My professor did the following proof in my functional analysis lecture: Check that $(C[0, 1], \|\cdot\|_1)$ is not a Banach space, where $\|f\|_1 = \int_0^1 |f(x)|\,{\rm d}x.$
Proof: taking $(f_n)_{n \geq 1}$ a sequence in $C[0, 1]$ given by: $$f_n(x) = \begin{cases} \frac{1}{\sqrt x}, &\text{ if } \frac{1}{n} \le x \le 1 \\ \sqrt n, &\text{ if } 0 \le x \le \frac{1}{n} \end{cases} $$
If $n > m$, I computed: $$\|f_m - f_n\|_1 = \frac{1}{\sqrt m} - \frac{1}{\sqrt n}$$
Sending $m,n$ to $+\infty $ , $$\|f_m - f_n\|_1 \rightarrow 0$$ which proves that the sequence is Cauchy
Then if $\exists f \in C[0, 1]$ such that when $ n\rightarrow \infty , \|f_n - f\|_1\rightarrow 0 $ now when $ k\rightarrow \infty , f_{n_k} \rightarrow f $ almost everywhere in $[0,1]$, then $f(x) = \lim_{k \to \infty } f_{n_k}(x)= \frac{1}{\sqrt x}$ almost everywhere in $[0,1]$
but no continuous function $f$ in $[0,1]$ exists that coincides almost everywhere with the function $\frac{1}{\sqrt x}, x \in [0,1]$ , therefore $C[0, 1]$ is not complete
I don't understand the part after Cauchyness of $(f_n)$ is proven Where did that $k$ variable come from? Looks like it has to do with a subsequence but I don't get what they are doing and where did that "almost everywhere" come from too ? How did they get the function out of the 1-norm? From this, how does it follow $\lim_{k \to \infty } f_{n_k}(x)= \frac{1}{\sqrt x}$ almost everywhere in $[0,1]$ ? And then he writes "but no continuous function $f$ in $[0,1]$ exists that coincides almost everywhere with the function $\frac{1}{\sqrt x}, x \in [0,1]$", how can you know that if that is precisely what we are trying to prove here ? Can someone help me clear this up?
Note: I am just starting with measure theory, so I know "almost everywhere" means "except in a zero-measure set", but I have no experience, so please be explicit if using some known property or theorem
Thanks.
The norm that is used is the $L^1([0,1])$ norm.
It is known in measure theory, that if a sequence of functions converges in $L^1([0,1])$, that there exists a subsequence that converges pointwise almost everywhere to the same limit. This subsequence is most likely where the $k$ comes from.
The convergence $\lim_k f_{n_k}(x) = \frac1{\sqrt{x}}$ can be concluded from the pointwise convergence $\lim_n f_n(x)=\frac1{\sqrt{x}}$ for all $x\in (0,1]$.
No continuous function from $C[0,1]$ exists which coincides with $\frac1{\sqrt{x}}$ almost everywhere because all functions from $C[0,1]$ are bounded. But if a function is equal to $f(x)=\frac1{\sqrt{x}}$ almost everywhere, then it cannot be bounded by some $M>1$, because $f(x)>M$ holds for $x\in (0,1/M^2)$, and the set $(0,1/M^2)$ has positive measure.