If $X$ is a complete doubling metric space equipped with a complete probability measure $\mu$ such that all Borel sets are $\mu$-measurable, then $C_c(X)$ --- the continuous functions with compact support --- are dense in $L_1(\mu)$.

Question: What are the weakest conditions under which $C_c(X)$ is dense in $L_1(\mu)$ for non-doubling (i.e., infinite doubling dimensional) metric spaces?

  • $\begingroup$ Do you want to assume regularity of the measure? $\endgroup$
    – user1688
    Apr 20 '17 at 9:49
  • $\begingroup$ Do nice things follow from that assumption alone? What else does one need to assume? $\endgroup$ Apr 20 '17 at 9:52
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    $\begingroup$ I have the feeling that regularity might do the trick, however it might be to strong an assumption as I do not know your purpose. Have no time at the moment, shall think about this later. $\endgroup$
    – user1688
    Apr 20 '17 at 10:00
  • $\begingroup$ Great, very much looking forward! $\endgroup$ Apr 20 '17 at 10:01
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    $\begingroup$ @AryehKontorovich In an infinite dimensional Banach space, an open ball is not precompact, and the support of a nonzero continuous function contains some open ball, so it is not compact. I'm not sure what separable had to do with anything. $\endgroup$
    – Ian
    Apr 20 '17 at 14:14

Like in Nate's comment, you need locally compactness. Assuming that, if $\mu$ is regular, then $C_c(X)$ is dense in $L^1(\mu)$. For the proof: For a compact subset $K$ there exists a sequence $f_n$ in $C_c(X)$ with $f_n\ge 1_K$ and $\int f_n\,d\mu\to \mu(K)$. Therefore, $1_K$ lies in the closure of $C_c(X)$ in $L^1$. By another approximation, $1_A$ lies in this closure for every measurable $A$. This suffices.


I see the question got upvoted, so I'll post a brief note explaining how we got around the local compactness for our application, in case anyone else finds it useful. In Lemma A.1 of Hanneke, Kontorovich, Sabato, and Weiss - Universal Bayes consistency in metric spaces, we prove that for every metric probability space, the collection of Lipschitz functions is dense in $L_1$. The result may be known or folklore, but it was exactly the analogue of $C_c$ that we needed.


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