Is there a sequence of continous function pointwise convergent to Riemann function?

Question

I'm reading Baire's Category theory recently. One can find the following theorem in Chapter 4 of Stein's Functional Analysis:

"Suppose that $\{f_n\}$ is a sequence of continuous complex-valued functions on a complete metric space $X$, and $\{f_n\}$ convergent to $f$ pointwise for each $x\in X$.

$$\lim_{n\rightarrow+\infty}f_n(x)=f(x)$$

Then, the set of points where $f$ is continuous is a generic set in $X$. In other words, the set of points where $f$ is discontinuous is of the first category."

This theorem tells us that the possibly largest set of discontinuous points of a function obtained by pointwise convergence of a series of continous functions is of the first category. Then, a nature question is, (1) given a function $f$ defined on a complete metric space, say $[0,1]$, which is discontinuous on a first category set, is there a series of continous functions pointwise convergent to $f$?

In particular, the Riemann function is discontinuous on the set of rational number $Q\cap[0,1] $, a first category set. (2) Is there a sequence of continous function pointwise convergent to Riemann function?

Added: Questions for this post has been completely solved. For question (1), see David Mitra's comment, for a construction proof of question (2), see answer by John.

Answer 1

I do not have the answer for the general question. Let me just answer your last question.

By the Riemann function $T$ I assume you are referring to this function.

Now let $n\in \mathbb N$. Let $Q_n = \{ r= \frac{p}{q} \in [0,1]: q\le n\}$ and $\epsilon_n = \frac 14 \min_{x, y\in Q_n} |x-y|$. Define

$$f_n(x) = \max_{r\in Q_n} \left(T(r) \max\{0, 1-\frac{1}{\epsilon_n}|x- r|\}\right)$$

Then each $f_n$ are continuous and $f_n(x) = T(x)$ for all $x\in Q_n$. Thus

$$\lim_{n\to \infty} f_n(x)= T(x)$$

for all $x\in \mathbb Q \cap [0,1]$.

Now we deal with the case where $x$ is irrational. For each $n$, let $r_n = \frac{p_n}{q_n} \in Q_n$ so that

$$|x-r_n| = \min_{r\in Q_n} |x-r|.$$

Thus we have $f_n(x) \le T(r_n) = \frac{1}{q_n}$. But $q_n\to \infty$ as $n\to \infty$, so $f_n(x) \to 0 = T(x)$ as $n\to \infty$.