4
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • 5
    $\begingroup$ The keyword you want to search for is "Baire class 1" functions. See for example Kechris's Classical Descriptive Set Theory (Springer GTM 156), §24.B. One characterization is that $f\colon\mathbb{R}\to\mathbb{R}$ is pointwise limit of continuous functions iff $f^{-1}(U)$ is a countable union of closed sets for every open set $U$ (op. cit., 24.10). Its set of continuous points is then a comeager (hence dense) $G_\delta$ (op. cit., 24.14). $\endgroup$
    – Gro-Tsen
    Commented Feb 3, 2016 at 10:40

2 Answers 2

Reset to default
6
$\begingroup$

The point-wise limit $f$ is continuous in a dense $G_\delta$. For a proof see for example Real analysis by Bruckner, Bruckner & Thomson.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Would this hold for arb topologies? $\endgroup$
    – Daniel Donnelly
    Commented Oct 26 at 9:44
6
$\begingroup$

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 $(0,1)$.

Improve this answer
$\endgroup$
3
  • 6
    $\begingroup$ Amusingly, the function $\mathbf{1}_{\mathbb{Q}}$ taking the value $1$ on the rationals and $0$ on the irrationals, which is discontinuous everywhere, is not (as per smyrlis's answer) pointwise limit of continuous functions (=Baire class 1), but it is (Baire class 2) pointwise limit of functions like this one which are themselves pointwise limits of continuous functions. So pointwise limits of pointwise limits are not necessarily pointwise limits. $\endgroup$
    – Gro-Tsen
    Commented Feb 3, 2016 at 10:34
  • $\begingroup$ @Gro-Tsen +1 thank you. Very interesting comment. $\endgroup$
    – Ali Taghavi
    Commented Feb 3, 2016 at 10:40
  • 1
    $\begingroup$ I am grateful for all the precise references. The Ali Taghavi explicit example is illuminating. $\endgroup$
    – Bazin
    Commented Feb 3, 2016 at 16:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .