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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?

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  • $\begingroup$ Is this true for the Bochner Lp spaces? $\endgroup$
    – AIM_BLB
    Jun 20 '19 at 23:26
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Let $f\in\mathbb L^p$ and $\varepsilon\gt 0$. Choose $N$ such that $\left\lVert f-f\mathbf 1_{-N\leqslant f\leqslant N}\right\rVert_p\leqslant \varepsilon/2$. Let $f_N:=f\mathbf 1_{-N\leqslant f\leqslant N}$.

  • Lusin's theorem gives a closed set $F$ such that $[0,1]\setminus F$ has measure smaller than $2^{-p} \varepsilon^p/\left(2N\right)^p$, and $f_N$ restricted to $F$ is continuous.

  • Tietze extension theorem applied to $f_N$ and $F$ gives that the extension $g$ is still bounded by $N$.

    Consequently, $$\left\lVert f_N-g\right\rVert_p^p=\int_{[0,1]\setminus F} \left\lvert f_N-g\right\rvert_p^p\leqslant (2N)^p\lambda\left([0,1]\setminus F\right)\leqslant 2^{-p}\varepsilon^{-p}. $$ We thus got a continuous function $g$ such that $$\left\lVert f-g\right\rVert_p\leqslant \varepsilon,$$ which show that the set of continuous functions is dense in $\mathbb L^p$.

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  • $\begingroup$ Hello, I'm reading this post and I have a similar question asked here recently: math.stackexchange.com/q/3823641/792125. If I replace a single continuous function by sequence of continuous functions, is this argument still valid? I wish to adapt the proof using Lusin's theorem in my case, but I'm not sure whether I can use Tietze's extension. And I'd like to avoid using the fact that $C_c(X)$ is dense in $L^p(\mu)$ for $1\leq p<\infty$ and $X$ locally compact Hausdorff, given in some real analysis texts. Can you kindly provide a proof regarding my case? Thank you! $\endgroup$
    – Mike
    Sep 13 '20 at 2:28
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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 the Lebesgue measure there exists a closed set $F$ and an 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 the distance function.

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Fix $p\text{ , and }1\leq p\lt \infty.$

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 $[0,1]$.

Note that $f\equiv g$ on $F_\epsilon$, so we only need to take care of the integral on $[0,1]- F_\epsilon$.

Continuous function is always integrable on $[0,1]$, so $g^p$ is integrable on $[0,1]$.

Since $|f(x)-g(x)|^p \leq 2^p (|f(x)|^p + |g(x)|^p)$ and $f \in L^p[0,1],$

we know $\int_{[0,1]}|f(x)-g(x)|^p \lt \infty, i.e. |f(x)-g(x)|^p$ is integrable on $[0,1].$

By the proposition I post, $ \int_{[0,1]-F_\epsilon}|f(x)-g(x)|^p \to 0 $ when $m([0,1]- F_\epsilon)\to 0.$

Note that $\epsilon \to 0 \Rightarrow m([0,1]- F_\epsilon)\to 0$ (Since $m([0,1]- F_\epsilon \lt \epsilon$)

For each $\epsilon \gt 0$, we can find a corresponding continuous function $g_\epsilon, $ and $\Vert f-g_\epsilon \Vert \to 0$ when $\epsilon \to 0$.

So, $C([0,1])$ is dense in $L^p[0,1]$.

Reference: The proposition is from the Real Analysis,4th Ed, written by Royden and Fitzpatrick.

enter image description here

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  • $\begingroup$ Although my proof looks like a bit tedious, its center thought is simple. The most important estimate is done by the proposition, which is one of well-known properties of integrable functions. $\endgroup$
    – Sam Wong
    Mar 14 '18 at 17:58
  • $\begingroup$ Dear @SamWong, I couldn't get the inequality $|f(x)-g(x)|^p \leq 2^p (|f(x)|^p + |g(x)|^p)$. May you help me? $\endgroup$
    – rgm
    Apr 20 '18 at 13:32
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    $\begingroup$ @rgm Hint: For fixed $x$, $\vert f(x)-g(x)\vert \le 2max\{\vert f(x) \vert, \vert g(x) \vert\}$ and $2max\{\vert f(x) \vert, \vert g(x) \vert\}$ is $\{\vert f(x) \vert\}$ or $\{\vert g(x) \vert\}$. If you still can not figure it out, I will reply you an answer tomorrow. $\endgroup$
    – Sam Wong
    Apr 20 '18 at 16:35

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