Question regarding one approach to showing that the vector space $C[0, 1]$ is not complete w.r.t. the $L1$ integral norm

Question

First to clarify few things: I know that this question has been asked several times at this forum, take for example

1. What is the completion of $C[0,1]$ equipped with the integral norm?
2. The space $C[a,b]$ with $\|f\|:=\int_a^b |f(x)|d(x)$ is a normed space but not a Banach space
3. $C([0, 1])$ is not complete with respect to the norm $\lVert f\rVert _1 = \int_0^1 \lvert f (x) \rvert \,dx$

I am not, therefore, asking yet another proof for this question, but rather a sanity check/deepening explanation on an approach I took. I have had quite a brake on basic undergraduate level real analysis proofs, so my tool set is a bit rusty regarding convergence proofs.

(Preamble) My idea was to take the following modified sequence of sinc functions $f(n, x) = \begin{cases}\mathrm{sinc(n(x - 1/2))} &:x \in \left[\frac{1}{2} - \frac{-\pi}{n}, \frac{1}{2} + \frac{\pi}{n}\right]\\\ 0 &: \text{otherwise}\end{cases}$. It is quite easy to see that i.) for each fixed $n = 1, 2,\dots$ the sequence is continuous on $[0,1]$ w.r.t. $x$ ii.) the sequence is Cauchy w.r.t. the norm $||f|| = \int_{0}^1|f(x)|dx$, iii.) $\forall x \in [0, 1]:\lim_{n\to\infty}f(n, x) = \delta(x - 1/2)$ with the limit function certainly not being continuous on $[0,1]$.

(Question:) To my knowledge the usual punchline of any non-completeness proof is to argue that the limit is not in your space of interest. In our case this is quite evident. However, the one observation I cannot shake from my mind is the fact that when the Lebesgue measure is used in norm convergence, we do not care what happens in sets of measure zero. In particular, we also have that $\forall x \in [0,1]:\lim_{n\to\infty}||f(n,x) - 0|| = 0$ as the support of $f(n,-)$ shrinks to one point. So, as crazy as it sounds, to limit seems not to be unique, which is obviously nonsense. But why is this nonsense?

Edit: To be explicit, I am looking to understand 1.) Why does my approach seem not to work, i.e. has it something to do with this shrinking support of the sequence and 2.) is there any hope of recovering a valid proof?

Answer 1

I think the crux of the matter was pointed out by Kurt G. in the comments. Namely, $L^1$ convergence does not imply pointwise convergence to the same function.

Your sequence $(f_n)_n$ satisfies $f_n \to 0$ in $L^1$, but $f_n \to \mathbf{1}_\left\{\frac12\right\}$ pointwise and there is no contradiction here. Even though the pointwise limit is a discontinuous function, the sequence $(f_n)_n$ converges in $L^1$ so it does not provide a counterexample to the completeness of $(C[0,1],\|\cdot\|_1)$.

Perhaps those $\operatorname{sinc}$ functions are obfuscating things so take a look at this simple sequence $(g_n)_n$ in $C[0,1]$ given by $$g_n(t) = \begin{cases} 1-nt, &\text{ if }t \in \left[0,\frac1n\right],\\ 0, &\text{ if }t \in \left[\frac1n,1\right] \end{cases}$$ whose graph is a right triangle with legs $1$ and $\frac1n$. Then $\|g_n\|_1 = \frac1{2n} \to 0$ so $g_n \to 0$ in $L^1$ but pointwise we have $g_n \to \mathbf{1}_{\{0\}}$ and this is ok.

To answer your additional question, this has nothing to do with shrinking support of the functions. For example, the support of $g_n$ is $\left[0,\frac1n\right]$ which shrinks down to $\{0\}$ in the limit. On the other hand, the sequence $(n^2g_n)_n$ is unbounded in $L^1$-norm and hence cannot converge in $L^1$, but the supports again shrink down to $\{0\}$ in the limit. This sequence $(n^2g_n)_n$ incidentally does not converge pointwise to any function in $\Bbb{R}^{[0,1]}$ as the values at $0$ diverge.

Your thought "...the usual punchline of any non-completeness proof is to argue that the limit is not in your space of interest" is exactly right and can be applied here: select any function $f \in L^1[0,1]\setminus C[0,1]$ and approximate it w.r.t. $L^1$ norm with a sequence in $C[0,1]$. Such a sequence will be Cauchy but not convergent in $(C[0,1],\|\cdot\|_1)$. Your example is not of this form since $0 \in C[0,1]$.

Answer 2

We prove non-completness by constructing a non converging Cauchy sequence, and not by showing that the "limit" (where u implicitly assuming your initial space is embedded in some "bigger" space, completion, for example, containing "the limit") doesn't belong to the initial space because we are looking for the limit inside our space . Regarding your Question, the "limit" is Something relative, which depends on the topology, so unless we specify some topology(the sens in which the terms of the sequence "approach" the limit) , talking about the limit is meaninless. Now, even if we specify some topology, this still doesn't prevent the pathological "non-uniqueness" of limits. As a simple example, consider the set $\{0,1\}$ endowed with the following topology :the open subsets are :the empty set, the whole set and the singleton $\{1\} $,then the constant sequence $u_{n} =1$ converge to 0 and to 1.