Examples of function sequences in C[0,1] that are Cauchy but not convergent To better train my intuition, what are some illustrative examples of function sequences in C[0,1] that are Cauchy but do not converge under the integral norm?
 A: You can get examples by considering elements of $L^1[0,1]$ that are not equal almost everywhere to any continuous function, and considering sequences of continuous functions converging in the $L^1$ norm to these discontinuous functions.  Because convergent sequences are Cauchy and $L^1$ limits are unique up to equality almost everywhere, such sequences will be Cauchy and nonconvergent in $C[0,1]$.
E.g., let $f$ be $1$ on $[0,\frac{1}{2}]$ and $0$ elsewhere.  Let $f_n$ be the continuous function that is $1$ on $[0,\frac{1}{2}]$, $0$ on $[\frac{1}{2}+\frac{1}{n},1]$, and linear on $[\frac{1}{2},\frac{1}{2}+\frac{1}{n}]$.  Then because $f_n\to f$ in $L^1$, $(f_n)$ is Cauchy.  (The Cauchy-ness is also easy to verify directly.)  If there were a limit function $g\in C[0,1]$, you would have $g=f$ a.e..  But this is impossible, because the left-hand and right-hand limits at $\frac{1}{2}$ would not agree.  
More generally, a Cauchy sequence in a metric space $X$ with completion $\overline{X}$ that does not converge in $X$ is basically the same as a sequence in $X$ that converges to an element of $\overline{X}\setminus X$.  In a case like this where $X=C[0,1]$ with $L^1$ norm and $\overline{X}=L^1[0,1]$ have explicit descriptions, you can find examples by starting with an element of $\overline{X}\setminus X$, and find a sequence in $X$ converging to that element.  The same idea applies to demonstrating nonconvergent Cauchy sequences in $\mathbb{Q}$, where you can take any irrational number and consider the sequence of truncated decimal expansions.
A: By measure theory you can always pass to a subsequence that converges pointwise a.e. so it's usually quite easy to determine the only possible limit. Then you should try to build the examples in such a way that the pointwise (a.e.) limit has no chance of being continuous.
Here's an example of what I have in mind:
$f_{n}(x) = \begin{cases} 0 & 0 \leq x \leq \frac{1}{2} \\\ nx - \frac{n}{2} & \frac{1}{2} \leq x \leq \frac{1}{2} + \frac{1}{n} \\\ 1 & \frac{1}{2} + \frac{1}{n} \leq x \leq 1 \end{cases}$ defined for $n \geq 2$. It's easy to see that it converges in the $L^{1}$-norm to the characteristic function of $[\frac{1}{2},1]$.

Added: Since an $L^{1}$-Cauchy sequence will always converge ($L^{1}$ is complete by the theorem of Riesz-Fischer (or Fréchet-Riesz)), the only thing you have to do is to pick a function in $L^{1}$ that doesn't have a continuous representative. The easiest example is a step function (that's probably the reason why all three answerers came up with essentially the same example). This suffices, because every function in $L^{1}$ is the $L^1$-limit of a sequence of continuous (even smooth) functions, as can be seen by convolving with mollifiers, for example. In other words, $C[0,1]$ (or even $C^{\infty}[0,1]$) are dense in $L^{1}$.
A: How about the sequence of functions given by $f_n(x)=x^n$? This sequence is Cauchy in $(C([0,1]),\|\cdot\|_{L^1(0,1)})$ but does not converge in $C([0,1])$.
EDIT: This example doesn't work. Thanks for pointing that out. An example that does work is given by the sequence of continuous functions $f_n$ with
$f_n(x)=\frac{2}{n}(x-\frac{1}{2}+\frac{1}{n})$ for $|x-\frac{1}{2}|\leq \frac{1}{n}$, $f_n(x)=0$ for $x<\frac{1}{2}-\frac{1}{n}$ and $f_n(x)=1$ for $x>\frac{1}{2}+\frac{1}{n}$. That's basically a continuous approximation of the characteristic function of the interval $[\frac{1}{2},1]$, which is not continuous.
