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In class we proved that $C_c(X)$ is dense in $L^p$ where $X$ is a locally compact, $\sigma$-compact Hausdorff space either equipped with a Radon measure or equipped with a locally finite measure $\mu$ on the Borel sigma algebra.

The proofs, especially of the second variant are fairly long and I find it very hard to remember all the conditions (locally finite, $\sigma$-compact etc.). The first variant uses Lusin's theorem (among others). So I thought I might try to reduce this to a more specific case, the Lebesgue measure (which is Radon) and $X$ any subset of $\mathbb R$. (Do I need any assumption on $X$? Perhaps it needs to be measurable.)

Can you tell me if this is correct? Thank you.

Claim: $C_c(X)$ is dense in $L^p$ with the Lebesgue measure and $X \subset \mathbb R$.

Proof: We know that simple functions are dense in $L^p$. So if we can construct a function in $C_c$ that is $\varepsilon$-close (in $\|\cdot\|_p$) to $\chi_M$, the characteristic function of a measurable set $M$ then we're done. We know that $\mu$ is inner and outer regular so for $\varepsilon > 0$ we can find a compact set $K \subset M$ such that $\mu(M - K ) \leq \varepsilon$. Also, we can find an open set $O$ containing $M$. Let $f : K \sqcup O^c \to \mathbb R$ be the function that is $1$ on $K$ and $0$ on $O^c$. Then using Tietze we can continuously extend it to all of $X$. Then its extension $F \in C_c(X)$ and $$ \|F - \chi_M \|_p^p = \int_X |F - \chi_M|^p d \mu = 1 \cdot \mu(M -K)^p \leq \varepsilon^p$$

So I don't need Lusin's theorem. Right?

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    You just proved that $F\in C(X)$. Why $F$ is compactly supported?2012-07-13
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    I may be missing something but isn't this more or less what's done in quite some detail in the accepted answer to [your earlier question](http://math.stackexchange.com/q/66999/5363)? The objection Norbert raised is worked around as well in that answer and it is explicitly stated that you don't need Lusin there.2012-07-13
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    @t.b. You're missing nothing, of course. I was trying to avoid Lusin and I didn't remember that I'd asked this question before.2012-07-14
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    @Norbert Thank you, I made a mistake there. I was thinking that the support of $F$ is $K$ but that's clearly false.2012-07-14
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    Looks as if commenter stopped using the site in March. He also doesn't explain why $g$ (obtained from applying Urysohn) has compact support.2012-07-14
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    For each $x \in K$ choose an open neighborhood $U_x$ of $x$ with compact closure (by local compactness of $X$). Extract a finite subcover $U_1,\ldots,U_n$ of the cover $\{U_x\}_{x \in K}$ of $K$ (by compactness of $K$). Let $U = U_1 \cup \cdots \cup U_n$, note that $\overline{U} = \overline{U}_1 \cup \cdots \cup \overline{U}_n$ is compact. Set $g = 1$ on $K$ and $g = 0$ on $U^c$ and apply Urysohn to get a continuous function $g\colon X \to [0,1]$ with $g|_K \equiv 1$ and $g|_{U^c} \equiv 0$. In particular $\operatorname{supp}g\subset \overline{U}$, which is compact.2012-07-14
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    Note that a locally compact space need not be normal, but you can alwas separate disjoint closed sets $A$ and $B$, where $A$ is compact, with disjoint open sets $U \supset A$ and $V \supset B$ such that $\overline{U}$ is compact. Then the usual onion shell proof of Urysohn's lemma (which is usually given for normal spaces) yields a continuous function with compact support.2012-07-14
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    @t.b. Oh yes of course. Thank you.2012-07-14
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    I'm not using $\sigma$-compactness anywhere so that's a redundant assumption.2012-07-14
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    But I need to assume that $X$ is $T_4$ (for Urysohn).2012-07-14
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    No, as I outlined in my last comment, there's a version of Urysohn's lemma that holds for all locally compact Hausdorff spaces, provided one of the closed sets is compact, see e.g. [here](http://www.math.ethz.ch/~lanford/integrate.ps) Proposition 1.5. You might need the $\sigma$-compactness assumption for granting the existence of a non-trivial measure on $X$ satisfying the assumptions but a little care lets you get rid of $\sigma$-compactness entirely (if you need that for whatever reason).2012-07-14
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    @t.b. Ok, thanks. I'm very sorry. This sounds bad but: I'm not very good at remembering information that I didn't request. I don't know how else to put it. By that I don't mean that you should stop doing it though it's just an excuse for not remembering your penultimate comment.2012-07-14
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    Sorry, but you lost me... What else than a request for the information I gave you is "He also doesn't explain why $g$ (obtained from applying Urysohn) has compact support."?2012-07-14
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    @t.b. Yes, but you answer that in your first comment. Anyway, I'm very tired. Never mind. Just forget my last comment.2012-07-14
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    Ok ok, one really needs both of your comments. But the way I did it is: skim your first comment, think "yay now I know how it works", quickly skim your other comment, then sit down and try to do it and finally: realise that it doesn't all add up.2012-07-14
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    But you often provide additional information and then later when it turns out that you told me something before and I don't remember I feel bad because you wasted time on me.2012-07-14
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    Dear All, this question interests me a lot. Nonetheless through all the comments, I didn't get an clear answer. Could someone answer this question i.e. post a complete proof of why $C_c$ is dense in $L^p$? Thank you2016-08-08

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