Existence of a sequence of continuous functions convergent pointwise to the indicator function of irrational numbers

Question

Prove that there does not exist a sequence of continuous functions $ f_n :\left[ {0,1} \right] \to R $ such that converges pointwise, to the function $$f(x)= \begin{cases} 0 & \text{if $x$ is rational},\\\\ 1 & \text{otherwise}. \end{cases}. $$

I have no idea How can I prove this. Prove that there no exist such sequence if the convergence is uniform, it's easy, because the limit would be continuous, but here I don't know How can I do. I suppose that some "nice" properties are "preserved" in the limit, in this kind of convergence, but I don't know any of them.

Answer 1

The reason (given in comments) that $f$ is not a pointwise limit of continuous functions is that $f$ is discontinuous everywhere, while pointwise limits of continuous functions have a comeager set of points of continuity. The latter fact is proved here, additional details are given here, and a textbook reference is: Theorem 1.19 on page 20 of Real analysis by Bruckner, Bruckner & Thomson.

Answer 2

Here is a simple solution to this old posting.

Suppose $(f_n:n\in\mathbb{N})\subset \mathcal{C}(\mathbb{R})$ is such that $f_n\xrightarrow{n\rightarrow\infty}\mathbb{1}_{\mathbb{Q}}$ pointwise. Define the sets $$E_n:=\bigcap_{m\geq n}\{x\in\mathbb{R}:|f_m(x)|\leq\frac12\}$$ Each $E_n$, being the intersection of closed sets, is closed.

If $x\in E_n$, then $|f_m(x)|\leq\frac12$ for all $m\geq n$ and so, $\mathbb{1}_{\mathbb{Q}}(x)=\lim_nf_n(x)\leq\frac12<1$, each means that $x$ is irrational. Hence $E_n$ is a nowhere dense closed set.

Conversely, $\mathbb{R}\setminus\mathbb{Q}\subset \bigcup_nE_n$ since for each irrational $x$, $f_n(x)\xrightarrow{n\rightarrow\infty}0$. Hence $$\mathbb{R}\setminus\mathbb{Q}=\bigcup_nE_n$$ and so, $\mathbb{R}\setminus\mathbb{Q}$ is a set of first category. This implies that $\mathbb{R}=\big(\mathbb{R}\setminus\mathbb{Q}\big)\cup\mathbb{Q}$ is of first category. We know from Baire's theorem that $\mathbb{R}$ (with the usual topology) is of second category: contradiction!