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I have really no idea about this:

Problem: Show that there exists a function $f:[0,1]\rightarrow\mathbb{R}$ such that:

  1. $f$ is discontinuous in all $x\in \mathbb Q$.
  2. $f$ is increasing in $[0,1]$.
  3. $f$ is integrable.

EDIT: Sorry, it is not discontinuous in all $x\in \mathbb R \setminus \mathbb Q$, just in $\mathbb Q$.

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I think that $1$ and $3$ contradict itself, since if $f$ is almost everywhere not continuous it is not integrable –  Belgi Feb 6 '13 at 10:45
    
@Belgi: that depends on your measure... –  akkkk Feb 6 '13 at 10:54
    
@Belgi The indicator function of the rationals is not continuous anywhere, but is integrable on any measurable set. –  JSchlather Feb 6 '13 at 10:56

2 Answers 2

up vote 3 down vote accepted

In general for any countable set $C \subset \mathbb R$ you can find a monotone function that is discontinuous only on $C$. Your particular case has already been answered elsewhere on the site, see this answer of Brian Scott.

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You found it quicker than I did! –  Brian M. Scott Feb 6 '13 at 11:00
    
@BrianM.Scott I used google ;). –  JSchlather Feb 6 '13 at 11:01
    
I usually do too, but I couldn’t remember a good set of search terms for that one. –  Brian M. Scott Feb 6 '13 at 21:59

You can take $f_0(x)=x$ and a surjective map $q:\mathbb{N}\to\mathbb{Q}\cap[0,1]$, then take: $$ I_x^N = \{n\in\mathbb{N}:q(n)\leq x\}, $$ $$ f_N(x) = f_0(x)+\sum_{n\in I_x^N}\frac{1}{n^2+1}, $$ $$ f(x) = \sup_{N}\, f_N(x). $$

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