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?](https://math.stackexchange.com/questions/2356071/what-is-the-completion-of-c0-1-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](https://math.stackexchange.com/questions/1737562/the-space-ca-b-with-f-int-ab-fxdx-is-a-normed-space-but-not-a?noredirect=1&lq=1)
3. [$C([0, 1])$ is not complete with respect to the norm $\lVert f\rVert \_1 = \int\_0^1 \lvert f (x) \rvert \,dx$](https://math.stackexchange.com/questions/402841/c0-1-is-not-complete-with-respect-to-the-norm-lvert-f-rvert-1-int-0?noredirect=1&lq=1)

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?