# Is $C_0(\mathbb{R}^n)$ dense in $L^p(A)$ when $A\subset \mathbb{R}^n$ is measurable and $1\leq p \leq \infty$?

Define $C_0(\mathbb{R}^n)$ to be all continuous functions with compact support. I managed to show the claim in the case $1\leq p <\infty$ and $A=\mathbb{R}^n$. This was done in steps:

1. Use Urysohn's lemma to show that you can approximate characteristic functions of measurable sets by functions in $C_0(\mathbb{R}^n)$.

2. From 1) it follows that simple functions can be approximated by functions in $C_0(\mathbb{R}^n)$.

3. Show that simple functions are dense in $L^p(\mathbb{R}^n)$. From this it immediately follows along with 2) that $C_0(\mathbb{R}^n)\subset L^p(\mathbb{R}^n)$ is dense.

I also think that when $p=\infty$ the claim does not hold. This is because the uniform limit of continuous functions is continuous. Therefore one cannot approximate for example the step function $\chi_{[0, \infty)}$ by functions in $C_0(\mathbb{R}^n)$.

So my question is: can we use the same proof for arbitrary measurable set $A$? Can we just conclude that $C_0(A)\subset L^p(A)$ is dense and extend the functions in $C_0(A)$ by zero to $C_0(\mathbb{R}^n)$? Or am I missing something crucial here?

Thank you!

• Thank you, added it now. – peastick Sep 17 '18 at 7:18
• Well, continuous functions on an arbitrary set are not really useful. To use continuity (and differentiability, etc...), you need that the functions are defined on an open set. This said, the crucial point of the proof you sketched is Urysohn's lemma. You should try and see if it works on an arbitrary set. I bet it does not. – Giuseppe Negro Sep 17 '18 at 7:31
• The version of Urysohn's lemma I used is the following: Let $X$ be proper metric space, $K\subset U$ where $K$ is compact and $U$ open. Then there is $f\in C_0(X)$ s.t. $0\leq f\leq 1$, $f|K=1$ and $\text{spt}(f)\subset U$. I can't see why this would fail for a general subset of $\mathbb{R}^n$. – peastick Sep 17 '18 at 7:45
• Or wait a minute, I guess the problem comes from the fact that even if $X$ is proper then it does not necessarily hold that a general $A\subset X$ is proper. Am I right in this? – peastick Sep 17 '18 at 7:53
• What is "proper"? Closed balls are compact? Then, yes, a subset of $\mathbb R^n$ needs not be proper (for example, $\mathbb Q$ is not). – Giuseppe Negro Sep 17 '18 at 7:59

## 1 Answer

If $f\in L^{p}(A)$ then extend $f$ to $L^{p}(\mathbb R^{n})$ by making it $0$ outside $A$. Approximate $g$ by $C_0(\mathbb R^{n})$ function $h$. Then $h$ approximates $f$ on $A$. I don't think the question has anything to do with $C_0(A)$ (which does not even make sense for a general measurable set $A$).

• Oh, it was that easy. Thank you very much. – peastick Sep 17 '18 at 7:56