Let $f_n:[0,1]\rightarrow \mathbb R$ be a sequence of continuous functions converging pointwise, i.e. such that $\forall x\in [0,1]$, the sequence $(f_n(x))_{n\in \mathbb N}$ converges. We set $f(x)=\lim_nf_n(x)$.
Of course the function $f$ will fail in general to be continuous, due to the weakness of the pointwise convergence. I guess that it is possible to find an example of a sequence $f_n$ as above such that $f$ is discontinuous on a dense subset: is there a simple example? Is it possible for $f$ to be discontinuous everywhere?
Best Answer
There is sequence of continuous functions $f_{n}$ on the unit interval $[0,1]$ which converges to a function $f$ such that $f$ is discontinuous at rational points of $(0,1)$, a dense subset of the interval.
Let $f(x)=\begin{cases} 0& x \;\text{is irrational or } x\in \{0,1\}\\ 1/n & x=m/n,\;\;\;\;(m,n)=1 \end{cases}$
Let $\{r_{0},r_{1},\ldots, r_{n}\ldots, \}$ be the sequence of rational numbers in $[0,1]$.
Let $f_{n}$ be the unique continuous picewise linear function which satisfies $f_{n}(r_{j})=f(r_{j})\;\text{for}\;\;j=0,1,\ldots,n$ and vanishes at end points of the interval.
It is easy to show that $f_{n}$ converges to $f$ and $f$ is discountinuous at rational points of $(0,1)$.