Showing that the space $C[0,1]$ with the $L_1$ norm is incomplete

This example works to show $C[0,1]$ is not complete with respect to the $L^p$ norm for all $1\leq p < \infty.$

Consider the piecewise linear function described by $$f_n(x)=\begin{cases} 1,&\text{if }0\le x\le\frac12\\ 1-n(x-\frac{1}{2}),&\text{if }\frac12\le x\le \frac{1}{2}+\frac{1}{n}\\\\ 0,&\text{if }\frac{1}{2}+\frac{1}{n}\le x\le 1\;. \end{cases}$$

Then $$ \| f_n - f_m \|_p = \left( \int^{1/2+1/n}_{1/2} |f_n(x)-f_m(x)|^p dx\right)^{1/p} \leq \left( \frac{1}{n} \right)^{1/p} \to 0$$

so $f_n$ is Cauchy. Suppose $f_n$ has limit $f\in C[0,1].$ Then $$\int^{1/2}_0 |f(x)-f_n(x)|^p dx \leq \|f-f_n\|_p^p \to 0$$ so $f(x)=1$ on $[0,1/2].$ Similarly we see $f(x) = 0$ on $[1/2,1]$, which is a contradiction.

Of course, the calculations are easy to verify once you have the example, and no one actually remembers the explicit equations for such a thing. All you have to remember is that it is $1$ on $[0,1/2]$ then goes down very quickly to $0.$

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It's worth remembering that the continuous functions are dense in $L^1[0,1]$, so you can limit to any discontinuous function in $L^1[0,1]$ you desire. For a proof, see here.

For an explicit example, consider the sequence of functions defined for $x\in[0,1]$ by

$$f_n(x)=\begin{cases} \left(x+\frac{1}{2}\right)^n,&\text{if }0\le x<\frac{1}{2}\\\\ 1,&\text{if } \frac{1}{2}\le x\le 1\;. \end{cases}$$

One easily checks that these are continuous, but the sequence limits to a function $f$ defined by

$$f(x)=\begin{cases} 0,&\text{if }0\le x<\frac{1}{2}\\\\ 1,&\text{if } \frac{1}{2}\le x\le 1\;. \end{cases}$$

One can try to remember $f_n(x)=\dfrac1{\mathrm e^{n(2x-1)}+1}$, which converges in $L^1$ to $f(x)=\mathbf 1_{x\leqslant1/2}$.