# Continuous functions on $[0,1]$ is dense in $L^p[0,1]$ for $1\leq p< \infty$

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?

• Is this true for the Bochner Lp spaces? Jun 20 '19 at 23:26

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 $$\mathbb L^p$$.

• 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!
– Mike
Sep 13 '20 at 2:28

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.

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.

• 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. Mar 14 '18 at 17:58
• 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?
– rgm
Apr 20 '18 at 13:32
• @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. Apr 20 '18 at 16:35