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

I tried to show that the continuous functions on $[0,1]$ are dense in $L^p[0,1]$ for $ 1 \leq p< \infty $

by using Lusin's theorem.

I proceeded as follows..

By using Lusin's theorem, for any $f \in L^p[0,1]$, for any given $ \epsilon $ $ > $ 0, there exists a closed set $ F_\epsilon $ such that $ m([0,1]- F_\epsilon) < \epsilon$ and $f$ restricted to $F_\epsilon$ is continuous.

Using Tietze's extension theorem, extend $f$ to a continuous function $g$ on $[0,1]$. We claim that $\Vert f-g\Vert_p $ is sufficiently small.

$$ \Vert f-g\Vert_p ^p = \displaystyle \int_{[0,1]-F_\epsilon} |f(x)-g(x)|^p dx $$ $$ \leq \displaystyle \int_{[0,1]-F_\epsilon} 2^p (|f(x)|^p + |g(x)|^p) dx $$ now using properties of $L^p$ functions, we can make first part of our integral sufficiently small. furthermore, since $g$ is conti on $[0,1]$, $g$ has an upper bound $M$, so that second part of integration also become sufficiently small.

I thought I solved problem, but there was a serious problem.. our choice of g is dependent of $\epsilon$ , so constant $M$ is actually dependent of $\epsilon$, so it is not guaranteed that second part of integration becomes 0 as $\epsilon $ tends to 0.

I think if our choice of extension can be chosen further specifically, for example, by imposing $g \leq f$ such kind of argument would work. Can anyone help to complete my proof here?

share|cite|improve this question
I'm sorry for this.. I accepted all of my questions after I saw your comment. – Detectives Oct 31 '12 at 14:39
up vote 1 down vote accepted
  • We can assume $f$ bounded almost everywhere by $N$, has the sequence $\{f\chi_{-n\leq f\leq n}\}$ converges in $L^p$ to $f$.

  • Lusin's theorem gives a closed set $F_k$ such that $[0,1]\setminus F_k$ has measure $\leq k^{-1}$, and $f$ restricted to $F_k$ is continuous. So $f_{\mid F_k}$ is bounded by $M$.

  • Tietze extension theorem gives that the extension still is bounded by $M$.

The preceding point show the result when $f$ is bounded almost everywhere. So we can approximate an almost everywhere bounded function by a continuous one in $L^p$. Now we use a $2\varepsilon$ argument: fix $f\in L^p$, $\widetilde f$ approximating $f$ in $L^p$ by $\varepsilon$. Take $g$ continuous approximating $\widetilde f$ in $L^p$ up to $\varepsilon$. Then $\lVert f-g\rVert_{L^p}\leq 2\varepsilon$.

share|cite|improve this answer
but isn't it still true that M depends on k? what if M goes to infinity if k goes to infinity? – Detectives Oct 31 '12 at 14:41
I've added details. – Davide Giraudo Oct 31 '12 at 14:47

Since $L_p([0,1])=\mathrm{cl}(\mathrm{span}\{\chi_E:E\in\mathfrak{M}([0,1])\})$, it is enough to prove that $$ \forall\varepsilon>0\quad\forall E\in\mathfrak{M}([0,1])\quad\exists f\in C([0,1])\quad \Vert f-\chi_E\Vert<\varepsilon $$ Indeed by regularity of Lebesgue measure there exist closed set $F$ and open set $U$ such that $F\subset E\subset U$ with $\mu(U\setminus F)<\varepsilon$. The desired $f\in C([0,1])$ is $$ f(t)=\frac{d(t, [0,1]\setminus U)}{d(t, [0,1]\setminus U)+d(t,F)} $$ where $d(t, S)=\inf\{|t-s|:s\in S\}$ is a distance function.

share|cite|improve this answer

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.