1]$$ as convergence in the real numbers, with the absolute value for every fixed t ( $$|f_{n_k}(t) -f(t) | \text{ a.e in} [0,1]$$ ) ?

• $1/{\sqrt t}$ is not continuous on $[0,1]$. Jan 18, 2022 at 11:50
• Since it happens almost everywhere, you can find a sequence $t_n$ which tends to $0$ such that $f(t_n)=\frac{1}{\sqrt{t_n}}$ for all $n$. This means $f$ can't be continuous at $0$, a contradiction.
– Mark
Jan 18, 2022 at 11:52
• @DavidC.Ullrich But $1/\sqrt t$ is a.e continuous in[0,1], right?, which is what $\lim_{k \to \infty} f_{n_k}(t)= \frac{1}{\sqrt t} \text{ a.e in} [0,1]$ says. So the limit exists Jan 18, 2022 at 11:59
• It's not an element of $C[0,1]$! So the sequence is not convergent in $C[0,1]$. Jan 18, 2022 at 12:00
• If two continuous functions on $(0,1]$ are equal a.e. then they are equal on all of $(0,1]$. Jan 18, 2022 at 16:39

1. The limit function $$f$$ does not belong to $$\left(C([0, 1]), \|\cdot\|_{\ell^1}\right),$$ creating the contradiction since $$f$$ was supposed to belong to this space. More precisely, $$f$$ is equal to $$1/\sqrt{t}$$ a.e, but by continuity, $$f(0) = \lim_{t \to 0} f(t) = +\infty,$$ and therefore $$f$$ may not be continuous at $$0,$$ since continuous functions on $$[0,1]$$ are in particular finite at $$0,$$ a contradiction.