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

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.

• I'm not voting to close, but this question is essentially a duplicate of math.stackexchange.com/q/75192
– t.b.
Oct 30, 2011 at 21:06
• I said that I didn't vote to close and I meant it. Still, the point is that the set of discontinuity of a pointwise limit of continuous functions is a set of first category, while your function is discontinuous everywhere.
– t.b.
Oct 30, 2011 at 21:21
• Wow! That´s nice!!! where can I find a proof of this fact? Or at least the name of the theorem, to find it´s proof Oct 30, 2011 at 21:31
• It's a theorem of Baire. In his answer in the thread mentioned above LostInMath points to Theorem 1.19 on page 20 of Bruckner, Bruckner & Thomson Real analysis (if you can't access that page, changing google.com to google.cl may help)
– t.b.
Oct 30, 2011 at 21:36
• Jeeze. how can cambridge set something like this for an analysis II question... Jan 8, 2014 at 19:12

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.