Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\newcommand{\R}{\Bbb R}$ Consider the Lebesgue measure in $\R$ and the following proposition:

P. For each representative of a function class $f\in L^2[0,1]$ there is a sequence of continuous functions $(f_n)_{n\in\Bbb N}$ such that:

  1. $|f_n-f|$ is Riemann integrable on $[0,1]$, for all $n\in\Bbb N$.
  2. $\lim\limits_{n\to\infty} \int\limits_0^1 |f_n(x)-f(x)|^2\ \mathrm d x=0$.

There is no reason why this proposition should be true, but I cannot find a counterexample.

share|cite|improve this question
Maybe the Dirichlet function (characteristic function of $\mathbb{Q}$) could provide a counterexample for 1? – Jose27 Aug 1 '12 at 4:04
You require your function to be in $L^2$, but the approximation in 2 is in $L^1$. I guess you have to choose one of them. – Martin Argerami Aug 1 '12 at 5:09
I don't think this matters. As you can see in my answer below, the problem is that leo asks for approximation in the sense of Riemann integral (as $|f_n-f|$ is supposed to be R. integrable). – Stefan Geschke Aug 1 '12 at 5:47
@MartinArgerami I agree with Stefan. There is no problem because $f$ Riemann integrable implies $f^2$ Riemann integrable. However I have edited the question to be consistent. – leo Aug 1 '12 at 5:53
up vote 4 down vote accepted

This proposition is not true. Let $f:[0,1]\to\mathbb R$ be defined by $f(x)=1$ iff $x\in\mathbb Q$ and $f(x)=0$ otherwise. $f$ is Lebesgue integrable but not Riemann integrable. Suppose $|f_n-f|$ is Riemann integrable and $\int_0^1|f_n-f|dx<1/4$. Then there is a partition $0=t_0<t_1<\dots<t_n=1$ of the unit interval such that $$\sum_{i=1}^n(t_i-t_{i-1})\sup\{|f_n(x)-f(x)|:x\in[t_{i-1},t_i]\}<1/4.$$ Now there is $i\in\{1,\dots,n\}$ such that $\sup\{|f_n(x)-f(x)|:x\in[t_{i-1},t_i]\}<1/4$. The function $f_n$ is not continuous on $[t_{i-1},t_i]$ since on a dense subset of that interval, $f_n<1/4$ and on a dense set, $f_n>3/4$.

share|cite|improve this answer
How much close? For example $f_n$ equal to ? – leo Aug 1 '12 at 4:35
Got it. Thanks. – leo Aug 1 '12 at 5:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.